Difference between revisions of "009A Sample Final 1, Problem 9"

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<span class="exam">(e) Use the above information (a) to (d) to sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x)</math>.
 
<span class="exam">(e) Use the above information (a) to (d) to sketch the graph of &nbsp;<math style="vertical-align: -5px">y=f(x)</math>.
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<hr>
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!Foundations: &nbsp;
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[[009A Sample Final 1, Problem 9 Solution|'''<u>Solution</u>''']]
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|Recall:
 
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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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|'''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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|'''4.''' Inflection points occur when &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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'''Solution:'''
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[[009A Sample Final 1, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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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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|We start by taking the derivative of &nbsp;<math style="vertical-align: -5px">f(x).</math>&nbsp; We have &nbsp;<math style="vertical-align: -5px">f'(x)=3x^2-12x.</math>
 
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|Now, we set &nbsp;<math style="vertical-align: -5px">f'(x)=0.</math>&nbsp; So, we have &nbsp;<math style="vertical-align: -6px">0=3x(x-4).</math>
 
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|Hence, we have &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=4.</math>
 
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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: &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
 
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!Step 2: &nbsp;
 
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|To check whether the function is increasing or decreasing in these intervals, we use testpoints.
 
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|For &nbsp;<math style="vertical-align: -5px">x=-1,~f'(x)=15>0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=1,~f'(x)=-9<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=5,~f'(x)=15>0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is increasing on &nbsp;<math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty),</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(0,4).</math>
 
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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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|By the First Derivative Test, the local maximum occurs at &nbsp;<math style="vertical-align: -1px">x=0</math>&nbsp; and the local minimum occurs at &nbsp;<math style="vertical-align: -1px">x=4.</math>
 
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!Step 2: &nbsp;
 
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|So, the local maximum value is &nbsp;<math style="vertical-align: -5px">f(0)=5</math>&nbsp; and the local minimum value is &nbsp;<math style="vertical-align: -5px">f(4)=-27.</math>
 
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'''(c)'''
 
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!Step 1: &nbsp;
 
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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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|We have &nbsp;<math style="vertical-align: -5px">f''(x)=6x-12.</math>
 
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|We set &nbsp;<math style="vertical-align: -5px">f''(x)=0.</math>
 
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|So, we have &nbsp;<math style="vertical-align: -1px">0=6x-12.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: 0px">x=2.</math>
 
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|This value breaks up the number line into two intervals: &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
 
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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 two intervals.
 
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|For &nbsp;<math style="vertical-align: -5px">x=0,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=-12<0.</math>
 
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|For &nbsp;<math style="vertical-align: -5px">x=3,</math>&nbsp; we have &nbsp;<math style="vertical-align: -5px">f''(x)=6>0.</math>
 
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|Thus, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(2,\infty),</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,2).</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(d) &nbsp;
 
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| Using the information from part (c), there is one inflection point that occurs at &nbsp;<math style="vertical-align: 0px">x=2.</math>
 
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|Now, we have &nbsp;<math style="vertical-align: -5px">f(2)=8-24+5=-11.</math>
 
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|So, the inflection point is &nbsp;<math style="vertical-align: -5px">(2,-11).</math>
 
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!(e) &nbsp;
 
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| Insert sketch here.
 
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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">(-\infty,0),(4,\infty),</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(0,4).</math>
 
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|-
 
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|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;The local maximum value is &nbsp;<math style="vertical-align: -5px">f(0)=5,</math>&nbsp; and the local minimum value is &nbsp;<math style="vertical-align: -5px">f(4)=-27.</math>
 
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is concave up on the interval &nbsp;<math style="vertical-align: -5px">(2,\infty),</math>&nbsp; and concave down on the interval &nbsp;<math style="vertical-align: -5px">(-\infty,2).</math>
 
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|&nbsp; &nbsp;'''(d)''' &nbsp; &nbsp; <math style="vertical-align: -5px">(2,-11)</math>
 
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|&nbsp; &nbsp;'''(e)''' &nbsp; &nbsp; See graph above.
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 15:59, 2 December 2017

Given the function  ,

(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  .

Solution


Detailed Solution


Return to Sample Exam

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