Question

how do you find the local minimum and maximum points of f(x)=x^4-6x^2

Answer 1

1st step find first derivative
2nd step set first derivative=0 and solve for x
3rd step test to see if the function is changing from increasing to decreasing (=>local max) or decreasing to increasing (=>local min)
If neither of those happen, then you do not have a local min/local max (the slopes have to change sign at the critical number)

Answer 2

what is the derivative?

Answer 3

Introduce a new variable, X = x^2. Now you can rewrite f(x) as X^2 - 6X, since X^2 = (x^2)^2 = x^4.

Now you can easily find where the derivative of X^2 - 6X = 0. The solution is X=0, and X=6. Now change them back into x, using X=x^2 => sqrt(X) = x. So you have the critical points of f(x) are 0, and +/- sqrt(6). I think that's right... figure out which of those are minimum and maximum points by looking at second derivative.