# Functions: Give an appropriate domain for g(x) which enables g^-1(x) to exist?

## $g \left(x\right) = \frac{1}{x + 7} ^ 2$ ${g}^{-} 1 \left(x\right) = - \frac{1}{\sqrt{x}} - 7$ The questions asks to find a domain for $g \left(x\right)$ such that it also enables ${g}^{-} 1 \left(x\right)$ to exist. My first answer was: $\left\{y : y \ge 0\right\}$ but apparently this is wrong. Any help would be much appreciated!

Jan 10, 2018

The domain of $g \left(x\right)$ is $x \in \mathbb{R} | x \ne - 7$

#### Explanation:

The domain for $g \left(x\right)$ is $x \in \mathbb{R} | x \ne - 7$ because $x = - 7$ will cause division by 0 and this is not allowed.

Furthermore, the range for $g \left(x\right)$ is $y > 0$

The domain of an inverse function is the range of the function and the range of an inverse function is the domain of the function.

Please observe that there is no value of x that will make ${g}^{-} 1 \left(x\right) = - 7$ and $x > 0$ is the domain for ${g}^{-} 1 \left(x\right)$