# Question #7d077

Jan 11, 2018

See below.

#### Explanation:

A function is one to one if:

No two elements in the domain map on to the same element in the range.

Since we have a restricted domain, this criteria is met, so $f \left(x\right)$ is one to one

For an unrestricted domain $f \left(x\right) = {x}^{2}$ is a many to one function.

Inverse of $f \left(x\right)$

We need to express $x$ as a function of $y$:

$y = {x}^{2}$

$x = \pm \sqrt{y}$

Substituting $x = y$

$y = \pm \sqrt{x}$ $\therefore$ ${f}^{-} 1 \left(x\right) = \pm \sqrt{x}$

Domain of $f \left(x\right)$

For $x \le 0$

Domain is $\left\{x \in \mathbb{R} : - \infty < x \le 0\right\}$

Since ${x}^{2} \ge 0$ for all $\mathbb{R}$

Range is:

$\left\{y \in \mathbb{R} : 0 \le y < \infty\right\}$

Domain of ${f}^{-} 1 \left(x\right)$

Because the domain of $f \left(x\right)$ is $x \le 0$, only the inverse $y = - \sqrt{x}$ is needed. The domain of this will be the same as the range of $f \left(x\right)$ i.e.

$\left\{x \in \mathbb{R} : 0 \le x < \infty\right\}$

The range will be the same as the domain of $f \left(x\right)$ i.e.

$\left\{y \in \mathbb{R} : - \infty < y \le 0\right\}$

For one to one functions the range of $f \left(x\right)$ is the domain of ${f}^{-} 1 \left(x\right)$ and the domain of $f \left(x\right)$ is the range of ${f}^{-} 1 \left(x\right)$.