You cannot find a min and max from the first equation. If you take the left hand side as a function, we can find a min and max. First you set the derivative equal to 0. This gives a quadratic equation that you can solve to find the relative min/maxes.
Inflection points are points where the 2nd derivative changes sign. Thus the 2nd derivative is 0 at that point. So you have to check both solutions of your quadratic second derivative. Looking at a parabola, if there really are two 0s of the second derivative then they both represent sign changes of the second derivative.
The first equation is actually the first derivative from a quartic equation. So with how that first derivative is set up there is no way to find min/max? Is that because there is no way to get x by itself?
And the second derivative I just do the quadratic function solutions (both + and -) to find the inflection points?
Oh I see what you are saying. You are correct you have to solve that cubic equation. What I would do is have something like wolfram-alpha solve it for you.
And the second one, you are on the right track. If your parabola has 2 solutions, each is an inflection point. If it has 1 or 0 solutions, neither is an inflection point. This is because the concavity (second derivative) never changes signs for those case.
Apr 22nd, 2015
Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.