# How do you find the domain and range of sqrt(25-x^2) ?

Jan 26, 2017

The domain is $= \left[- 5 , 5\right]$
The range is $= \left[0 , 5\right]$

#### Explanation:

What is under the sqrt sign is $\ge 0$

Therefore,

$25 - {x}^{2} \ge 0$

$\left(5 + x\right) \left(5 - x\right) \ge 0$

Let $f \left(x\right) = \left(5 + x\right) \left(5 - x\right)$

We can now build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 5$$\textcolor{w h i t e}{a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$5 + x$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$5 - x$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left[- 5 , 5\right]$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \left[- 5 , 5\right]$

When $x = \pm 5$, $\sqrt{f \left(x\right)} = 0$

When $x = 0$, $\sqrt{f \left(x\right)} = 5$, this is the max. value

so,
The range is $R = \left[0 , 5\right]$