f(x) = ax^4 + bx^3 + cx^2 + dx + e

f '(x) = 4ax^3 + 3bx^2 + 2cx + d

0 <= x < 1

Ok, thats all you need to solve this problem. What you want to do is solve f '(x) for zero. Find all the x coordinate points where y equals zero in that equation and you'll have the x positions in the function f(x) at which the gradient is zero.

Mr F says: So you're proposing to langers2k that s/he solve a general cubic equation ......
In a quartic like f(x) the derivative will have 3 points where y=0, so 3 turning points in f(x), one will be the maximum, one will be the minimum, and another will be in between both.

Mr F says: Not true. For example, $\displaystyle {\color{red}y = x^4 - 2x^3 + \frac{3}{2} x^2 - \frac{1}{2} x - \frac{15}{16} = \left(x- \frac{1}{2}\right)^4 - 1}$ has only **one** (minimum) turning point.
Once you have the x values calculated from the derivative f '(x) you can sub the 3 values back in to the original function f(x) and find which of those is the minimum and thus which value of x is the minimum.

But!.... Since you have been given a range for x, you might find that 2 of those values of x you found in the derivative function are outside those bounds, so you can exclude those straight away.

For example:

x1 = -12.3 , x2 = 0.4 , x3 = 54.6

In this case, both x1 and x3 lie outside the bounds of zero and one (0<=x<1) so the answer would be x2.

Identify the appropriate x value, then sub it back in to the original equation and give the corresponding y value for the minimum.

If this solved your problem, stop reading here. If you still haven't solved it, read on.

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You might find that none of the 3 calculated x values lie within the boundaries specified, so simply take the lower limit (in this case "0") and sub that in as the x value in the original function f(x). Then sub in the upper limit (in this case "1") in f(x) and find which is smaller, and that will be your answer.

I haven't told you how to solve for x in this :P I can't do everything for you now can I?

Mr F wryly observes: Very glib. The devil is in the detail. Which would be why there is none .......
If you need help with that, let me know and I'll step you through solving for x in a cubic derivative.

Mr F remarks: I think you're biting off more than you can chew here, sport.
Hope it helps, and good luck.