ok, you have the mean slope, which is 0
So, where is f'(x) = 0?
f' = -12x
so f'(0) = 0
This should all have been clear, since f(x) is a parabola, with x=0 as the axis of symmetry, and the interval [-3,3] has its center there.
Consider the function f (x)=7−6x^2 on the interval [ − 3 , 3 ] . Find the average or mean slope of the function on this interval, i.e.
f(3)-f(-3) / 3-(-3)=
By the Mean Value Theorem, we know there exists a c in the open interval (−3,3) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
=
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