Find all relative and absolute extrema in the function, k(x) = (x^2 - 4)^-2, on the interval [-3, 3].
k'(x) = [(-4x)] / [(x^2 - 4)^3]
Critical Points: x = 2, x = -2
Since there are vertical asymptotes at x = +/- 2, there are no relative extrema.
There is no absolute maximum because the function has no upper bound.
When x = -3, k(-3) = .04
When x = 3, k(3) = .04
Therefore, the function has absolute minimum of k(-3) = k(3) = .04
Basically there are:
No relative extrema
No absolute maximum
Absolute minimum at k(-3) and k(3)
Is this correct?
Any help is appreciated