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May 25, 2015

Homework Help: Pre-cal

Posted by Lisa on Friday, February 16, 2007 at 7:32pm.

Hi, I don't understand this problem also.
Find the the relative maximmum and minimum of x^3 - 6x^2 + 15.
I'm supposed to do it using a graphing calculator but I still don't see any "lows" or "highs".

Relative max and mins will occur when the derivative is equal to zero. The derivative is 3x^2 - 12x. It equals 0 at x = 0 or x = 4. So they are your max and mins. The second derivative (deriv of the deriv) is 6x-12, which is negative at x=0 and positive at x=4. So since second derivative is negative at x = 0 it is a max, and since positive at x=4, it is a min. I hope derivatives were taught by this point in Pre calc, that is the only way I can explain it.

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