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