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: [![enter image description here](https://i.stack.imgur.com/odWRV.png.)](https://i.stack.imgur.com/odWRV.png.)

5 Part c. Where Theorem 1 is given to be: 
[![enter image description here](https://i.stack.imgur.com/a7AZs.png)](https://i.stack.imgur.com/a7AZs.png)