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| | <span class="exam">on the interval <math style="vertical-align: -5px">[0,2].</math> | | <span class="exam">on the interval <math style="vertical-align: -5px">[0,2].</math> |
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| | + | [[009A Sample Final 2, 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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| − | |'''1.''' 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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| − | |'''2.''' 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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| − | |}
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| | + | [[009A Sample Final 2, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
| − | !Step 1:
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| − | |To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[0,2],</math>
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| − | |we need to find the critical points of <math style="vertical-align: -5px">f(x).</math>
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| − | |Using the Quotient Rule, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\frac{(1+x)(1-x)'-(1-x)(1+x)'}{(1+x)^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-2}{(1+x)^2}.}
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| − | \end{array}</math>
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| − | |We notice that <math style="vertical-align: -6px">f'(x)\ne 0</math> for any <math style="vertical-align: 0px">x.</math>
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| − | |So, there are no critical points in the interval <math style="vertical-align: -5px">[0,2].</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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| − | |Now, we have
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| − | | <math>f(0)=1,~f(2)=\frac{-1}{3}.</math>
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| − | |Therefore, the absolute maximum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -1px">1</math>
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| − | |and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -15px">\frac{-1}{3}.</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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| − | | The absolute maximum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -1px">1</math> and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -15px">\frac{-1}{3}.</math>
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| − | |}
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| − | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
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![{\displaystyle [0,2].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/468478827a88ca588290bff8eb1d2e1256a03848)