# How to find domain for f(x) = sqrt(25 - x^2)?

Sep 29, 2015

$\left[- 5 , 5\right]$

#### Explanation:

So that the answer won't be imaginary, $25 - {x}^{2}$ must be greater than or equal to $0$.

$\left\{x | 25 - {x}^{2} \ge 0\right\}$
$\left\{x | 25 - {x}^{2} - 25 \ge 0 - 25\right\}$
$\left\{x | - {x}^{2} \ge - 25\right\}$
$\left\{x | \left(- 1\right) \left(- {x}^{2}\right) \le \left(- 1\right) \left(- 25\right)\right\}$
$\left\{x | {x}^{2} \le 25\right\}$

This will then be separated to two cases:
If $x$ is positive:
$\left\{x | {x}^{2} \le 25\right\}$
$\left\{x | x \le 5\right\}$
$\textcolor{b l u e}{\left(- \infty , 5\right]}$

If $x$ is negative:
$\left\{x | {x}^{2} \le 25\right\}$
$\left\{x | x \ge - 5\right\}$
$\textcolor{red}{\left[- 5 , \infty\right)}$

We will then get the intersection of the two intervals:
$\textcolor{b l u e}{\left(- \infty , 5\right]} \cap \textcolor{red}{\left[- 5 , \infty\right)}$
color(magenta)([-5,5]