009A Sample Final 2, Problem 6

Find the absolute maximum and absolute minimum values of the function

$f(x)={\frac {1-x}{1+x}}$

on the interval  $[0,2].$

ExpandFoundations:
1. To find the absolute maximum and minimum of  $f(x)$  on an interval  $[a,b],$
we need to compare the  $y$  values of our critical points with  $f(a)$  and  $f(b).$
2. To find the critical points for  $f(x),$  we set  $f'(x)=0$  and solve for  $x.$
Also, we include the values of  $x$  where  $f'(x)$  is undefined.

Solution:

ExpandStep 1:
To find the absolute maximum and minimum of  $f(x)$  on the interval  $[0,2],$
we need to find the critical points of  $f(x).$
Using the Quotient Rule, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {(1+x)(1-x)'-(1-x)(1+x)'}{(1+x)^{2}}}\\&&\\&=&\displaystyle {\frac {(1+x)(-1)-(1-x)(1)}{(1+x)^{2}}}\\&&\\&=&\displaystyle {{\frac {-2}{(1+x)^{2}}}.}\end{array}}$
We notice that  $f'(x)\neq 0$  for any  $x.$
So, there are no critical points in the interval  $[0,2].$

ExpandStep 2:
Now, we have
$f(0)=1,~f(2)=-{\frac {1}{3}}.$
Therefore, the absolute maximum value for  $f(x)$  is  $1$
and the absolute minimum value for  $f(x)$  is  $-{\frac {1}{3}}.$

ExpandFinal Answer:
The absolute maximum value for  $f(x)$  is  $1$  and the absolute minimum value for  $f(x)$  is  $-{\frac {1}{3}}.$

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