Find the absolute minimum value of f(x)=x^2 - 3x^4 on the interval [-1,1]?

There is always a "minimum value" if the curve is defined and the graph is not linear and parallel to the x axis.

first of all differentiate to get 2x - 12x^3. Set equal to 0, and get 2x(1 - 6x^2) = 0.

so x = 0, and x = sqrt(1/6)

NOW, there are several possibilities:

1. at the point -1 (endpoint), there is a maximum

2. at the point -1 (endpoint), there is a minimum

3. at the point +1 (endpoint) there is a maximum

4. at the point +1 (endpoint) there is a maximum

5. at the point 0, there is a maximum

6. at the point 0, there is a minimum

5. at the point sqrt(1/6), there is a maximum

6. at the point sqrt(1/6), there is a minimum

SO, we evaluate the function at each of the points

at -1, the value of the function is 1 -3 = -2

at +1, the value of the function is 1 - 3 = -2

at 0, the value of the function is 0

at 1/6, the value of the function is 1/36 - 3/1296 (greater than -2, less than 0)

There are two minimums--at +1 and -1 (endpoints)

I suspect the purpose of this problem was to demonstrate that while you can differentiate and find a (local) maxima/minima, if the solution is defined over a range, you should check the end points to make sure that you have achieved the true maxima/minima over the range of concern.

Personally, I think the problem is a bit "tricky" in that it has two solutions, and thus is overly complicated. There was also no need for the fractions.

always,

tony

Glipp was thinking like a caveman and didn't check the endpoints !

f(- 1) = f(1) = - 2 which IS the absolute minimum of the function on the interval.

f'(x) = 2x - 12x^3 = 2x(1 - 6x^2) = 0

x = 0 or x = ±1/√6

f(0) = 0

f(±1/√6) = 1/6 - 3/36 = 3/36

absolute minimum is at x = 0 where f(x) = 0