## Precalculus (6th Edition) Blitzer

The $f\left( g\left( x \right) \right)$ and $g\left( f\left( x \right) \right)$ of the function $f\left( x \right)=2-5x$ and $g\left( x \right)=\frac{2-x}{5}$ is $f\left( g\left( x \right) \right)=x\text{ and }g\left( f\left( x \right) \right)=x$ respectively, and f and g are inverses of each other.
Now, consider the function, \begin{align} & f\left( g\left( x \right) \right)=f\left( \frac{2-x}{5} \right) \\ & =2-5\left( \frac{2-x}{5} \right) \\ & =2-2+x \\ & =x \end{align} Next consider the function, \begin{align} & g\left( f\left( x \right) \right)=g\left( 2-5x \right) \\ & =\frac{2-\left( 2-5x \right)}{5} \\ & =\frac{5x}{x} \\ & =x \end{align} This shows that $f\text{ and }g$ are inverses of each other. Therefore, the $f\left( g\left( x \right) \right)$ and $g\left( f\left( x \right) \right)$ of the function $f\left( x \right)=2-5x$ and $g\left( x \right)=\frac{2-x}{5}$ is $f\left( g\left( x \right) \right)=x\text{ and }g\left( f\left( x \right) \right)=x$ respectively, and f and g are inverses of each other.