$f(x)=\sqrt{36-4x^{2}}$ here $-3\leq x\leq3$
$f(x)=(36-4x^{2})^{\frac{1}{2}}$------------>$(1)$
differentiate both sides with respect to $'x'$
$f'(x)=\frac{1}{2}(36-4x^{2})^{(\frac{1}{2}-1)}.(0-4.2x)$
$f'(x)=\frac{1}{2}(36-4x^{2})^{\frac{-1}{2}}.(-8x)$
$f'(x)=\frac{1}{2}[\frac{1}{(36-4x^{2})^{\frac{1}{2}}}].(-8x)$
$f'(x)=\frac{-4x}{(36-4x^{2})^{\frac{1}{2}}}$
$f'(x)=\frac{-4x}{2.(9-x^{2})^{\frac{1}{2}}}$
$f'(x)=\frac{-2x}{(9-x^{2})^{\frac{1}{2}}}$------------->$(2)$
for statinary points $f'(x)=0$
$\Rightarrow \frac{-2x}{(9-x^{2})^{\frac{1}{2}}}=0$
$\Rightarrow x=0$
$(or)$
$(9 - x^{2})^{-1/2} = 0.$
we can write it like
$0^{-2} = ( (9- x^{2})^{-1/2})^{-2}$
$9 - x^{2} = 0$
$x = -3,3$
So, we got three stationary points or critical points $x=0,-3,3$
now,$f(-3)=\sqrt{(36-4(-3)^{2})}=\sqrt{(36-36)}=0$
$f(0)=\sqrt{(36-4(0)^{2})}=\sqrt{36}=6$
$f(3)=\sqrt{(36-4(3)^{2})}=\sqrt{(36-36)}=0$
Absolute minimum $f(x)=0$ at $x=-3,3$
Absolute maximum $f(x)=6$ at $x=0$
