View source for 009A Sample Final 1, Problem 10

<span class="exam">Consider the following continuous function:
::::::<math>f(x)=x^{1/3}(x-8)</math>

<span class="exam">defined on the closed, bounded interval <math style="vertical-align: -5px">[-8,8]</math>.

::<span class="exam">a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>.

::<span class="exam">b) Determine the absolute maximum and absolute minimum values for <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[-8,8]</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations: &nbsp; 
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|Recall:
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::'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <math style="vertical-align: -1px">x.</math>
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::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined. 
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::'''2.''' To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on an interval <math>[a,b],</math>
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::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
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'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1: &nbsp; 
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|To find the critical points, first we need to find <math style="vertical-align: -5px">f'(x).</math>
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|Using the Product Rule, we have
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::<math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
&&\\
& = & \displaystyle{\frac{x-8}{3x^{\frac{2}{3}}}+x^{\frac{1}{3}}.}\\
\end{array}</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2: &nbsp;
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|Notice <math style="vertical-align: -5px">f'(x)</math> is undefined when <math style="vertical-align: -1px">x=0.</math>
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|Now, we need to set <math style="vertical-align: -5px">f'(x)=0.</math>
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|So, we get 
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::<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
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|We cross multiply to get 
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::<math style="vertical-align: 1px">-3x=x-8.</math>
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|Solving, we get <math style="vertical-align: -1px">x=2.</math>
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|Thus, the critical points for <math style="vertical-align: -5px">f(x)</math> are <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1: &nbsp; 
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|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>&thinsp; at the critical points and at the endpoints of the interval. 
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|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>&thinsp; and <math style="vertical-align: -5px">f(8)=0.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2: &nbsp;
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|Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -1px">32</math> 
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|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer: &nbsp; 
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|'''(a)'''&thinsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
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|'''(b)'''&thinsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
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