Question

The range of the function $f\left(x\right)=\sqrt{9-x^2}$ is

• A. $(0,3]$
• B. $[0,3)$
• C. $(0,3)$
• D. $[0,3]$

Answer 1

Answer: (D)

$[0,3]$

$f\left(x\right)=\sqrt{9-x^2}$
Let $y=f\left(x\right)$
$y=\sqrt{9-x^2}$
Squaring both sides we get $y^2=9-x^2$
$x^2=9-y^2$
$x=\sqrt{9-y^2}$

it is only defined if
$9-y^2\ge 0$
$y^2-9\le 0$

$(y+3)(y-3)\le 0$
$-3\le y\le 3$ $\dots \left(1\right)$
But $y=\sqrt{9-x^2}$
$\therefore y\ge 0$ $\dots \left(2\right)$

from $\left(1\right)$ and $\left(2\right)$
$\therefore y\in [0,3]$
Range of function f(x) is $[0,3]$