## College Algebra (10th Edition)

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