# 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  $[0,2].$ Foundations:
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:

Step 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  $[0,2].$ Step 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  $-{\frac {1}{3}}.$ The absolute maximum value for  $f(x)$ is  $1$ and the absolute minimum value for  $f(x)$ is  $-{\frac {1}{3}}.$ 