## Calculus: Early Transcendentals (2nd Edition)

The solution is $$f(x)=3x-2.$$
The right side is a polynomial of the $1st$ degree so we will ask that $f(x)$ be a polynomial of the first degree i.e. it will be of the form of $f(x)=ax+b$. Then we have $$f(f(x))=f(ax+b)=a(ax+b)+b=a^2x+ab+b=9x-8.$$ This means that $a^2=9\rightarrow a=3$ and $ab+b=-8\Rightarrow3b+b=-8\Rightarrow4b=-8\Rightarrow b=-2.$ So finally $f(x)=3x-2.$