absolute extrema problem.... My calc skills are kind of rusty.  Was wondering if I could get an assist on this one perhaps?  I am looking for the absolute extrema if they exist and all the x values they occur at in the domain.  Can someone explain this to me with a graph and the steps so I can see the process?
$$
f(x)=x\sqrt{4-x^2}, ~~~[-2,2]
$$
I have worked out how to find the derivative I think. I just wanted some one to double check my math, and steps if possible?
$$\begin{align}
f(x)&=x\sqrt{4-x^2}\\
&=\frac d{dx}(x)\left(\sqrt{4-x^2}\right)+x\cdot\frac d{dx}\left(4-x^2\right)\cdot x\\
&=1\left(\sqrt{4-x^2}\right)+\frac1{2\sqrt{4-x^2}}\left(4-x^2\right)(x)\\
&=\sqrt{4-x^2}+\frac{-\frac d{dx}\left(x^2\right)\cdot x}{2\sqrt{4-x^2}}\\
&=\sqrt{4-x^2}-\frac{2x\cdot x}{2\cdot\sqrt{4-x^2}}\\
(x)&=\sqrt{4-x^2}-\frac {x^2}{\sqrt{4-x^2}}
\end{align}$$
OK...
So I have taken the time to do the other steps and wrote them out.  I wanted to see if you guys think I am on the right track with how I came to the solution.  Thanks for your help thus far!  This has helped!  Sorry about the chicken scratch again!
I first solved the equation for zero with the following algebra:

Then I considered the original function and plugged those values back into the function and came up with the following:

I think that this is correct because my graphing calculator confirms this, but I just need confirmation of my steps, thank you!
 A: Ideas: 


*

*A continuous function on a closed bounded interval $[a,b]$ achieves a maximum/minimum.

*There are two kinds of points in $[a,b]$, interior points $(a,b)$ and boundary points $\{a,b\}$.

*For $f$ differentiable, if an extremum occurs at an interior point, it must be that $f'(x)=0$ at that point.
Thus for differentiable functions on $[a,b]$, to find the minimum and maximum we know that it occurs either somewhere that $f'(x)=0$ or maybe at $a$ or $b$ (regardless of whether $f'(a)=0$ or $f'(b)=0$).
Thus we should calculate $f'(x)$ and solve for where it is zero to get possible points $x_1,\ldots,x_n$ where the max/min may occur. Then we should compare the values $f(x_1),\ldots,f(x_n),f(a),f(b)$. Then the biggest of these numbers is the maximum of $f$ and the smallest is the minimum of $f$.
I'll help you get started, write $f(x)=x(4-x^2)^{1/2}$, then by the product rule (and chain rule) 
$$
f'(x) = \sqrt{4-x^2} + x\cdot \frac12(4-x^2)^{-1/2}\cdot (-2x) = \cdots
$$
Can you finish?
