Question

If h(x)=2*x^3-12*x^2-30*x+8, where does h(x) have a local maximum? (I know the critical points are x=-1, x=5, however, I am not sure how to show x=-1 is the local maximum)

Answer 1

Check the second-derivative, that is, how the slope of the function (the derivative) changes with respect to x. If this value is positive, it means the slope is increasing around the point, if this point is a critical point, how would that look? How would it look for a decreasing slope, i.e. a negative second derivative?

Answer 2

You also know that the point in question is a critical point, so right at the point the slope is zero, but the second derivative tells you how this slope is changing around the point. If it's negative (slope decreasing) it means the slope must have been positive right before the critical point, since it's zero at the point, and must become negative right after the critical point.
In other words, if the second-derivative is negative the curve locally looks like this: /‾\

Answer 3

Thank you very much, this makes much more sense now!