Maximizing by setting Derivative equal to 0. Stuck (struggling)

Question

I'm having immense difficulty doing this question.

Want to maximize (respect to x): $f(x)$=$(x-x_0)(x-x_1)(x-x_2)(x-x_3)$ where

1. $x_1-x_0=h$

2. $x_2-x_1=h$

3. $x_3-x_2=h$.

Want to get the value of x in terms of $x_1$ and $h$.

First I take the derivative and get:

$(x-x_0)(x-x_1)(x-x_2)$ +$(x-x_0)(x-x_1)(x-x_3)+(x-x_0)(x-x_2)(x-x_3)+ (x-x_1)(x-x_2)(x-x_3)=0$

Now using the facts 1,2, and 3, I substitute for $x_0, x_2$, and $x_3 $:(Need to write everything in terms of x1 and h)

$x_0$=$x_1-h$,

$x_1=x_1$,

$x_2=x_1+h$,

$x3=h+x2=h+(x_1)+h=2h+x_1$

Then we have for the derivative:

$(x-x_1+h)(x-x_1)(x-(x_1+h))$ + $(x-x_1+h)(x-x1)(x-(2h+x_1))$+

$(x-x_1+h)(x-x_1-h)(x-(2h+x_1))$+$(x-x_1)(x-x_1-h)(x-(2h+x_1))$=0.

Then we have:

$[(x-x_1)^2+h(x-x_1)]((x-(x_1+h))$+$[(x-x_1)^2+h(x-x_1)]((x-(2h+x_1))$+

$[(x-x_1)^2-h^2](x-(2h+x_1))$+$[(x-x_1)^2-h(x-x_1)]((x-(2h+x_1))$=0

This is all I have but I can't solve for x still. Wolfram Alpha won't give me answer. Any help would be much appreciated thank you. I believe the max value of f(x) should be (like max y value): $f(x)=9/16$. or $(f(x)=72/128)$ Thank you. I have been working on this problem for over three hours so any help would really be appreciated.

Here is the actual question I'm working on for reference:

5 Part c. Where Theorem 1 is given to be:

Answer 1

let $a = \frac {x_1+x_2}{2}$ then

$x_1 = a - \frac {h}{2}\\ x_0 = a - 3\frac {h}{2}\\ x_2 = a + \frac {h}{2}\\ x_3 = a + 3\frac {h}{2}$

This will take advantage of some of the symmetry of our fuction

$f(x) = (x-a - \frac {3h}{2})(x-a + \frac {3h}{2})(x-a - \frac {h}{2})(x-a + \frac {h}{2})\\ ((x-a)^2 - \frac {9h^2}{4})((x-a)^2 - \frac {1h^2}{4})\\ (x-a)^4 - \frac {10h^2}{4}(x-a)^2 + \frac {9h^4}{16}$

$f'(x) = 4(x-a)^3 - \frac {10h^2}{2} (x-a) = 0\\ (2(x-a) - \frac {h\sqrt {10}}{2})(x-a)(2(x-a) + \frac {h\sqrt {10}}{2})=0$

We have minima at $x = \frac{h\sqrt{10}}{4} + x_1+\frac {h}{2}, -\frac{h\sqrt{10}}{4} + x_1 +\frac {h}{2}$

And a maximum at $x = a = x_1 + \frac {h}{2}$

$f(a) = \frac {9h^4}{16}$

Answer 2

Improving your method, you should group efficiently: $$\begin{align}f'(x)=&\color{red}{(x-x_0)(x-x_1)(x-x_2)}+\color{blue}{(x-x_0)(x-x_1)(x-x_3)}+\\ +&\color{blue}{(x-x_0)(x-x_2)(x-x_3)}+ \color{red}{(x-x_1)(x-x_2)(x-x_3)}=0 \Rightarrow \\ &\color{red}{(x-x_1)(x-x_2)(2x-(x_0+x_3))}+\color{blue}{(x-x_0)(x-x_3)(2x-(x_1+x_2))}=0 \Rightarrow \\ &\color{red}{(x-x_1)(x-x_2)(2x-(2x_1+h))}+\color{blue}{(x-x_0)(x-x_3)(2x-(2x_1+h))}=0 \Rightarrow \\ &(2x-2x_1-h)[(x-x_1)(x-x_2)+(x-x_0)(x-x_3)]=0 \Rightarrow \\ &2x-2x_1-h=0 \Rightarrow x=x_1+\frac h2 \Rightarrow \\ f(x_1+\frac h2)=&(x_1+\frac h2-x_1+h)(x_1+\frac h2-x_1)(x_1+\frac h2-x_1-h)(x_1+\frac h2-x_1-2h)=\\ =&\frac{3h}{2}\cdot \frac h2\cdot \left(-\frac h2\right)\cdot \left(-\frac{3h}{2}\right)=\frac{9h^4}{16} \ \text{(max)}, \ \text{because} \\ f''(x_1+\frac h2)=&-5h^2<0.\end{align}$$