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| − | <span class="exam">Consider the following continuous function: | + | <span class="exam">If a resistor of <math style="vertical-align: 0px">R</math> ohms is connected across a battery of <math style="vertical-align: 0px">E</math> volts with internal resistance <math style="vertical-align: 0px">r</math> ohms, then the power (in watts) in the external resistor is |
| − | ::::::<math>f(x)=x^{1/3}(x-8)</math>
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| − | <span class="exam">defined on the closed, bounded interval <math style="vertical-align: -3px">[-8,8]</math>. | + | ::<math>P=\frac{E^2R}{(R+r)^2}.</math> |
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| − | <span class="exam">a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>. | + | <span class="exam">If <math style="vertical-align: 0px">E</math> and <math style="vertical-align: 0px">r</math> are fixed but <math style="vertical-align: 0px">R</math> varies, what is the maximum value of the power? |
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| − | <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: -3px">[-8,8]</math>. | + | <hr> |
| | + | [[009A Sample Final 1, Problem 10 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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| − | '''Solution:''' | + | [[009A Sample Final 1, Problem 10 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:
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| − | |To find the critical point, 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}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{x-8}{3x^{\frac{2}{3}}}+x^{\frac{1}{3}}}\\
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| − | \end{array}</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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| − | |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 <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: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</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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| − | |We need to compare the values of <math style="vertical-align: -5px">f(x)</math> 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> and <math style="vertical-align: -5px">f(8)=0</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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| − | |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> 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> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6)</math>.
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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 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)''' 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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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
