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

From Grad Wiki
Jump to navigation Jump to search
 
(3 intermediate revisions by the same user not shown)
Line 9: Line 9:
  
 
<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>
 +
<hr>
 +
[[009A Sample Final 2, Problem 10 Solution|'''<u>Solution</u>''']]
  
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Foundations: &nbsp;
 
−
|-
 
−
|'''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>
 
−
|-
 
−
|'''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>&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>
 
−
|-
 
−
|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
−
|}
 
  
 +
[[009A Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
−
'''Solution:'''
 
  
−
'''(a)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp;
 
−
|-
 
−
|Using the Quotient Rule, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
−
\displaystyle{f'(x)} & = & \displaystyle{\frac{(x^2+1)(4x)'-(4x)(x^2+1)'}{(x^2+1)^2}}\\
 
−
&&\\
 
−
& = & \displaystyle{\frac{(x^2+1)(4)-(4x)(2x)}{(x^2+1)^2}}\\
 
−
&&\\
 
−
& = & \displaystyle{\frac{-4(x^2-1)}{(x^2+1)^2}.}
 
−
\end{array}</math>
 
−
|-
 
−
|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp;
 
−
|-
 
−
|So, we have &nbsp;<math style="vertical-align: -6px">-4(x^2-1)=0.</math>
 
−
|-
 
−
|Hence, we have &nbsp;<math style="vertical-align: 0px">x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 
−
|-
 
−
|So, these values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; break up the number line into 3 intervals:
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,-1),(-1,1),(1,\infty).</math>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
−
|-
 
−
|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 
−
|-
 
−
|For &nbsp;<math style="vertical-align: -15px">x=-2,~f'(x)=\frac{-12}{25}<0.</math>
 
−
|-
 
−
|For &nbsp;<math style="vertical-align: -5px">x=0,~f'(x)=4>0.</math>
 
−
|-
 
−
|For &nbsp;<math style="vertical-align: -15px">x=2,~f'(x)=\frac{-12}{25}<0.</math>
 
−
|-
 
−
|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>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 3: &nbsp;
 
−
|-
 
−
|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;
 
−
|-
 
−
|Thus, the local maximum and local minimum values of &nbsp;<math style="vertical-align: -4px">f</math>&nbsp; are
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(1)=2,~f(-1)=-2.</math>
 
−
|-
 
−
|
 
−
|}
 
−
 
−
'''(b)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|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>
 
−
|-
 
−
|We have &nbsp;<math style="vertical-align: -5px">f''(x)=6x-12.</math>
 
−
|-
 
−
|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
−
|-
 
−
|So, we have &nbsp;<math style="vertical-align: -1px">0=6x-12.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=2.</math>
 
−
|-
 
−
|This value breaks up the number line into two intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
 
−
|-
 
−
|
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
−
|-
 
−
|
 
−
|}
 
−
 
−
'''(c)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|First, we note that the degree of the numerator is &nbsp;<math style="vertical-align: -1px">1</math>&nbsp; and
 
−
|-
 
−
|the degree of the denominator is &nbsp;<math style="vertical-align: 0px">2.</math>&nbsp;
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 2: &nbsp;
 
−
|-
 
−
|Since the degree of the denominator is greater than the degree of the numerator,
 
−
|-
 
−
|<math style="vertical-align: -5px">f(x)</math>&nbsp; has a horizontal asymptote
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=0.</math>
 
−
|}
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!(d): &nbsp;
 
−
|-
 
−
|Insert sketch
 
−
|-
 
−
|
 
−
|-
 
−
|
 
−
|-
 
−
|
 
−
|}
 
−
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Final Answer: &nbsp;
 
−
|-
 
−
|&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>
 
−
|-
 
−
|&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;
 
−
|-
 
−
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
 
−
|-
 
−
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>y=0</math>
 
−
|-
 
−
|&nbsp; &nbsp;'''(d)'''&nbsp; &nbsp; See above
 
−
|}
 
 
[[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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,}   find all intervals where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   is increasing and all intervals where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   is decreasing.

(b) Find all inflection points of the function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,}   find all intervals where the function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   is concave upward and all intervals where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}   is concave downward.

(c) Find all horizontal asymptotes of the graph  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=f(x).}

(d) Sketch the graph of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=f(x).}

Solution


Detailed Solution


Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009A_Sample_Final_2,_Problem_10&oldid=4253"