Question

Find the range and domain of following function
$f\left(x\right)=\sqrt{9-x^2}$.

Answer 1

We have, $f\left(x\right)=\sqrt{9-x^2}$

We observe that $f\left(x\right)$ is defined for all $x$ satisfying

$9-x^2\ge 0$

$\Rightarrow x^2-9\le 0$

$\Rightarrow (x-3)(x+3)\le 0$

$\Rightarrow -3\le x\le 3$

$\Rightarrow x\in [-3,3]$

Hence domain $\left(f\right)=[-3,3]$

let $y=f\left(x\right)$ Then

$y=\sqrt{9-x^2}$

$\Rightarrow y^2=9-x^2$

$\Rightarrow x^2=9-y^2$

$\Rightarrow x=\sqrt{9-y^2}$

Clearly $x$ will take real values, if

$9-y^2\ge 0$

$\Rightarrow y^2-9\le 0$

$\Rightarrow (y-3)(y+3)\le 0$

$\Rightarrow -3\le y\le 3$

$\Rightarrow y\in [-3,3]$

Also, $y=\sqrt{9-x^2}\ge 0$ for all $x\in [-3,3]$

$\therefore y\in [0,3]$ for all $x\in [-3,3]$

Hence, Range $\left(f\right)=[0,3]$