Question

Find the domain and range of
$f\left(x\right)=\sqrt{16-x^2}$

Answer 1

Consider the given function.

$f\left(x\right)=\sqrt{16-x^2}$

For domain under root should not be negative quantity,

$16-x^2\ge 0$

$16\ge x^2$

Therefore,

$x\le 4$ or $x\ge -4$

Then,

The domain$[-4,4]$

$Range$: $f\left(x\right)$ is maximum at $x=0,f\left(x\right)=4$

And $f\left(x\right)$ is minimum at $x=4,f\left(x\right)=0$

$Range$

$[0,4]$

Hence, this is the answer.