009A Sample Final 2, Problem 6 - Revision history

MathAdmin: Created page with " Find the absolute maximum and absolute minimum values of the function :: $f(x)=\frac{1-x}{1+x}$ on the interval
2017-04-10T16:24:18Z

Created page with "<span class="exam"> Find the absolute maximum and absolute minimum values of the function ::<math>f(x)=\frac{1-x}{1+x}</math> <span class="exam">on the interval <math..."

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<span class="exam"> Find the absolute maximum and absolute minimum values of the function

::<math>f(x)=\frac{1-x}{1+x}</math>

<span class="exam">on the interval  <math style="vertical-align: -5px">[0,2].</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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}
\displaystyle{f'(x)} & = & \displaystyle{\frac{(1+x)(1-x)'-(1-x)(1+x)'}{(1+x)^2}}\\
&&\\
& = & \displaystyle{\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}}\\
&&\\
& = & \displaystyle{\frac{-2}{(1+x)^2}.}
\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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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>
|}
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]