# Let f(x) = x^2 - 16 how do you find f^-1(x)?

Mar 21, 2018

This is a way to express finding the inverse function of $f \left(x\right) = {x}^{2} - 16$

#### Explanation:

First, write the function as $y = {x}^{2} - 16$.

Next, switch the $y$ and $x$ positions.

$x = {y}^{2} - 16 \rightarrow$ Solve for $y$ in terms of $x$

$x + 16 = {y}^{2}$

$y = \sqrt{x + 16}$

The inverse function should be ${f}^{-} 1 \left(x\right) = \sqrt{x + 16}$

Mar 21, 2018

#### Explanation:

Suppose that, $f : \mathbb{R} \to \mathbb{R} : f \left(x\right) = {x}^{2} - 16$.

Observe that, $f \left(1\right) = 1 - 16 = - 15 , \mathmr{and} , f \left(- 1\right) = - 15$.

$\therefore f \left(1\right) = f \left(- 1\right)$.

$\therefore f \text{ is not injective, or, } 1 - 1$.

$\therefore {f}^{-} 1$ does not exist.

However, if $f$ is defined on a suitable domain, e.g.,

${\mathbb{R}}^{+}$, then ${f}^{-} 1$ exists as Respected Serena D. has shown.