## Functions Modeling Change: A Preparation for Calculus, 5th Edition

Use $f(x)=x+1$ as a counterexample. $x=f(x)-1$, so $f^{-1}(x)=x-1$. Solving for $(f(x))^{-1}$, $$(f(x))^{-1}=\frac{1}{x+1}$$ It can been seen that these two functions aren't the same, so this statement must be false.