# If f is a one-to-one function such that f(2)=9, what is f^-1(9)?

Mar 31, 2016

${f}^{- 1} \left(9\right) = {f}^{- 1} \left(f \left(2\right)\right) = 2$

#### Explanation:

If $f$ is a one-to-one function, then its inverse function, ${f}^{- 1}$, is well-defined.

What does the inverse do ? Exactly what it is called.
Suppose, for example :

$f : \mathbb{R} \setminus \rightarrow \mathbb{R}$
$x \setminus \mapsto f \left(x\right) = y$

Then ${f}^{- 1}$ do the opposite/reverse :

${f}^{-} 1 : \mathbb{R} \setminus \rightarrow \mathbb{R}$
$y \setminus \mapsto {f}^{- 1} \left(y\right) = x$

Thus, if $f \left(x\right) = y$, then ${f}^{- 1} \left(f \left(x\right)\right) = {f}^{- 1} \left(y\right) = x$.

Therefore, if $f \left(2\right) = 9$, you apply ${f}^{- 1}$ to both sides and you get :
${f}^{- 1} \left(f \left(2\right)\right) = {f}^{- 1} \left(9\right) = 2$.