## Precalculus (6th Edition) Blitzer

The required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=3x+8$ and $g\left( x \right)=\frac{x-8}{3}$ are inverses of each other.
Consider the functions: $f\left( x \right)=3x+8$ and $g\left( x \right)=\frac{x-8}{3}$ The equation for $f$ is given as: $f\left( x \right)=3x+8$ Replace $x$ with $g\left( x \right)$ \begin{align} & f\left( g\left( x \right) \right)=3g\left( x \right)+8 \\ & =3\left( \frac{x-8}{3} \right)+8 \\ & =x-8+8 \\ & =x \end{align} Now, to find $g\left( f\left( x \right) \right)$ Consider the function $g\left( x \right)$: $g\left( x \right)=\frac{x-8}{3}$ Replace $x$ with $f\left( x \right)$ \begin{align} & g\left( f\left( x \right) \right)=\frac{f\left( x \right)-8}{3} \\ & =\frac{\left( 3x+8 \right)-8}{3} \\ & =\frac{3x}{3} \\ & =x \end{align} Because $g$ is inverse of $f$ (and vice-versa), the inverse notation can be used: $f\left( x \right)=3x+8$ and $\text{ }{{f}^{-1}}\left( x \right)=\frac{x-8}{3}$ It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ by subtracting $8$ and dividing by $3$ . \begin{align} & \text{ }{{f}^{-1}}\left( x \right)=\frac{x-8}{3} \\ & =g\left( x \right) \end{align} Hence, the required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=3x+8$ and $g\left( x \right)=\frac{x-8}{3}$ are inverses of each other.