View source for 009A Sample Final 1, Problem 9

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

<span class="exam">(a) Find the intervals in which the function increases or decreases.

<span class="exam">(b) Find the local maximum and local minimum values.

<span class="exam">(c) Find the intervals in which the function concaves upward or concaves downward.

<span class="exam">(d) Find the inflection point(s).

<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>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations: &nbsp; 
|-
|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>
|}


'''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; 
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|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 
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|&nbsp; &nbsp; &nbsp; &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: 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\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. 
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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>
|-
|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>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1: &nbsp; 
|-
|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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2: &nbsp;
|-
|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>
|}

'''(c)'''

{| 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>
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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 
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|&nbsp; &nbsp; &nbsp; &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: 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">(-\infty,2),(2,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2: &nbsp;
|-
|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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(d) &nbsp; 
|-
| 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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(e) &nbsp; 
|-
|[[File: 9AF1_9_GP.png|700px|center]]
|}

{| 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">(-\infty,0),(4,\infty),</math>&nbsp; and decreasing on &nbsp;<math style="vertical-align: -5px">(0,4).</math>
|-
|&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>
|-
|&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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