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?

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?
