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.
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