Asked 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 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.
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.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.