Precal
posted by Lisa .
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 6x12, 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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