Calculus Question: Relative Max/Min and Inflection Points?

For the function y = f(x), the solutions to the equation 0 = f'(x) are x = -3,0,2 and the solutions to the equation 0 - f"(x) are x= -1, 2. Furthermore, f"(-2) = 7, f"(1) = -9, and f"(3) = 4.

a. Where does y = f(x) have a relative maximum and where does y = f(x) have a relative minimum?

b. Where does y= f(x) have an inflection point?

Would love some help with this, I don't even know where to begin!

Re: Calculus Question: Relative Max/Min and Inflection Points?

The sign of the tells you if the function is increasing or decreasing in the neighborhood of . Note that on left of the relative maximum must be an increasing function, and on the right of the relative maximum it must be a decreasing function. It is the other way around in the neighborhood of a relative minimum.

And bear in mind that what tells you about , tells you about .

Solution to the is the candidate for the inflection point, and it actually is an inflection point if on the left and on the right of it is of the opposite sign.

Re: Calculus Question: Relative Max/Min and Inflection Points?

So how do I go about answering this? Where is the starting point?

Thanks for your reply.

Re: Calculus Question: Relative Max/Min and Inflection Points?

First, and are inflection points for function since is of the opposite sign on the opposite sides of either of those two numbers.

Re: Calculus Question: Relative Max/Min and Inflection Points?

So to answer B, the inflection points are x=-1 and x=2? I do not understand how to get relative maximum and minimum since an equation was never given (or I don't know how to get an equation from the derivative values). My instructor has never presented us with a problem like this. I would appreciate being walked through it step-by-step since I have an exam over it tomorrow. If that would be at all possible, I would appreciate it.

Re: Calculus Question: Relative Max/Min and Inflection Points?

Those inflection points split up the domain (suppose ) into intervals: , , . On each of those intervals you know the sign of since it can change sign only in inflection points.

*If you know that than is convex on . Otherwise is said to be concave.*

Since you know that and then , therefore is convex on . Therefore the stationary point can only be relative minimum.

Since you know that and then , therefore is concave on . Therefore the stationary point can only be relative maximum.

Re: Calculus Question: Relative Max/Min and Inflection Points?

Do the last one yourself.