Find all relative and absolute extrema in the function, k(x) = (x^2 - 4)^-2, on the interval [-3, 3].

My work:

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