Thread: Relative and Absolute Extrema? Did I solve correctly?

1. Relative and Absolute Extrema? Did I solve correctly?

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

2. You're almost completely correct. For the very fact that there are vertical asymptotes (which, by definition, go to infinity), there are neither absolute min's or max's.

However, consider what a relative min/max is compared to a absolute min/max.

3. A relative min/max must have lower/higher values on each side, whereas a absolute min/max can be an endpoint. Does this mean that there are no relative or absolute extrema in this function?
Thank you so much for answering!

4. There are no absolute max points because the peak of the graph occurs at x=2, which is not defined (as you found). There are also no absolute min points because it has a horizontal asymptote, causing the graph to get closer and closer to 0 for infinity. However, since you have the bounds of [-3,3], there is almost certainly either a relative max, min, or both. Here's a hint - you found it already, you just mislabeled it.

Remember, 95% of the time, if there are bounds given, there will be a relative max, min, or both.

5. Oh ok,
So .04 is the relative minimum at x = +/- 3? I didn't think that relative extrema could exist on endpoints?
Thanks again

6. Yeah, they can. Think about x^2 on [0,5]. The relative min would be 0, since that is the min of the parabola.