Mean value free response?

Question

The continuous function f is defined on the closed interval [-5,5]. The graph of f consists of a parabola and two line segments. Let g be a function such that g'(x)=f(x).

There is no value of x in the open interval (-1,3) at which f(3)-f(1)/3-(-1). Explain why this does not violate the Mean Value Theorem.

Answer 1

The requirement of the MVT is that the function be differentiable on the interval (a,b). Continuity is not enough.

To help understand why this is the case, consider the continuous function f(x) = |x| on the interval [-1 , 1]. Note that this function is not differentiable at x = 0 (the graph has a sharp corner, when its slope jumps from -1 to 1).

The average rate of change on this interval is 0, because f(1) = f(-1). However, the instantaneous rate of change, f ' (x), ≠ 0 anywhere on this interval.