Posted by Brit on .
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 ______?

CALCULUS! 
Anonymous,
f(2)=32+4=28
f(2)=324=28
f'(c)=(28(28))/(2(2))=14
f'(c)=12c^22=14
c^2=16/12=4/3
c=+2/sqrt(3)