Answer
$g(2)=-\frac{1}{3}$
$g(-1)=$ undefined
$g( \frac{1}{2})=\frac{1}{3}$
$g(a)=\frac{1-a}{1+a}$
$g(a-1)=\frac{2-a}{a}$
$g(x^{2}-1)=\frac{2-x^{2}}{x^{2}}$
Work Step by Step
We are given:
$g(x)= \frac{1-x}{1+x}$
We evaluate:
$g(2)=\frac{1-(2)}{1+(2)}=\frac{-1}{3}=-\frac{1}{3}$
$g(-1)=\frac{1-(-1)}{1+(-1)}=\frac{2}{0}=$ undefined
$g( \frac{1}{2})=\frac{1-(\frac{1}{2})}{1+(\frac{1}{2})}=\frac{\frac{2}{2}-\frac{1}{2}}{\frac{2}{2}+\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{3}$
$g(a)=\frac{1-(a)}{1+(a)}=\frac{1-a}{1+a}$
$g(a-1)=\frac{1-(a-1)}{1+(a-1)}=\frac{1-a+1}{1+a-1}=\frac{2-a}{a}$
$g(x^{2}-1)=\frac{1-(x^{2}-1)}{1+(x^{2}-1)}=\frac{2-x^{2}}{x^{2}}$
