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Question
Find the critical numbers of f(if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema.
f(x)=x^3-6x^2+15
To find the critical values, take the first derivative. That is, f'(x) = 3x² - 12x.
This factors to f'(x) = 3x(x-4). The critical points are where f'(x) = 0.
These are the points x=0 and x=4.
Note that if x is negative, then 3x and x-4 are both negative, and the product of two negatives is a positive. So for x<0, the derivative is positive. Also, if x>4, both 3x and x-4 are both positive, and the product of two positives is a positive. This means the derivative is positive. That is where the function is increasing, is on (-∞,0) and (4,∞).
When x is between 0 and 4, 3x is positive and x-4 is negative, so the derivative is negative.
This means on the interval (0,4), the function is decreasing.
The extreme point are the points in the middle of the intervals we just found.
They are at x=0 and x=4.
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