# 009A Sample Final 2, Problem 6

Find the absolute maximum and absolute minimum values of the function

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

on the interval  ${\displaystyle [0,2].}$

Foundations:
1. To find the absolute maximum and minimum of  ${\displaystyle f(x)}$  on an interval  ${\displaystyle [a,b],}$

we need to compare the  ${\displaystyle y}$  values of our critical points with  ${\displaystyle f(a)}$  and  ${\displaystyle f(b).}$

2. To find the critical points for  ${\displaystyle f(x),}$  we set  ${\displaystyle f'(x)=0}$  and solve for  ${\displaystyle x.}$

Also, we include the values of  ${\displaystyle x}$  where  ${\displaystyle f'(x)}$  is undefined.

Solution:

Step 1:
To find the absolute maximum and minimum of  ${\displaystyle f(x)}$  on the interval  ${\displaystyle [0,2],}$
we need to find the critical points of  ${\displaystyle f(x).}$
Using the Quotient Rule, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {(1+x)(1-x)'-(1-x)(1+x)'}{(1+x)^{2}}}\\&&\\&=&\displaystyle {\frac {(1+x)(-1)-(1-x)(1)}{(1+x)^{2}}}\\&&\\&=&\displaystyle {{\frac {-2}{(1+x)^{2}}}.}\end{array}}}$

We notice that  ${\displaystyle f'(x)\neq 0}$  for any  ${\displaystyle x.}$
So, there are no critical points in the interval  ${\displaystyle [0,2].}$
Step 2:
Now, we have
${\displaystyle f(0)=1,~f(2)=-{\frac {1}{3}}.}$
Therefore, the absolute maximum value for  ${\displaystyle f(x)}$  is  ${\displaystyle 1}$
and the absolute minimum value for  ${\displaystyle f(x)}$  is  ${\displaystyle -{\frac {1}{3}}.}$

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