## Calculus (3rd Edition)

$f^{-1}(x)= \frac{x+4}{7}$
y= 7x-4 or x= $\frac{y+4}{7}$. Thus the inverse is $f^{-1}(y)= \frac{y+4}{7}$. Now, $f^{-1}(f(x))=f^{-1}(7x-4)= \frac{(7x-4)+4}{7}=x$ and $f(f^{-1}(x))=f(\frac{x+4}{7})=7(\frac{x+4}{7})-4=x$ which implies that f is invertible and the inverse is defined as $f^{-1}(x)= \frac{x+4}{7}$.