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.
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.
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.
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.
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? If you need help with that, let me know and I'll step you through solving for x in a cubic derivative.
Hope it helps, and good luck.