Difference between revisions of "009A Sample Final 2, Problem 10"

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<span class="exam">(d) Sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
 
<span class="exam">(d) Sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x).</math>
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[[009A Sample Final 2, Problem 10 Solution|'''<u>Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Foundations: &nbsp;
 
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|-
 
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|'''1.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing when &nbsp;<math style="vertical-align: -5px">f'(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is decreasing when &nbsp;<math style="vertical-align: -5px">f'(x)<0.</math>
 
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|-
 
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|'''2. The First Derivative Test''' tells us when we have a local maximum or local minimum.
 
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|'''3.''' <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up when &nbsp;<math style="vertical-align: -5px">f''(x)>0</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave down when &nbsp;<math style="vertical-align: -5px">f''(x)<0.</math>
 
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|-
 
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|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|}
 
  
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[[009A Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''Solution:'''
 
  
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'''(a)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 
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|-
 
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|Using the Quotient Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f'(x)} & = & \displaystyle{\frac{(x^2+1)(4x)'-(4x)(x^2+1)'}{(x^2+1)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)(4)-(4x)(2x)}{(x^2+1)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{-4(x^2-1)}{(x^2+1)^2}.}
 
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\end{array}</math>
 
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|-
 
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|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp;
 
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|-
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -6px">-4(x^2-1)=0.</math>
 
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|-
 
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|Hence, we have &nbsp;<math style="vertical-align: 0px">x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 
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|-
 
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|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-1),(-1,1),(1,\infty).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 
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|-
 
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|For &nbsp;<math style="vertical-align: -15px">x=-2,~f'(x)=\frac{-12}{25}<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=0,~f'(x)=4>0.</math>
 
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|For &nbsp;<math style="vertical-align: -15px">x=2,~f'(x)=\frac{-12}{25}<0.</math>
 
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|-
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-1,1)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Using the First Derivative Test, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; has a local minimum at &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; and a local maximum at &nbsp;<math style="vertical-align: 0px">x=1.</math>&nbsp;
 
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|-
 
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|Thus, the local maximum and local minimum values of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; are
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(1)=2,~f(-1)=-2.</math>
 
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|
 
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|}
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|To find the intervals when the function is concave up or concave down, we need to find &nbsp;<math style="vertical-align: -5px">f''(x).</math>
 
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|-
 
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|Using the Quotient Rule and Chain Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f''(x)} & = & \displaystyle{\frac{(x^2+1)^2(-4(x^2-1))'+4(x^2-1)((x^2+1)^2)'}{((x^2+1)^2)^2}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(x^2+1)'}{(x^2+1)^4}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)^2(-8x)+4(x^2-1)2(x^2+1)(2x)}{(x^2+1)^4}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{(x^2+1)(-8x)+16(x^2-1)x}{(x^2+1)^3}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{8x^3-24x}{(x^2+1)^3}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{8x(x^2-3)}{(x^2+1)^3}.}
 
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\end{array}</math>
 
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|-
 
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|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">8x(x^2-3)=0.</math>&nbsp;
 
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|Hence,
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
 
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|This value breaks up the number line into four intervals:
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3}),(-\sqrt{3},0),(0,\sqrt{3}),(\sqrt{3},\infty).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|Again, we use test points in these four intervals.
 
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|For &nbsp;<math style="vertical-align: -5px">x=-2,</math>&nbsp; we have &nbsp;<math style="vertical-align: -15px">f''(x)=\frac{-16}{125}<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=-1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=2>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=1,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-2<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=2,</math>&nbsp; we have &nbsp;<math style="vertical-align: -15px">f''(x)=\frac{16}{125}>0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on &nbsp;<math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>&nbsp; and concave down on &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|The inflection points occur at
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=0,~x=-\sqrt{3},~x=\sqrt{3}.</math>
 
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|Plugging these into &nbsp;<math style="vertical-align: -5px">f(x),</math> we get the inflection points
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</math>
 
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|}
 
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'''(c)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|By L'Hopital's Rule, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow \infty } f(x)} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{4x}{x^2+1}}\\
 
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&&\\
 
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& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{4}{2x}}\\
 
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&&\\
 
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& = & \displaystyle{0.}
 
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\end{array}</math>
 
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|Similarly, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{x\rightarrow -\infty } f(x)} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{4x}{x^2+1}}\\
 
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&&\\
 
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& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow -\infty} \frac{4}{2x}}\\
 
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&&\\
 
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& = & \displaystyle{0.}
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|Since
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\displaystyle{\lim_{x\rightarrow -\infty } f(x)=\lim_{x\rightarrow \infty } f(x)=0,}</math>
 
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|<math style="vertical-align: -5px">f(x)</math>&nbsp; has a horizontal asymptote
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=0.</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(d): &nbsp;
 
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|Insert sketch
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-1,1)</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(-\infty,-1)\cup (1,\infty).</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;The local maximum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">2</math>&nbsp; and the local minimum value of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">-2</math>&nbsp;
 
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on &nbsp;<math style="vertical-align: -5px">(-\sqrt{3},0)\cup(\sqrt{3},\infty),</math>&nbsp; and concave down on &nbsp;<math style="vertical-align: -5px">(-\infty,-\sqrt{3})\cup(0,\sqrt{3}).</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; The inflection points are &nbsp;<math style="vertical-align: -5px">(0,0),(-\sqrt{3},-\sqrt{3}),(\sqrt{3},\sqrt{3}).</math>
 
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>y=0</math>
 
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|&nbsp; &nbsp;'''(d)'''&nbsp; &nbsp; See above
 
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|}
 
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 10:29, 1 December 2017

Let

(a) Find all local maximum and local minimum values of    find all intervals where    is increasing and all intervals where    is decreasing.

(b) Find all inflection points of the function    find all intervals where the function    is concave upward and all intervals where    is concave downward.

(c) Find all horizontal asymptotes of the graph  

(d) Sketch the graph of  

Solution


Detailed Solution


Return to Sample Exam

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