Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\frac{-x}{2}}{\sqrt{5-x}}$$
$f'(x) = 0$, when $x = 0, $ and is undefined when x= 5
Plugging in the roots of the derivative function and the end points of the interval into the original function:
\begin{align*}
f(0) & = 0\\
f(5) & = 0\\
f(-6) & = -6\sqrt{11}\\
f(4) & = 4 \cdot 1 = 4
\end{align*}
So why is the 4 not the absolute maximum value?
p.s. I assumed the first term goes to zero when taking a derivative by the product rule. I confused d/dx x = 1, with any number d/dd = 0

Your derivative is wrong. Use the product rule to separate the $x$ from the $\sqrt{5-x}$.
$$\frac{d}{dx}\left(x\sqrt{5-x} \right)
  =x\frac{d}{dx}\left(\sqrt{5-x} \right)+\sqrt{5-x}\frac{d}{dx}\left(x \right)$$
$$=x\frac{1}{2\sqrt{5-x}}(-1)+\sqrt{5-x}\cdot 1$$
$$=\frac{-x}{2\sqrt{5-x}}+\frac{2(5-x)}{2\sqrt{5-x}}$$
$$=\frac{10-3x}{2\sqrt{5-x}}$$
Setting this equal to zero, you get
$$x=\frac{10}3$$
This is undefined only at $x\ge 5$, which is not in your interval of interest.
You should be able to continue from here.
