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| | <span class="exam">(d) Sketch the shape of the graph of <math style="vertical-align: -4px">f.</math> | | <span class="exam">(d) Sketch the shape of the graph of <math style="vertical-align: -4px">f.</math> |
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| | + | [[009A Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''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>
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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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| − | |-
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| − | |'''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>
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| − | |}
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| | + | [[009A Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| | | | |
| − | '''(a)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math>
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| − | |-
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| − | |We have <math style="vertical-align: -5px">f'(x)=24x^2-4x^3.</math>
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| − | |-
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| − | |Now, we set <math style="vertical-align: -5px">f'(x)=0.</math> So, we have
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| − | |-
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| − | | <math style="vertical-align: -6px">0=4x^2(6-x).</math>
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| − | |-
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| − | |Hence, we have <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: -1px">x=6.</math>
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| − | |-
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| − | |So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals:
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| − | |-
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| − | | <math style="vertical-align: -5px">(-\infty,0),(0,6),(6,\infty).</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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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 <math style="vertical-align: -5px">x=-1,~f'(x)=28>0.</math>
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| − | |-
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| − | |For <math style="vertical-align: -5px">x=1,~f'(x)=20>0.</math>
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| − | |-
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| − | |For <math style="vertical-align: -5px">x=7,~f'(x)=-196<0.</math>
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| − | |-
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| − | |Thus, <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,6)</math> and decreasing on <math style="vertical-align: -5px">(6,\infty).</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |The critical points of <math style="vertical-align: -5px">f(x)</math> occur at <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: 0px">x=6.</math>
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| − | |-
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| − | |Plugging these values into <math style="vertical-align: -5px">f(x),</math> we get the critical points
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| − | |-
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| − | | <math style="vertical-align: -4px">(0,4)</math> and <math style="vertical-align: -4px">(6,436).</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Using the first derivative test and the information from part (a),
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| − | |-
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| − | | <math style="vertical-align: -4px">(0,4)</math> is not a local minimum or local maximum and
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| − | |-
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| − | | <math style="vertical-align: -4px">(6,436)</math> is a local maximum.
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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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 <math style="vertical-align: -5px">f''(x).</math>
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| − | |-
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| − | |We have <math style="vertical-align: -5px">f''(x)=48x-12x^2.</math>
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| − | |-
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| − | |We set <math style="vertical-align: -5px">f''(x)=0.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | | <math style="vertical-align: -1px">0=12x(4-x).</math>
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| − | |-
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| − | |Hence, <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: 0px">x=4</math>.
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| − | |-
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| − | |This value breaks up the number line into three intervals:
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| − | |-
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| − | | <math style="vertical-align: -5px">(-\infty,0),(0,4),(4,\infty).</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Again, we use test points in these three intervals.
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| − | |-
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| − | |For <math style="vertical-align: -5px">x=-1,</math> we have <math style="vertical-align: -5px">f''(x)=-60<0.</math>
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| − | |-
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| − | |For <math style="vertical-align: -5px">x=1,</math> we have <math style="vertical-align: -5px">f''(x)=48>0.</math>
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| − | |-
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| − | |For <math style="vertical-align: -5px">x=5,</math> we have <math style="vertical-align: -5px">f''(x)=-60<0.</math>
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| − | |-
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| − | |Thus, <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(0,4)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty).</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(d):
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| − | |-
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| − | |Insert graph
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| − | |}
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| − |
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math style="vertical-align: -5px">f(x)</math> is increasing on <math style="vertical-align: -5px">(-\infty,6)</math> and decreasing on <math style="vertical-align: -5px">(6,\infty).</math>
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| − | |-
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| − | | '''(b)''' The critical points are <math style="vertical-align: -4px">(0,4)</math> and <math style="vertical-align: -4px">(6,436).</math>
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| − | |-
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| − | | <math style="vertical-align: -4px">(0,4)</math> is not a local minimum or local maximum and <math style="vertical-align: -5px">(6,436)</math> is a local maximum.
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| − | |-
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| − | | '''(c)''' <math style="vertical-align: -5px">f(x)</math> is concave up on the interval <math style="vertical-align: -5px">(0,4)</math> and concave down on the interval <math style="vertical-align: -5px">(-\infty,0)\cup (4,\infty).</math>
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| − | |-
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| − | | '''(d)''' See above
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| − | |}
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| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
