Inactive Tutor answered 12/04/21
f(-2) = 4(-8)+6 = -26
f(2) = 4(8)-6 = 26
f'(c) = (f(b)-f(a))/(b-a)
(f(2)-f(-2))/(2- -2) = 52/4 = 13
f'(c) = 12c2-3 = 13
c = ±√(16/12) = ±2√3/3
The smaller one is -2√3/3
The larger one is 2√3/3
James H.
asked 12/04/21Consider the function f(x) = 4x3 − 3x on the interval [− 2, 2]. Find the average or mean slope of the function on this interval. By the Mean Value Theorem, we know there exists at least one c in the open interval (− 2, 2) such that f'(c) is equal to this mean slope.For this problem, there are two values of c that work.The smaller one is and the larger one is
Inactive Tutor answered 12/04/21
f(-2) = 4(-8)+6 = -26
f(2) = 4(8)-6 = 26
f'(c) = (f(b)-f(a))/(b-a)
(f(2)-f(-2))/(2- -2) = 52/4 = 13
f'(c) = 12c2-3 = 13
c = ±√(16/12) = ±2√3/3
The smaller one is -2√3/3
The larger one is 2√3/3
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