f(-4) = 68
f(4) = 28
average slope = (28-68)/(4+4) = -5
f'(x) = 6x-5
f'(c) = 6c - 5 = -5
c = 0
I only get one value.
Consider the function f(x)=3x^2 – 5x on the interval [-4,4]. Find the average slope of the function on this interval.
By the mean value theorem, we know there exists at least one c in the open interval (-4,4) such that f’(c) is equal to this mean slope. For this problem, there are 2 values of c that work. The smaller one and the larger one are?
1 answer