I want to ask a question looks like a little bit trivial:
Given that $x_i = x_0 + ih$, $for$ $i = 1, 2, 3$, and $h>0$, at which point the function $f(x)=\prod_{i=0}^3(x-x_i)$ where $x\in[x_0,x_3] $will attain its maximum?
I know compute its derivatives and check the global minimum/maximum points is a way. I computed the maximum of $g(x)=\prod_{i=0}^2(x-x_i)$ and the process is a little bit complicated. I wonder know is there any other way to compute? I noticed that the space of any adjacent points is the same.
From Rolle's theorem we can see that $f'$ has a zero in each interval $I_1 = (x_0, x_1)$, $I_2 = (x_1, x_2)$, and $I_3 = (x_2, x_3)$, and there can be no more zeros of $f'$ since $\operatorname{deg} f' = 3$.
$f$ is negative in $I_1$ and $I_3$ and positive in $I_2$, therefore the maximum must occur at the zero of $f'$ in $I_2$.
Finally, $f$ is symmetric with respect to $x_s = \frac{x_1+x_2}{2}$: $$ f(x_s -h) = f(x_s + h) \implies f'(x_s) = 0 $$ so that the maximum of $f$ on $[x_0, x_3]$ is attained at $\frac{x_1+x_2}{2}$.
More generally, you can use the symmetry of $f$ to express it as $$ f(x) = (x-x_s - \frac{3h}{2})(x-x_s - \frac{h}{2})(x-x_s + \frac{h}{2})(x-x_s + \frac{3h}{2}) \\ = \left( (x-x_s)^2 - \frac{9h^2}{4}\right)\left( (x-x_s)^2 - \frac{h^2}{4}\right) \\ = g\left( (x-x_s)^2 \right) $$ where $g$ is defined as $$ g(t) = \left( t - \frac{h^2}{4}\right)\left( t - \frac{9h^2}{4}\right) \, . $$ $g$ is a parabola and its derivative is zero at the vertex $$ t = \frac{1}{2}\left(\frac{h^2}{4} + \frac{9h^2}{4} \right) = \frac{5h^2}{4} $$ It follows that $$ f'(x) = 2(x-x_s) g'\left( (x-x_s)^2 \right) $$ is zero if $x=x_s$ or if $(x-x_s)^2 = \frac{5h^2}{4}$, i.e. at the points $$ x_s \, , \, x_s - \frac{\sqrt 5 h}{2} \, , \, x_s + \frac{\sqrt 5 h}{2} $$ so that are the points where $f$ attains the maximum and the two minima in $[x_0, x_3]$.