Ok I have to identify the extram i.e.the maximum and minimum points of this function and...i'm kinda confused$\displaystyle f(x)=x(\sqrt{4-{x}^{2 } } ) $

$\displaystyle f'(x)=\frac{4-2{x}^{2 } }{\sqrt{4-{x}^{2 } } } $

Thats what i got for my derivative so i just factor out numerator and denominator right?

To get:

$\displaystyle 4-2{x}^{ 2}= 0 $

$\displaystyle \sqrt{4-{x}^{2 } } = 0$

as my zeroes