## Precalculus (6th Edition) Blitzer

The required values are $f\left( g\left( x \right) \right)=\frac{3x-40}{7}$ and $g\left( f\left( x \right) \right)=\frac{3x-4}{7}$. And the functions $f\left( x \right)=3x-7$ and $g\left( x \right)=\frac{x+3}{7}$ are not inverses of each other.
Consider the functions: $f\left( x \right)=3x-7$ and $g\left( x \right)=\frac{x+3}{7}$ The equation for $f$ is given as: $f\left( x \right)=3x-7$ Replace $x$ with $g\left( x \right)$ \begin{align} & f\left( g\left( x \right) \right)=3g\left( x \right)-7 \\ & =3\left( \frac{x+3}{7} \right)-7 \\ & =\frac{3x+9-49}{7} \\ & =\frac{3x-40}{7} \end{align} Now, to find $g\left( f\left( x \right) \right)$ Consider the function $g\left( x \right)$: $g\left( x \right)=\frac{x+3}{7}$ Replace $x$ with $f\left( x \right)$ \begin{align} & g\left( f\left( x \right) \right)=\frac{f\left( x \right)+3}{7} \\ & =\frac{\left( 3x-7 \right)+3}{7} \\ & =\frac{3x-4}{7} \end{align} Thus, $f\left( g\left( x \right) \right)\ne x$ and $g\left( f\left( x \right) \right)\ne x$ Therefore, the required values are $f\left( g\left( x \right) \right)=\frac{3x-40}{7}$ and $g\left( f\left( x \right) \right)=\frac{3x-4}{7}$. And the functions $f\left( x \right)=3x-7$ and $g\left( x \right)=\frac{x+3}{7}$ are not inverses of each other.