Finding local max or local min values

Determine where f(x)=8x^{3}+18x^{2}-24x-9 has any local maximum or local minimum values.

What I did:

f'(x)=24x^{2}+36x-24

0=24x^{2}+36x-24

0=12(x+2)(2x-1)

x=-2, (1/2)

After getting the critical numbers, I'm not really sure where to go on from there because of the answers given.

Can someone explain to me about finding local max and min a bit.

**From the multiple choice answers that were given, it was either**

a. local max at x=3/4, local min at x=-4/3

b. local max at x=-3/4, local min at x=4/3

c. local max at x=4/3, local x=-3/4

d. local max at x=-4/3, local min at x=3/4

Re: Finding local max or local min values

Try checking the second derivative at each of the critical points. If it's negative at that point you have a maximum, if it's positive at that point you have a minimum. If it's 0 at that point the test is inconclusive...

Re: Finding local max or local min values

Quote:

Originally Posted by

**Prove It** Try checking the second derivative at each of the critical points. If it's negative at that point you have a maximum, if it's positive at that point you have a minimum. If it's 0 at that point the test is inconclusive...

Ok, so the second derivative would be like: f''(x)=58x+36

f''(-1)=58(-1)+36=-58+36=-22 max

f''(1)=58(1)+36=58+36=94 min

But I was wondering, isn't the second derivative test used to find max extremas and min extremas? Or I'm thinking of something different. Or is it the same as local maximum and minimum.

But anyway, so since I know what's the min and max, would I just plug the critical numbers back into the original equation right?

But... when I plug in 1, and -1, the numbers don't come out like the fractions in one of the multiple choice answers. So I'm not sure.

Re: Finding local max or local min values

Why are you evaluating the second derivative at 1 and -1? Your critical points are where x = 1/2 and x = -2...