Answer
$\displaystyle g(x)=\frac{x-1}{x+1}$
Work Step by Step
We are given:
$f(x)=\displaystyle \frac{1}{x}$
and:
$(\displaystyle \frac{f}{g})(x)=\frac{x+1}{x^{2}-x}$
Thus:
$g(x)=\frac{f(x)}{(\displaystyle \frac{f}{g})(x)}$
$\displaystyle g(x)=\frac{\frac{1}{x}}{\frac{x+1}{x^{2}-x}}$
$\displaystyle g(x)=\frac{1*(x^2-x)}{x(x+1)}$
$\displaystyle g(x)=\frac{(x-1)x}{x(x+1)}$
$\displaystyle g(x)=\frac{x-1}{x+1}$
