# Math Help - Applications of Derivatives

1. ## Applications of Derivatives

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

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

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

To get:

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

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

as my zeroes

2. ok ... solve for x in each case to determine critical values.

$4-2x^2 = 2(2-x^2) = 2(\sqrt{2}-x)(\sqrt{2}+x) = 0$

$\sqrt{4-x^2} = \sqrt{(2-x)(2+x)} = 0$

Now what? ... btw, pay attention to the domain of the original function.

3. for the numerator i got $\pm 2$and the denominator $\pm \sqrt{2}$

but how u solved the numerator i thought if you can't factor of x's you would just say $-2{x}^{2 } = -4$

and then solve from there

4. Originally Posted by purplec16
for the numerator i got $\pm 2$and the denominator $\pm \sqrt{2}$

but how u solved the numerator i thought if you can't factor of x's you would just say $-2{x}^{2 } = -4$

... but you can factor $4 - 2x^2$ ,just as I presented it.
you've got it backwards ... roots of the numerator are $x = \pm \sqrt{2}$ , denominator ... $x = \pm 2$

one more time ... domain of f(x) is important in this problem.

5. Remember! Division by zero is not defined.

skeeter got the roots excluding $2$ and by separating the contents using the formula for the difference of two squares,
that is $4-2x^2 = 2(2-x^2) = 2(\sqrt{2} - x)(\sqrt{2} + x)$, just as he had written, because $(a^2 - b^2) =(a-b)(a+b)$.

Now, considering that division by zero isn't defined, what has to be zero for the fraction to equal zero?