## Algebra and Trigonometry 10th Edition

$f^{-1} (x)=5x+4$
We have $f^{-1} (x) =5x+4$ and $f(f^{-1} (x))=f(5x+4)= \dfrac{(5x+4)-4}{5}=\dfrac{5x}{5}=x$ Now, $f(f^{-1} (x))=f(\dfrac{x-4}{5})= 5(\dfrac{x-4}{5})+4=x-4+4=x$ Thus, we have $f^{-1} (x)=5x+4$