## Precalculus (6th Edition)

$f$ and $g$ are not inverses of each other.
RECALL: (1) $(f \circ g)(x) = f\left(g(x)\right)$ (2) The function $g(x)$ is the inverse of function of a one-to-one function $(x)$ if for every element of the domain, $(f \circ g)(x) =x$ and $(g \circ f)(x)=x$ Find $(f \circ g)(x)$ by substituting $-\frac{1}{3}x-12$ to $x$ in $f(x)$: $(f\circ g)(x) \\=f\left(g(x)\right) \\=-3\left(-\frac{1}{3}x-12\right)+12 \\=-3(-\frac{1}{3}x)-12(-3)+12 \\=x+36+12 \\=x+48$ Since $(f\circ g)(x)\ne x$, then $f$ and $g$ are not inverses of each other.