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Product Rule and Quotient Rule

2017-10-05T05:46:23Z

Grad: /* Introduction */

==Introduction==
Taking the derivatives of simple functions (i.e. polynomials) is easy using the power rule.

For example, if  <math style="vertical-align: -5px">f(x)=x^3+2x^2+5x+3,</math>  then  <math style="vertical-align: -5px">f'(x)=3x^2+4x+5.</math>

But, what about more complicated functions?

For example, what is  <math style="vertical-align: -5px">f'(x)</math>  when  <math style="vertical-align: -5px">f(x)=\sin x \cos x?</math>

Or what about  <math style="vertical-align: -5px">g'(x)</math>  when  <math style="vertical-align: -15px">g(x)=\frac{x}{x+1}?</math>

Notice  <math style="vertical-align: -5px">f(x)</math>  is a product, and  <math style="vertical-align: -5px">g(x)</math>  is a quotient. So, to answer the question of how to calculate these derivatives, we look to the Product Rule and the Quotient Rule. The Product Rule and the Quotient Rule give us formulas for calculating these derivatives.

'''Product Rule'''

Let  <math style="vertical-align: -5px">h(x)=f(x)g(x).</math>  Then,

::<math>h'(x)=f(x)g'(x)+f'(x)g(x).</math>

'''Quotient Rule'''

Let  <math style="vertical-align: -19px">h(x)=\frac{f(x)}{g(x)}.</math>  Then,

::<math>h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}.</math>

==Warm-Up==
Calculate  <math style="vertical-align: -5px">f'(x).</math>

'''1)'''   <math style="vertical-align: -7px">f(x)=(x^2+x+1)(x^3+2x^2+4)</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Product Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(x^3+2x^2+4)'+(x^2+x+1)'(x^3+2x^2+4).</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4).</math>
|-
|-
|NOTE: It is not necessary to use the Product Rule to calculate the derivative of this function.
|-
|You can distribute the terms and then use the Power Rule.
|-
|In this case, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{(x^2+x+1)(x^3+2x^2+4)}\\
&&\\
& = & \displaystyle{x^2(x^3+2x^2+4)+x(x^3+2x^2+4)+1(x^3+2x^2+4)}\\
&&\\
& = & \displaystyle{x^5+2x^4+4x^2+x^4+2x^3+4x+x^3+2x^2+4} \\
&&\\
& = & \displaystyle{x^5+3x^4+3x^3+6x^2+4x+4.}
\end{array}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=5x^4+12x^3+9x^2+12x+4.</math>
|-
|In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
|-
|or equivalently
|-
|       <math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
|}

'''2)'''   <math style="vertical-align: -14px">f(x)=\frac{x^2+x^3}{x}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|
Using the Quotient Rule, we have
|-
|
::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
|-
|Then, using the Power Rule, we have
|-
|
::<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)(1)}{x^2}.</math>
|-
|NOTE: It is not necessary to use the Quotient Rule to calculate the derivative of this function.
|-
|You can divide and then use the Power Rule.
|-
|In this case, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{\frac{x^2+x^3}{x}}\\
&&\\
& = & \displaystyle{\frac{x^2}{x}+\frac{x^3}{x}}\\
&&\\
& = & \displaystyle{x+x^2.} \\
\end{array}</math>
|-
|Now, using the Power Rule, we get
|-
|
::<math>f'(x)=1+2x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
||       <math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
|-
|or equivalently
|-
|       <math>f'(x)=1+2x</math>

|-
|}

'''3)'''   <math style="vertical-align: -14px">f(x)=\frac{\sin x}{\cos x}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Using the Quotient Rule, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{\cos x(\sin x)'-\sin x (\cos x)'}{(\cos x)^2}}\\
&&\\
& = & \displaystyle{\frac{\cos x(\cos x)-\sin x (-\sin x)}{(\cos x)^2}}\\
&&\\
& = & \displaystyle{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}} \\
&&\\
& = & \displaystyle{\frac{1}{\cos^2 x}}\\
&&\\
& = & \displaystyle{\sec^2 x}
\end{array}</math>
|-
|since  <math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>  and  <math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
|-
|Since  <math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>  we have
|-
|
::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>f'(x)=\sec^2 x</math>
|-
|}

== Exercise 1 ==

Calculate the derivative of  <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}(\csc x-4).</math>

First, we need to know the derivative of  <math style="vertical-align: 0px">\csc x.</math>  Recall

::<math>\csc x =\frac{1}{\sin x}.</math>

Now, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
&&\\
& = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\
&&\\
& = & \displaystyle{-\csc x \cot x.}
\end{array}</math>

Using the Product Rule and Power Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
&&\\
& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
&&\\
& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>

== Exercise 2 ==

Calculate the derivative of  <math style="vertical-align: -5px">g(x)=2x\sin x \sec x.</math>

Notice that the function  <math style="vertical-align: -5px">g(x)</math>  is the product of three functions.

We start by grouping two of the functions together. So, we have  <math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>

Using the Product Rule, we get

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
&&\\
& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
\end{array}</math>

Now, we need to use the Product Rule again. So,

::<math>\begin{array}{rcl}
\displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
\end{array}</math>

So, we have
::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>

But, there is another way to do this problem. Notice

::<math>\begin{array}{rcl}
\displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\
&&\\
& = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\
&&\\
& = & \displaystyle{2x\tan x.}
\end{array}</math>

Now, you would only need to use the Product Rule once instead of twice.

== Exercise 3 ==

Calculate the derivative of  <math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>

Using the Quotient Rule, we have

::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\
&&\\
& = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
\end{array}</math>

So, we get
::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.</math>

== Exercise 4 ==

Calculate the derivative of  <math style="vertical-align: -14px">f(x)=\frac{e^x}{x^2\sin x}.</math>

First, using the Quotient Rule, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
\end{array}</math>

Now, we need to use the Product Rule. So, we have

::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
&&\\
& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
\end{array}</math>

So, we have
::<math>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>

009C Sample Final 3, Problem 9

2017-03-19T18:29:14Z

Grad:

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Many concepts in physics involve the notion of a relative frame. For example, if I'm in a box dropped from an airplane, I won't be moving relative to the box. However, I'm still heading towards the ground with acceleration <math style="vertical-align: -5px">10\,\textrm{m/sec}^{2}. </math>  Say it drops for 5 seconds, so the box is going <math style="vertical-align: -5px">50\,\textrm{m/sec} </math> when it hits the ground. Even if I jump with all my might and pull off something like <math style="vertical-align: -5px">5\,\textrm{m/sec} </math> of upward velocity, I'll still feel the impact of hitting the ground at

     <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>

Essentially, equations of motion can often be broken into parts, and
then added up.
|-
|
|-
|
|-
|
|-
|
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|If a wheel of radius one is resting at the origin, its axis will be at the point <math style="vertical-align: -5px">(1,0). </math> For this solution, we will assume the point <math style="vertical-align: 0px">P </math> is below the axle, although the problem does not state the position of <math style="vertical-align: 0px">P </math>. When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle <math style="vertical-align: 0px">\theta </math> on a circle of radius <math style="vertical-align: 0px">R </math> is <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math>  the wheel will roll forward the length of the arc, which is just <math style="vertical-align: 0px">\theta. </math>

Moreover, the axle's <math style="vertical-align: 0px">x </math> position will change in the same manner,
while the height of the axle will always be fixed at <math style="vertical-align: -4px">y=1 </math>. This
means we can describe the position of the axle as a function of <math style="vertical-align: -4px">\theta, </math>
or

:: <math>a(\theta)\,=\,(\theta,1). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the wheel is rotating, we also know that our point <math style="vertical-align: 0px">P </math> will rotate around the axle. As described in the problem, it is halfway
between the rim and the center/axle, so it is <math style="vertical-align: -4px">1/2 </math> unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of <math style="vertical-align: 0px">P </math> relative to the axle can be described as

:: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>

Notice that when <math style="vertical-align: -4px">\theta=0, </math> the point would be at the position

:: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>

which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at <math style="vertical-align: -5px">\theta=\pi/4. </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We therefore have a frame (the axle) that is moving, and a point <math style="vertical-align: 0px">P </math> that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find

:: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>

|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:: <math>P(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
|  
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 4

2017-03-12T16:56:31Z

Grad:

(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x)^{n+1}}{(n+1)}\cdot\frac{n}{(-1)^n(x)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n}{n+1}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n}{n+1}}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
&&\\
& = & \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -1px">R=1.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -15px">b_n=\frac{1}{n}.</math>.
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+1}<\frac{1}{n}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{n}.</math>
|-
|This is a  <math>p</math>-series with  <math>p=1.</math>  Hence, the series diverges.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=-1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -1px">R=1.</math>
|-
|    '''(b)'''     <math>(-1,1]</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 3

2017-03-12T16:55:15Z

Grad:

Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=0}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''2.''' If a series absolutely converges, then it also converges.
|-
|'''3.''' '''Alternating Series Test'''
|-
|        Let  <math>\{a_n\}</math>  be a positive, decreasing sequence where  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
|-
|        Then,  <math>\sum_{n=1}^\infty (-1)^na_n</math>  and  <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
|-
|        converge.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using the Ratio Test.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!} \cdot\frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,</math>
|-
|the series is absolutely convergent by the Ratio Test.
|-
|Therefore, the series converges.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|For
|-
|        <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math>
|-
|The sequence  <math style="vertical-align: -5px">\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n+1}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    converges
|-
|   '''(b)'''    converges
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 2

2017-03-12T16:54:15Z

Grad:

For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>
|-
|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series
|-
|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.
|-
|'''2.''' The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
|-
|
        <math>s_n=\sum_{i=1}^n a_i.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -3px">a_n</math>  be the  <math style="vertical-align: 0px">n</math>th term of this sum.
|-
|We notice that
|-
|        <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>  and  <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
|-
|So, this is a geometric series with  <math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
|-
|Since  <math style="vertical-align: -5px">|r|<1,</math>  this series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, the sum of this geometric series is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
&&\\
& = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
&&\\
& = & \displaystyle{\frac{8}{3}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using partial fraction decomposition. Let
|-
|       <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
|-
|If we multiply this equation by  <math style="vertical-align: -5px">(2x-1)(2x+1),</math>  we get
|-
|       <math>1=A(2x+1)+B(2x-1).</math>
|-
|If we let  <math style="vertical-align: -14px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">A=\frac{1}{2}.</math>
|-
|If we let  <math style="vertical-align: -14px">x=-\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">B=-\frac{1}{2}.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{-\frac{1}{2}}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we look at the partial sums,  <math>s_n</math>  of this series.
|-
|First, we have
|-
|        <math>s_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).}
\end{array}</math>
|-
|If we compare  <math>s_1,s_2,s_3,</math>  we notice a pattern.
|-
|We have
|-
|       <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, to calculate the sum of this series we need to calculate
|-
|       <math>\lim_{n\rightarrow \infty} s_n.</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since the partial sums converge, the series converges and the sum of the series is  <math style="vertical-align: -15px">\frac{1}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{8}{3}</math>
|-
|   '''(b)'''    <math>\frac{1}{2}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2

2017-03-12T16:38:53Z

Grad: /*  Problem 4  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 2,_Problem_1| Problem 1 ]] ==
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

== [[009C_Sample Final 2,_Problem_2| Problem 2 ]] ==
 For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

== [[009C_Sample Final 2,_Problem_3| Problem 3 ]] ==
Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=0}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}</math>

== [[009C_Sample Final 2,_Problem_4| Problem 4 ]] ==
(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

== [[009C_Sample Final 2,_Problem_5| Problem 5 ]] ==
 Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

== [[009C_Sample Final 2,_Problem_6| Problem 6 ]] ==
(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

== [[009C_Sample Final 2,_Problem_7| Problem 7 ]] ==

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

== [[009C_Sample Final 2,_Problem_8| Problem 8 ]] ==

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

== [[009C_Sample Final 2,_Problem_9| Problem 9 ]] ==

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>

== [[009C_Sample Final 2,_Problem_10| Problem 10 ]] ==

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

009C Sample Final 2

2017-03-12T16:38:30Z

Grad: /*  Problem 3  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 2,_Problem_1| Problem 1 ]] ==
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

== [[009C_Sample Final 2,_Problem_2| Problem 2 ]] ==
 For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

== [[009C_Sample Final 2,_Problem_3| Problem 3 ]] ==
Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=0}^{\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+1}</math>

== [[009C_Sample Final 2,_Problem_4| Problem 4 ]] ==
(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{+\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

== [[009C_Sample Final 2,_Problem_5| Problem 5 ]] ==
 Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

== [[009C_Sample Final 2,_Problem_6| Problem 6 ]] ==
(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

== [[009C_Sample Final 2,_Problem_7| Problem 7 ]] ==

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

== [[009C_Sample Final 2,_Problem_8| Problem 8 ]] ==

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

== [[009C_Sample Final 2,_Problem_9| Problem 9 ]] ==

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>

== [[009C_Sample Final 2,_Problem_10| Problem 10 ]] ==

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

009C Sample Final 2

2017-03-12T16:38:15Z

Grad: /*  Problem 2  */

'''This is a sample, and is meant to represent the material usually covered in Math 9C for the final. An actual test may or may not be similar.'''

'''Click on the  boxed problem numbers  to go to a solution.'''
<div class="noautonum">__TOC__</div>

== [[009C_Sample Final 2,_Problem_1| Problem 1 ]] ==
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

== [[009C_Sample Final 2,_Problem_2| Problem 2 ]] ==
 For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>

== [[009C_Sample Final 2,_Problem_3| Problem 3 ]] ==
Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=0}^{+\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=0}^{+\infty} (-1)^n \frac{1}{n+1}</math>

== [[009C_Sample Final 2,_Problem_4| Problem 4 ]] ==
(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{+\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

== [[009C_Sample Final 2,_Problem_5| Problem 5 ]] ==
 Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

== [[009C_Sample Final 2,_Problem_6| Problem 6 ]] ==
(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

== [[009C_Sample Final 2,_Problem_7| Problem 7 ]] ==

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

== [[009C_Sample Final 2,_Problem_8| Problem 8 ]] ==

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

== [[009C_Sample Final 2,_Problem_9| Problem 9 ]] ==

A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>

== [[009C_Sample Final 2,_Problem_10| Problem 10 ]] ==

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

009C Sample Final 2, Problem 7

2017-03-12T16:37:53Z

Grad:

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The Taylor polynomial of   <math style="vertical-align: -5px">f(x)</math>   at   <math style="vertical-align: -1px">a</math>   is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|-
| '''2.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by finding the coefficients of the Maclaurin series for  <math>f(x)=\frac{1}{(1-\frac{1}{2}x)^2}.</math>
|-
|We make a table to find the coefficients of the Maclaurin series.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(0) </math></td>
<td align = "center"><math> \frac{f^{(n)}(0)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^2} </math></td>
<td align = "center"><math> 1</math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^3}</math></td>
<td align = "center"><math> 1 </math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> \frac{\frac{3}{2}}{(1-\frac{1}{2}x)^4} </math></td>
<td align = "center"><math> \frac{3}{2} </math></td>
<td align = "center"><math> \frac{3}{4} </math></td>
</tr>
</table>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the first three terms of the Binomial Series is
|-
|        <math>1+x+\frac{3}{4}x^2.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By taking the derivative of the known series
|-
|   <math>\frac{1}{1-x}\,=\,1+x+x^2+\cdots,</math>
|-
|we find that the Maclaurin series of  <math>\frac{1}{(1-x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)x^n.</math>
|-
|Letting <math style="vertical-align: -5px"> x/2 </math>   play the role of <math style="vertical-align: -4px">x,</math> the Maclaurin series of  <math>\frac{1}{(1-\frac{1}{2}x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)\bigg(\frac{1}{2}x\bigg)^n=\sum_{n=0}^\infty \frac{(n+1)x^n}{2^n}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+2)x^{n+1}}{2^{n+1}} \frac{2^n}{(n+1)x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{|x|}{2} \frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}.}
\end{array}</math>
|-
|Now, the Ratio Test says this series converges if  <math style="vertical-align: -14px">\frac{|x|}{2}<1.</math>  So,  <math style="vertical-align: -6px">|x|<2.</math>
|-
|Hence, the radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>1+x+\frac{3}{4}x^2</math>
|-
|   '''(b)'''    The radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 2

2017-03-11T00:24:27Z

Grad:

For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>
|-
|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series
|-
|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.
|-
|'''2.''' The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
|-
|
        <math>s_n=\sum_{i=1}^n a_i.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -3px">a_n</math>  be the  <math style="vertical-align: 0px">n</math>th term of this sum.
|-
|We notice that
|-
|        <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>  and  <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
|-
|So, this is a geometric series with  <math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
|-
|Since  <math style="vertical-align: -5px">|r|<1,</math>  this series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, the sum of this geometric series is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
&&\\
& = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
&&\\
& = & \displaystyle{\frac{8}{3}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using partial fraction decomposition. Let
|-
|       <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
|-
|If we multiply this equation by  <math style="vertical-align: -5px">(2x-1)(2x+1),</math>  we get
|-
|       <math>1=A(2x+1)+B(2x-1).</math>
|-
|If we let  <math style="vertical-align: -14px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">A=\frac{1}{2}.</math>
|-
|If we let  <math style="vertical-align: -14px">x=-\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">B=-\frac{1}{2}.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{-\frac{1}{2}}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we look at the partial sums,  <math>s_n</math>  of this series.
|-
|First, we have
|-
|        <math>s_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).}
\end{array}</math>
|-
|If we compare  <math>s_1,s_2,s_3,</math>  we notice a pattern.
|-
|We have
|-
|       <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, to calculate the sum of this series we need to calculate
|-
|       <math>\lim_{n\rightarrow \infty} s_n.</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since the partial sums converge, the series converges and the sum of the series is  <math style="vertical-align: -15px">\frac{1}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{8}{3}</math>
|-
|   '''(b)'''    <math>\frac{1}{2}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 5

2017-03-11T00:21:54Z

Grad:

Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The Taylor polynomial of   <math style="vertical-align: -5px">f(x)</math>   at   <math style="vertical-align: -1px">a</math>   is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -14px">a=\frac{\pi}{4}.</math>
|-
|First, we make a table to find the coefficients of the Taylor polynomial.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(a) </math></td>
<td align = "center"><math> \frac{f^{(n)}(a)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \cos x </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2}</math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2}</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> -\sin x</math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> -\cos x </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{4} </math></td>
</tr>
<tr>
<td align = "center"><math>3</math></td>
<td align = "center"><math> \sin x </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{12}</math></td>
</tr>
</table>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -4px">T_n</math>  be the Taylor polynomial of order  <math>n.</math>
|-
|Since   <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!},</math>  we have
|-
| 
|-
|       <math>T_0=\frac{\sqrt{2}}{2}</math>
|-
|       <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
|-
|       <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
|-
|       <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3.</math>
|-
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       Let  <math style="vertical-align: -4px">T_n</math>  be the Taylor polynomial of order  <math>n.</math>
|-
| 
|-
|       <math>T_0=\frac{\sqrt{2}}{2}</math>
|-
|       <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
|-
|       <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
|-
|       <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 4

2017-03-11T00:20:59Z

Grad:

(a) Find the radius of convergence for the power series

::<math>\sum_{n=1}^{+\infty} (-1)^n \frac{x^n}{n}.</math>

(b) Find the interval of convergence of the above series.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x)^{n+1}}{(n+1)}\cdot\frac{n}{(-1)^n(x)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n}{n+1}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n}{n+1}}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
&&\\
& = & \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -1px">R=1.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -15px">b_n=\frac{1}{n}.</math>.
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+1}<\frac{1}{n}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{n}.</math>
|-
|This is a  <math>p</math>-series with  <math>p=1.</math>  Hence, the series diverges.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=-1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -1px">R=1.</math>
|-
|    '''(b)'''     <math>(-1,1]</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 3

2017-03-11T00:19:44Z

Grad:

Determine if the following series converges or diverges. Please give your reason(s).

(a)  <math>\sum_{n=0}^{+\infty} \frac{n!}{(2n)!}</math>

(b)  <math>\sum_{n=0}^{+\infty} (-1)^n \frac{1}{n+1}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''2.''' If a series absolutely converges, then it also converges.
|-
|'''3.''' '''Alternating Series Test'''
|-
|        Let  <math>\{a_n\}</math>  be a positive, decreasing sequence where  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
|-
|        Then,  <math>\sum_{n=1}^\infty (-1)^na_n</math>  and  <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
|-
|        converge.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using the Ratio Test.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!} \cdot\frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|        <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,</math>
|-
|the series is absolutely convergent by the Ratio Test.
|-
|Therefore, the series converges.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|For
|-
|        <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math>
|-
|The sequence  <math style="vertical-align: -5px">\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n+1}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    converges
|-
|   '''(b)'''    converges
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 5

2017-03-11T00:11:38Z

Grad:

Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  at  <math style="vertical-align: -14px">x=\frac{\pi}{4}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The Taylor polynomial of   <math style="vertical-align: -5px">f(x)</math>   at   <math style="vertical-align: -1px">a</math>   is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -14px">a=\frac{\pi}{4}.</math>
|-
|First, we make a table to find the coefficients of the Taylor polynomial.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(a) </math></td>
<td align = "center"><math> \frac{f^{(n)}(a)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \cos x </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2}</math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2}</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> -\sin x</math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> -\cos x </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> -\frac{\sqrt{2}}{4} </math></td>
</tr>
<tr>
<td align = "center"><math>3</math></td>
<td align = "center"><math> \sin x </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{2} </math></td>
<td align = "center"><math> \frac{\sqrt{2}}{12}</math></td>
</tr>
</table>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -4px">T_n</math>  be the Taylor polynomial of order  <math>n.</math>
|-
|Since   <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!},</math>  we have
|-
|       <math>T_0=\frac{\sqrt{2}}{2}</math>
|-
|       <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
|-
|       <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
|-
|       <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3.</math>
|-
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       Let  <math style="vertical-align: -4px">T_n</math>  be the Taylor polynomial of order  <math>n.</math>
|-
| 
|-
|       <math>T_0=\frac{\sqrt{2}}{2}</math>
|-
|       <math>T_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)</math>
|-
|       <math>T_2=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2</math>
|-
|       <math>T_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{4}\bigg)-\frac{\sqrt{2}}{4}\bigg(x-\frac{\pi}{4}\bigg)^2+\frac{\sqrt{2}}{12}\bigg(x-\frac{\pi}{4}\bigg)^3</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 6

2017-03-11T00:09:27Z

Grad:

(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the power series of  <math style="vertical-align: -1px">\sin x?</math>
|-
|        The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|-
|So, the power series of  <math style="vertical-align: -5px">\sin(x^2)</math>   is
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, to express the indefinite integral as a power series, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \int \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.}
\end{array}</math>
|-
| 
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From part (a), we have
|-
|        <math>\int \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.</math>
|-
|Now, we have
|-
|        <math>\int_0^1 \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}\bigg|_0^1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}-0}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}</math>
|-
|    '''(b)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 7

2017-03-11T00:02:29Z

Grad:

(a) Consider the function  <math style="vertical-align: -16px">f(x)=\bigg(1-\frac{1}{2}x\bigg)^{-2}.</math>  Find the first three terms of its Binomial Series.

(b) Find its radius of convergence.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The Taylor polynomial of   <math style="vertical-align: -5px">f(x)</math>   at   <math style="vertical-align: -1px">a</math>   is
|-
|
       <math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
|-
| '''2.''' '''Ratio Test'''
|-
|        Let  <math style="vertical-align: -7px">\sum a_n</math>  be a series and  <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
|-
|        Then,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by finding the coefficients of the Maclaurin series for  <math>f(x)=\frac{1}{(1-\frac{1}{2}x)^2}.</math>
|-
|We make a table to find the coefficients of the Maclaurin series.
|-
|
<table border="1" cellspacing="0" cellpadding="11" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> f^{(n)}(x) </math></td>
<td align = "center"><math> f^{(n)}(0) </math></td>
<td align = "center"><math> \frac{f^{(n)}(0)}{n!} </math></td>
</tr>
<tr>
<td align = "center"><math>0</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^2} </math></td>
<td align = "center"><math> 1</math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> \frac{1}{(1-\frac{1}{2}x)^3}</math></td>
<td align = "center"><math> 1 </math></td>
<td align = "center"><math> 1</math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> \frac{\frac{3}{2}}{(1-\frac{1}{2}x)^4} </math></td>
<td align = "center"><math> \frac{3}{2} </math></td>
<td align = "center"><math> \frac{3}{4} </math></td>
</tr>
</table>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the first three terms of the Binomial Series is
|-
|        <math>1+x+\frac{3}{4}x^2.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The Maclaurin series of  <math>\frac{1}{(1-x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)x^n.</math>
|-
|So, the Maclaurin series of  <math>\frac{1}{(1-\frac{1}{2}x)^2}</math>  is
|-
|       <math>\sum_{n=0}^\infty (n+1)\bigg(\frac{1}{2}x\bigg)^n=\sum_{n=0}^\infty \frac{(n+1)x^n}{2^n}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Ratio Test to determine the radius of convergence of this power series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+2)x^{n+1}}{2^{n+1}} \frac{2^n}{(n+1)x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{|x|}{2} \frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{n+2}{n+1}}\\
&&\\
& = & \displaystyle{\frac{|x|}{2}.}
\end{array}</math>
|-
|Now, the Ratio Test says this series converges if  <math style="vertical-align: -14px">\frac{|x|}{2}<1.</math>  So,  <math style="vertical-align: -6px">|x|<2.</math>
|-
|Hence, the radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>1+x+\frac{3}{4}x^2</math>
|-
|   '''(b)'''    The radius of convergence is  <math style="vertical-align: 0px">R=2.</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 8

2017-03-10T23:56:48Z

Grad:

Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Taylor's Theorem'''
|-
|        Let  <math style="vertical-align: -5px">f</math>  be a function whose  <math style="vertical-align: -4px">(n+1)^{\mathrm{th}}</math> derivative exists on an interval  <math style="vertical-align: 0px">I</math>,  and let  <math style="vertical-align: 0px">c</math>  be in  <math style="vertical-align: 0px">I.</math>
|-
|        Then, for each  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -4px">I,</math>  there exists  <math style="vertical-align: -3px">z_x</math>  between  <math style="vertical-align: 0px">x</math>  and  <math style="vertical-align: 0px">c</math>  such that
|-
|        <math>f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+\cdots+\frac{f^{(n)}(c)}{n!}(x-c)^n+R_n(x),</math>
|-
|        where  <math style="vertical-align: -18px">R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
|-
|        Also,  <math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using Taylor's Theorem, we have that the error in approximating  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  with
|-
|the Maclaurin polynomial of degree  <math style="vertical-align: 0px">n</math>  is  <math style="vertical-align: -16px">R_n\bigg(\frac{\pi}{3}\bigg)</math>  where
|-
|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}\bigg|\bigg(\frac{\pi}{3}-0\bigg)^{n+1}\bigg|.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We note that
|-
|       <math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>  or  <math style="vertical-align: -5px">|f^{n+1}(z)|=|\sin(z)|\le 1.</math>
|-
|Therefore, we have
|-
|        <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}.</math>
|-
|Now, we have the following table.
|-
|<table border="1" cellspacing="0" cellpadding="6" align = "center">
<tr>
<td align = "center"><math> n</math></td>
<td align = "center"><math> \approx\frac{1}{(n+1)!}\bigg(\frac{\pi}{3}\bigg)^{n+1}</math></td>
</tr>
<tr>
<td align = "center"><math>1</math></td>
<td align = "center"><math> 0.548311 </math></td>
</tr>
<tr>
<td align = "center"><math>2</math></td>
<td align = "center"><math> 0.191396</math></td>
</tr>
<tr>
<td align = "center"><math>3</math></td>
<td align = "center"><math> 0.050107 </math></td>
</tr>
<tr>
<td align = "center"><math>4</math></td>
<td align = "center"><math> 0.01049 </math></td>
</tr>
<tr>
<td align = "center"><math>5</math></td>
<td align = "center"><math> 0.00183 </math></td>
</tr>
<tr>
<td align = "center"><math>6</math></td>
<td align = "center"><math> 0.000274 </math></td>
</tr>
<tr>
<td align = "center"><math>7</math></td>
<td align = "center"><math> 0.0000358 </math></td>
</tr>
</table>
|-
|So,  <math style="vertical-align: 0px">n=7</math>  is the smallest value of  <math style="vertical-align: 0px">n</math>  where the error is less than or equal to 0.0001.
|-
|Therefore, for  <math style="vertical-align: 0px">n=7</math>  the Maclaurin polynomial approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>n=7</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 10

2017-03-10T23:52:10Z

Grad:

Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The formula for the arc length  <math style="vertical-align: 0px">L</math>  of a parametric curve with  <math style="vertical-align: -4px">\alpha \leq t \leq \beta </math>  is
|-
|
       <math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to calculate  <math style="vertical-align: -14px">\frac{dx}{dt}</math>  and  <math style="vertical-align: -14px">\frac{dy}{dt}.</math>
|-
|Since  <math style="vertical-align: -14px">x=t^2,~\frac{dx}{dt}=2t.</math>
|-
|Since  <math style="vertical-align: -14px">y=t^3,~\frac{dy}{dt}=3t^2.</math>
|-
|Using the formula in Foundations, we have
|-
|
       <math>L=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{L} & = & \displaystyle{\int_1^2 \sqrt{4t^2+9t^4}~dt}\\
&&\\
& = & \displaystyle{\int_1^2 \sqrt{t^2(4+9t^2)}~dt}\\
&&\\
& = & \displaystyle{\int_1^2 t\sqrt{4+9t^2}~dt.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we use  <math>u</math>-substitution.
|-
|Let  <math style="vertical-align: -2px">u=4+9t^2.</math>
|-
|Then,  <math style="vertical-align: -2px">du=18tdt</math>  and  <math style="vertical-align: -15px">\frac{du}{18}=tdt.</math>
|-
|Also, since this is a definite integral, we need to change the bounds of integration.
|-
|We have
|-
|       <math style="vertical-align: -5px">u_1=4+9(1)^2=13</math>  and  <math style="vertical-align: -5px">u_2=4+9(2)^2=40.</math>
|-
|Hence,
|-
|       <math>\begin{array}{rcl}
\displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{18}\cdot\frac{2}{3} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\
&&\\
& = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 2

2017-03-10T22:54:14Z

Grad:

For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

(b)  <math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>
|-
|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series
|-
|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.
|-
|'''2.''' The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
|-
|
        <math>s_n=\sum_{i=1}^n a_i.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -3px">a_n</math>  be the  <math style="vertical-align: 0px">n</math>th term of this sum.
|-
|We notice that
|-
|        <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>  and  <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
|-
|So, this is a geometric series with  <math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
|-
|Since  <math style="vertical-align: -5px">|r|<1,</math>  this series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, the sum of this geometric series is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
&&\\
& = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
&&\\
& = & \displaystyle{\frac{8}{3}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using partial fraction decomposition. Let
|-
|       <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
|-
|If we multiply this equation by  <math style="vertical-align: -5px">(2x-1)(2x+1),</math>  we get
|-
|       <math>1=A(2x+1)+B(2x-1).</math>
|-
|If we let  <math style="vertical-align: -14px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">A=\frac{1}{2}.</math>
|-
|If we let  <math style="vertical-align: -14px">x=\frac{-1}{2},</math>  we get  <math style="vertical-align: -14px">B=\frac{-1}{2}.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{\frac{-1}{2}}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we look at the partial sums,  <math>s_n</math>  of this series.
|-
|First, we have
|-
|        <math>s_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).}
\end{array}</math>
|-
|If we compare  <math>s_1,s_2,s_3,</math>  we notice a pattern.
|-
|We have
|-
|       <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, to calculate the sum of this series we need to calculate
|-
|       <math>\lim_{n\rightarrow \infty} s_n.</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since the partial sums converge, the series converges and the sum of the series is  <math style="vertical-align: -15px">\frac{1}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{8}{3}</math>
|-
|   '''(b)'''    <math>\frac{1}{2}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 5

2016-03-05T04:31:14Z

Grad:

A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing

 when 50 (meters) of the string has been let out?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''The Pythagorean Theorem:''' For a right triangle with side lengths <math style="vertical-align: -4px">a,b,c</math>, where <math style="vertical-align: 0px">c</math> is the length of the
|-
|
::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Insert diagram.
|-
|From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
|-
|Taking derivatives, we get
|-
|
::<math>2hh'=2ss'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If  <math style="vertical-align: -4px">s=50,</math> then  <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
|-
|So, we have  <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
|-
|Solving for  <math style="vertical-align: -5px">s',</math> we get  <math style="vertical-align: -14px">s'=\frac{24}{5}</math>  m/s.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:<math style="vertical-align: -14px">s'=\frac{24}{5}</math>  m/s
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:08:34Z

Grad:

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d)  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(e)  
|-
| Insert sketch here.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty),</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>  and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See graph in '''(e)'''.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:07:56Z

Grad: /* 7 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

== 5 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d)  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

== 6 ==

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(e)  
|-
| Insert sketch here.
|}

== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty),</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5,</math>  and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See graph in '''(e)'''.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:06:06Z

Grad: /* 5 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

== 5 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d)  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

== 6 ==

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(e)  
|-
| Insert sketch here.
|}

== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:05:19Z

Grad: /* 6 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

== 6 ==

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(e)  
|-
| Insert sketch here.
|}

== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:04:45Z

Grad: /* 5 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: 0px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is  <math style="vertical-align: -5px">(2,-11).</math>
|}

== 6 ==
'''(e)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Insert sketch here.
|}
== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:04:07Z

Grad: /* 4 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: 0px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -5px">x=0,</math>  we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -5px">x=3,</math>  we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is concave up on the interval <math style="vertical-align: -5px">(2,\infty),</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}

== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: -1px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is <math style="vertical-align: -5px">(2,-11).</math>
|}
== 6 ==
'''(e)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Insert sketch here.
|}
== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T21:01:34Z

Grad: /* 2 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>  We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math>  So, we have  <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math><math style="vertical-align: -6px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: 0px">x=0</math>  and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:  <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math>  is increasing on <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>  and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: -1px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -1px">x=0,</math> we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -1px">x=3,</math> we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}
== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: -1px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is <math style="vertical-align: -5px">(2,-11).</math>
|}
== 6 ==
'''(e)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Insert sketch here.
|}
== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T20:56:08Z

Grad: /* 1 */

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math>  is increasing when <math style="vertical-align: -5px">f'(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math>  is concave up when <math style="vertical-align: -5px">f''(x)>0</math>  and <math style="vertical-align: -5px">f(x)</math>  is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math> We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math> So, we have <math style="vertical-align: -5px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: -1px">x=0</math> and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: -1px">x</math> break up the number line into 3 intervals: <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: -1px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -1px">x=0,</math> we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -1px">x=3,</math> we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}
== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: -1px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is <math style="vertical-align: -5px">(2,-11).</math>
|}
== 6 ==
'''(e)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Insert sketch here.
|}
== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 9

2016-03-04T20:52:40Z

Grad:

Given the function <math style="vertical-align: -5px">f(x)=x^3-6x^2+5</math>,

a) Find the intervals in which the function increases or decreases.

b) Find the local maximum and local minimum values.

c) Find the intervals in which the function concaves upward or concaves downward.

d) Find the inflection point(s).

e) Use the above information (a) to (d) to sketch the graph of <math style="vertical-align: -5px">y=f(x)</math>.

== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1.''' <math style="vertical-align: -5px">f(x)</math> is increasing when <math style="vertical-align: -5px">f'(x)>0</math> and <math style="vertical-align: -5px">f(x)</math> is decreasing when <math style="vertical-align: -5px">f'(x)<0.</math>
|-
|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
|-
|'''3.''' <math style="vertical-align: -5px">f(x)</math> is concave up when <math style="vertical-align: -5px">f''(x)>0</math> and <math style="vertical-align: -5px">f(x)</math> is concave down when <math style="vertical-align: -5px">f''(x)<0.</math>
|-
|'''4.''' Inflection points occur when <math style="vertical-align: -5px">f''(x)=0.</math>
|}

'''Solution:'''
== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math> We have <math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
|-
|Now, we set <math style="vertical-align: -5px">f'(x)=0.</math> So, we have <math style="vertical-align: -5px">0=3x(x-4).</math>
|-
|Hence, we have <math style="vertical-align: -1px">x=0</math> and <math style="vertical-align: -1px">x=4.</math>
|-
|So, these values of <math style="vertical-align: -1px">x</math> break up the number line into 3 intervals: <math style="vertical-align: -3px">(-\infty,0),(0,4),(4,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
|-
|For <math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
|-
|For <math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
|-
|For <math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|By the First Derivative Test, the local maximum occurs at <math style="vertical-align: -1px">x=0</math> and the local minimum occurs at <math style="vertical-align: -1px">x=4.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|}
== 4 ==
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the intervals when the function is concave up or concave down, we need to find <math style="vertical-align: -5px">f''(x).</math>
|-
|We have <math style="vertical-align: -5px">f''(x)=6x-12.</math>
|-
|We set <math style="vertical-align: -5px">f''(x)=0.</math>
|-
|So, we have <math style="vertical-align: -1px">0=6x-12.</math> Hence, <math style="vertical-align: -1px">x=2.</math>
|-
|This value breaks up the number line into two intervals: <math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Again, we use test points in these two intervals.
|-
|For <math style="vertical-align: -1px">x=0,</math> we have <math style="vertical-align: -5px">f''(x)=-12<0.</math>
|-
|For <math style="vertical-align: -1px">x=3,</math> we have <math style="vertical-align: -5px">f''(x)=6>0.</math>
|-
|Thus, <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|}
== 5 ==
'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Using the information from part '''(c)''', there is one inflection point that occurs at <math style="vertical-align: -1px">x=2.</math>
|-
|Now, we have <math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
|-
|So, the inflection point is <math style="vertical-align: -5px">(2,-11).</math>
|}
== 6 ==
'''(e)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
| Insert sketch here.
|}
== 7 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,0),(4,\infty)</math> and decreasing on <math style="vertical-align: -5px">(0,4).</math>
|-
|'''(b)''' The local maximum value is <math style="vertical-align: -5px">f(0)=5</math> and the local minimum value is <math style="vertical-align: -5px">f(4)=-27.</math>
|-
|'''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(2,\infty)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,2).</math>
|-
|'''(d)''' <math style="vertical-align: -5px">(2,-11)</math>
|-
|'''(e)''' See Step 1 for graph.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:51:36Z

Grad:

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: 0px">dx</math>. We have  <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to  <math style="vertical-align: 0px">2^3</math>  to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -5px">dy=12\,dx</math>
|-
|'''(b)'''  <math style="vertical-align: -1px">6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:51:08Z

Grad: /* 4 */

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: 0px">dx</math>. We have  <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to  <math style="vertical-align: 0px">2^3</math>  to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
|}

== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -5px">dy=12\,dx</math>
|-
|'''(b)'''  <math style="vertical-align: -1px">6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:49:54Z

Grad: /* 4 */

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: 0px">dx</math>. We have  <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to  <math style="vertical-align: 0px">2^3</math>  to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
|}

== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>dy=12dx</math>
|-
|'''(b)'''  <math>6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:49:24Z

Grad: /* 3 */

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: 0px">dx</math>. We have  <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to  <math style="vertical-align: 0px">2^3</math>  to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
|}

== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>dy=12dx</math>
|-
|'''(b)''' <math>6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:47:44Z

Grad: /* 2 */

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: -1px">dx</math>. We have <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy=12(-0.1)=-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to <math style="vertical-align: 0px">2^3</math> to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\approx 2^3+-1.2=6.8.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>dy=12dx</math>
|-
|'''(b)''' <math>6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:45:06Z

Grad: /* 1 */

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math>dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math> we have
|-
|
::<math>dy=3x^2dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in <math style="vertical-align: -1px">x=2</math> into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy=3(2)^2dx=12dx.</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: -1px">dx</math>. We have <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy=12(-0.1)=-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to <math style="vertical-align: 0px">2^3</math> to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\approx 2^3+-1.2=6.8.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>dy=12dx</math>
|-
|'''(b)''' <math>6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 8

2016-03-04T20:33:37Z

Grad:

Let

::::::<math>y=x^3.</math>

a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.

b) Use differentials to find an approximate value for <math style="vertical-align: -2px">1.9^3</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
|-
|
::Since <math style="vertical-align: -1px">x=1,</math> the differential is <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
|}

'''Solution:'''
== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the differential <math>dy.</math>
|-
|Since <math style="vertical-align: -5px">y=x^3,</math> we have
|-
|
::<math>dy=3x^2dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in <math style="vertical-align: -1px">x=2</math> into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy=3(2)^2dx=12dx.</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: -1px">dx</math>. We have <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
|-
|Then, we plug this into the differential from part '''(a)'''.
|-
|So, we have
|-
|
::<math>dy=12(-0.1)=-1.2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to <math style="vertical-align: 0px">2^3</math> to get an
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
|-
|Hence, we have
|-
|
::<math>1.9^3\approx 2^3+-1.2=6.8.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>dy=12dx</math>
|-
|'''(b)''' <math>6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T20:32:12Z

Grad:

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get  <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point  <math style="vertical-align: -5px">(3,3).</math>
|-
|We plug <math style="vertical-align: -5px">(3,3)</math>  into the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  is
|-
|
::<math>y\,=\,-1(x-3)+3.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)'''  <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T20:31:51Z

Grad: /* 4 */

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get  <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point  <math style="vertical-align: -5px">(3,3).</math>
|-
|We plug <math style="vertical-align: -5px">(3,3)</math>  into the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  is
|-
|
::<math>y\,=\,-1(x-3)+3.</math>
|}

== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)'''  <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T20:31:21Z

Grad: /* 3 */

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get  <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point  <math style="vertical-align: -5px">(3,3).</math>
|-
|We plug <math style="vertical-align: -5px">(3,3)</math>  into the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  is
|-
|
::<math>y\,=\,-1(x-3)+3.</math>
|}

== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)''' <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T20:29:26Z

Grad: /* 2 */

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get  <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point <math style="vertical-align: -4px">(3,3).</math>
|-
|We plug in <math style="vertical-align: -4px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -4px">(3,3)</math> and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -4px">(3,3)</math> is
|-
|
::<math>y=-1(x-3)+3.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)''' <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T20:28:02Z

Grad: /* 1 */

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the <math style="vertical-align: -12px">\frac{dy}{dx}</math> terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get <math style="vertical-align: -12px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point <math style="vertical-align: -4px">(3,3).</math>
|-
|We plug in <math style="vertical-align: -4px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -4px">(3,3)</math> and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -4px">(3,3)</math> is
|-
|
::<math>y=-1(x-3)+3.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)''' <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T19:21:43Z

Grad:

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
== 1 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math> by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''
== 2 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the <math style="vertical-align: -12px">\frac{dy}{dx}</math> terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get <math style="vertical-align: -12px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}
== 3 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point <math style="vertical-align: -4px">(3,3).</math>
|-
|We plug in <math style="vertical-align: -4px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -4px">(3,3)</math> and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -4px">(3,3)</math> is
|-
|
::<math>y=-1(x-3)+3.</math>
|}
== 4 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)''' <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 7

2016-03-04T19:16:23Z

Grad:

A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math> by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the <math style="vertical-align: -12px">\frac{dy}{dx}</math> terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get <math style="vertical-align: -12px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point <math style="vertical-align: -4px">(3,3).</math>
|-
|We plug in <math style="vertical-align: -4px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -4px">(3,3)</math> and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -4px">(3,3)</math> is
|-
|
::<math>y=-1(x-3)+3.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)''' <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T19:15:25Z

Grad:

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that  <math style="vertical-align: -5px">f(0)=7.</math>
|-
|Also,  <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since  <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus,  <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>  and hence  <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that  <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with  <math style="vertical-align: 0px">a<c<b</math> such that  <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>  Since  <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>  So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>  has at most one zero.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T19:15:07Z

Grad: /* 2 */

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''
== 1 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that  <math style="vertical-align: -5px">f(0)=7.</math>
|-
|Also,  <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since  <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus,  <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>  and hence  <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|}

== 2 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exist <math style="vertical-align: -4px">a,b</math> such that  <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: 0px">c</math> with  <math style="vertical-align: 0px">a<c<b</math> such that  <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math>  Since  <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math>  So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math>  has at most one zero.
|}

== 3 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T19:12:51Z

Grad: /* 3 */

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''
== 1 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that  <math style="vertical-align: -5px">f(0)=7.</math>
|-
|Also,  <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since  <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus,  <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>  and hence  <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|}

== 2 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math> Since <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math> So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math> has at most one zero.
|}
== 3 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T19:11:15Z

Grad: /* 1 */

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''
== 1 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that  <math style="vertical-align: -5px">f(0)=7.</math>
|-
|Also,  <math style="vertical-align: -5px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since  <math style="vertical-align: -5px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus,  <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math>  and hence  <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math>  and  <math style="vertical-align: -5px">f(0)>0,</math>  there exists <math style="vertical-align: 0px">x</math> with  <math style="vertical-align: 0px">-5<x<0</math>  such that
|-
|<math style="vertical-align: -5px">f(x)=0</math>  by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math>  has at least one zero.
|}

== 2 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math> Since <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math> So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math> has at most one zero.
|}
== 3 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math> and <math style="vertical-align: -5px">f(0)>0,</math> there exists <math style="vertical-align: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
|-
|<math style="vertical-align: -5px">f(x)=0</math> by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math> has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T18:55:39Z

Grad:

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''
== 1 ==
'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that <math style="vertical-align: -3px">f(0)=7.</math>
|-
|Also, <math style="vertical-align: -3px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since <math style="vertical-align: -3px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus, <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math> and hence <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math> and <math style="vertical-align: -5px">f(0)>0,</math> there exists <math style="vertical-align: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
|-
|<math style="vertical-align: -5px">f(x)=0</math> by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math> has at least one zero.
|}
== 2 ==
'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math> Since <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math> So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math> has at most one zero.
|}
== 3 ==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math> and <math style="vertical-align: -5px">f(0)>0,</math> there exists <math style="vertical-align: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
|-
|<math style="vertical-align: -5px">f(x)=0</math> by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math> has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 6

2016-03-04T18:54:22Z

Grad:

Consider the following function:

::::::<math>f(x)=3x-2\sin x+7</math>

a) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at least one zero.

b) Use the Mean Value Theorem to show that <math style="vertical-align: -5px">f(x)</math>  has at most one zero.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''1. Intermediate Value Theorem:''' If <math style="vertical-align: -5px">f(x)</math>  is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math> and <math style="vertical-align: 0px">c</math> is any number
|-
|
::between <math style="vertical-align: -5px">f(a)</math>  and <math style="vertical-align: -5px">f(b)</math>, then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
|-
|'''2. Mean Value Theorem:''' Suppose <math style="vertical-align: -5px">f(x)</math>  is a function that satisfies the following:
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is continuous on the closed interval  <math style="vertical-align: -5px">[a,b].</math>
|-
|
::<math style="vertical-align: -5px">f(x)</math>  is differentiable on the open interval <math style="vertical-align: -5px">(a,b).</math>
|-
|
::Then, there is a number <math style="vertical-align: 0px">c</math> such that  <math style="vertical-align: 0px">a<c<b</math>  and <math style="vertical-align: -14px">f'(c)=\frac{f(b)-f(a)}{b-a}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First note that <math style="vertical-align: -3px">f(0)=7.</math>
|-
|Also, <math style="vertical-align: -3px">f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).</math>
|-
|Since <math style="vertical-align: -3px">-1\leq \sin(x) \leq 1,</math>
|-
|
::<math>-2\leq -2\sin(x) \leq 2.</math>
|-
|Thus, <math style="vertical-align: -5px">-10\leq f(-5) \leq -6</math> and hence <math style="vertical-align: -5px">f(-5)<0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -5px">f(-5)<0</math> and <math style="vertical-align: -5px">f(0)>0,</math> there exists <math style="vertical-align: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
|-
|<math style="vertical-align: -5px">f(x)=0</math> by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math> has at least one zero.
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Suppose that <math style="vertical-align: -5px">f(x)</math> has more than one zero. So, there exists <math style="vertical-align: -5px">a,b</math> such that <math style="vertical-align: -5px">f(a)=f(b)=0.</math>
|-
|Then, by the Mean Value Theorem, there exists <math style="vertical-align: -1px">c</math> with <math style="vertical-align: -1px">a<c<b</math> such that <math style="vertical-align: -5px">f'(c)=0.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We have <math style="vertical-align: -5px">f'(x)=3-2\cos(x).</math> Since <math style="vertical-align: -5px">-1\leq \cos(x)\leq 1,</math>
|-
|<math style="vertical-align: -5px">-2 \leq -2\cos(x)\leq 2.</math> So, <math style="vertical-align: -5px">1\leq f'(x) \leq 5,</math>
|-
|which contradicts <math style="vertical-align: -5px">f'(c)=0.</math> Thus, <math style="vertical-align: -5px">f(x)</math> has at most one zero.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' Since <math style="vertical-align: -5px">f(-5)<0</math> and <math style="vertical-align: -5px">f(0)>0,</math> there exists <math style="vertical-align: -1px">x</math> with <math style="vertical-align: -1px">-5<x<0</math> such that
|-
|<math style="vertical-align: -5px">f(x)=0</math> by the Intermediate Value Theorem. Hence, <math style="vertical-align: -5px">f(x)</math> has at least one zero.
|-
|'''(b)''' See '''Step 1''' and '''Step 2''' above.
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

009A Sample Final 1, Problem 5

2016-03-04T18:49:34Z

Grad:

A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing

 when 50 (meters) of the string has been let out?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''The Pythagorean Theorem:''' For a right triangle with side lengths <math style="vertical-align: -4px">a,b,c</math>, where <math style="vertical-align: 0px">c</math> is the length of the
|-
|
::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Insert diagram.
|-
|From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
|-
|Taking derivatives, we get
|-
|
::<math>2hh'=2ss'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If  <math style="vertical-align: -4px">s=50,</math> then  <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
|-
|So, we have  <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
|-
|Solving for  <math style="vertical-align: -5px">s',</math> we get  <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:<math>s'=\frac{24}{5}</math> m/s
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]