## Calculus: Early Transcendentals 8th Edition

For example, let $f(x)=y=x$ Making x the subject, $f^{-1}(y)=x=y$ $\therefore f^{-1}(x)=x$ LHS= $f^{-1}(x)=x$ RHS= $\frac{1}{x}$ Since RHS$\neq$LHS, the statement is false