calculus
posted by dee .
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
(3x2)(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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@Steve RE: Relative Extrema: AP Calculus
Thanks so much for all your help! That was way simpler than I thought, I think I was just overthinking it. Do you think you could also help me find the concave up at x = −2, x = −1, and x = 2 and show f(x) is concave down …