Maxima and minima of $f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$ 
Determine the local minima and maxima of the function $f:[0, \infty) \to \mathbb{R}$ $$f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$$
Does $f$ have a maximum or minimum?

Computing the derivative gets me here:
$f'(x) = \frac14(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}),$
but this is not an easy function to find the roots and then determine the maxima and minima of $f$. Is there an alternative approach here, or how should I proceed?
 A: Hint:
By factoring, using a common denominator and combining terms, you get
$$\begin{equation}\begin{aligned}
f'(x) & = \frac{1}{4}\left(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(2\sqrt{x}+\frac{x-5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{4x}{2\sqrt{x}}+\frac{x-5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{4x + x - 5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{5x - 5}{2\sqrt{x}}\right) \\
& = \frac{5(x-5)(x-1)}{8\sqrt{x}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Can you finish the rest yourself?
A: $$\begin{align}f'(x)& = \frac14(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}})\\&=\frac1{8\sqrt x}[4(x-5)x+(x-5)^2]\\&=\frac 5{8\sqrt x}(x-5)(x-1)\end{align}$$
A: Note that $f(x)\ge0$, so the extrema of $f(x)$ occur at the same $x$ values 
as the extrema of $g(x)=f(x)^2=\dfrac{x(x-5)^4}{16}.$
It is easy to compute that
$g'(x)=\dfrac{(x-5)^4+4x(x-5)^3}{16}=\dfrac{(x-5)^3(x-5+4x)}{16}=\dfrac{5(x-5)^3(x-1)}{16}.$
Can you take it from here?
A: It's easy to see that no maximum value can exist since the function grows without bound. Also, since the function is never negative, it follows that the possible minimum is $0,$ which is attained when $x=0.$
A way to handle it the way you want is to observe that it is a nonnegative quintic in $\sqrt x,$ on the same domain $[0,+\infty).$
Then the observations above become even clearer.
