Question

Find the range of $f\left(x\right)=\sqrt{9-x^2}$, range is [a,b], then b-a= ?

Answer 1

For real values of $x$
$9-x^2\ge 0$
$x^2\le 9$
$|x|\le 3$
$-3\le x\le 3$
Hence,
$xϵ[-3,3]$
Thus range of f(x) is $y_{max}=3$ while $y_{min}=0$
Hence range is $[0,3]$
Therefore, $b=3$ and $a=0$
Hence, $b-a=3-0$ $=3$.