f(-2)=-32+4=-28
f(2)=32-4=28
f'(c)=(28-(-28))/(2-(-2))=14
f'(c)=12c^2-2=14
c^2=16/12=4/3
c=+-2/sqrt(3)
Consider the function f(x)=4x^3–2x on the interval [–2,2]. Find the average or mean slope of the function on this interval. __14__
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 ______?
1 answer
1 answer