Posted by
**dee** on
.

relative extrema

x^3 + 1/2x^2 - 2x + 5

Ok I have

3x^2 + x - 2 = 0

Am I on the right track?

I think i know why I kept on asking for help with these pribelms. I forgot the step of factoring.

Please check...

3x^2 + x - 2 = 0

(3x-2)(x+1)=0

x=2/3, x=-1

Correct. Nice job.

If you calculate the second derivative (6x + 1) at those points, you can separate maxima from minima. The second derivative is negative at x = -1, making it a relative maximum.