# How do you find the domain and range of f(x)= -x^2 - 1?

Jul 20, 2016

Domain is $\mathbb{R}$

Range of is $\left(- \infty , - 1\right]$.

#### Explanation:

$f \left(x\right) = - \left({x}^{2} + 1\right) , \mathmr{and} , \left({x}^{2} + 1\right) = - f \left(x\right)$

Let us note that there is no restriction on $x$ as to what and what not values it should take from $\mathbb{R}$, clearly, the Domain ${D}_{f}$ is $\mathbb{R}$

Next, we have, $\forall x \in \mathbb{R} , {x}^{2} \ge 0 , s o , \left({x}^{2} + 1\right) \ge 1$, i.e., -f(x)>=1,#

& hence, $- 1 \ge f \left(x\right) \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Recall that the Range of $f$, i.e., ${R}_{f} = \left\{f \left(x\right) : x \in {D}_{f}\right\}$

$\left(1\right)$ means that ${R}_{f} \subset \left(- \infty , - 1\right] \ldots \ldots \ldots . . \left(2\right)$

Our claim is ${R}_{f} = \left(- \infty , - 1\right]$

To prove this claim; because of $\left(2\right)$, we need show that

$\left(- \infty , - 1\right] \subset {R}_{f} \ldots \ldots \ldots . . \left(3\right)$

For this, let $y \in \left(- \infty , - 1\right]$, y arbitrary. Then, to show $\left(3\right)$, we need show that $y \in {R}_{f}$, i.e., we need show that $\exists$ at least one $x \in {D}_{f}$, such that, $y = f \left(x\right)$

To this end, we define $x = \sqrt{- 1 - y}$, & for this $x$, we will show that $y = f \left(x\right)$

$y \in \left(- \infty , - 1\right] \Rightarrow y \le - 1 \Rightarrow y + 1 \le 0$

$\Rightarrow - \left(y + 1\right) = - 1 - y \ge 0 \Rightarrow \sqrt{- 1 - y} = x$ is defined. And, for this $x$, we have, $f \left(x\right) = - {x}^{2} - 1 = - {\left(\sqrt{- 1 - y}\right)}^{2} - 1 = - \left(- 1 - y\right) - 1 = 1 + y - 1 = y$

This proves $\left(3\right)$ and hence ${R}_{f} = \left(- \infty , - 1\right]$

A bit lengthy, but Enjoyable! Enjoy Maths.!