Question

Find the range of the function $\sqrt{16-x^2}$.

Answer 1

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

and it is maximum at x = 0 wher $f\left(x\right)$ = 16

So, the range of $f\left(x\right)$ is [0,4]