Answer
Evaluate $g(x) = \frac{1-x}{1+x}$:
$g(2) = \frac{-1}{3}$
$g(-1) =\frac{2}{0} = undefined $
$g(\frac{1}{2}) =\frac{1}{3}$
$g(a) = \frac{1-a}{1+a}$
$g(a-1) = \frac{-a+2}{a}$
$g(x^2-1) = \frac{-x^2+2}{x^2}$
Work Step by Step
For x = 2
$g(2) = \frac{1-2}{1+2}$ ... subtract numerator and add denominator
$= \frac{-1}{3}$
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For x = -1
$g(-1) = \frac{1-(-1)}{1+(-1)}$ ... subtract numerator and add denominator
$= \frac{2}{0} = undefined$
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For x = $\frac{1}{2}$
$g(\frac{1}{2}) = \frac{1-\frac{1}{2}}{1+\frac{1}{2}}$ ... convert whole numbers to fractions with common denominator
$= \frac{\frac{2}{2}-\frac{1}{2}}{\frac{2}{2}+\frac{1}{2}}$...subtract numerator and add denominator
$= \frac{\frac{1}{2}}{\frac{3}{2}}$...multiply top and bottom by 2
$=\frac{1}{3}$
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For x = a
$g(a) = \frac{1-a}{1+a}$
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For x = a - 1
$g(a-1) = \frac{1-(a-1)}{1+(a-1)}$... simplify top and bottom
$=\frac{1-a+1}{1+a-1}=\frac{-a+2}{a}$
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For x = $ x^2 - 1$
$g( x^2 - 1) = \frac{1-( x^2 - 1)}{1+( x^2 - 1)}$... simplify top and bottom
$=\frac{1- x^2 + 1}{1+ x^2 - 1}=\frac{-x^2+2}{x^2}$
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