Today’s question (problem submitted by a reader) covers the mean value theorem from Calc. I. This one is a bit subtle in the sense that it raises visibility to the existence of a derivative in an interval, even when the function is continuous. The problem was find the point, c, in the interval [-8,8] for the function f(x) = x2/3 that satisifes the mean value theorem.
For review, the mean value theorem states that if a function, f(x), is continuous in [a,b] and differentiable in (a,b), then there exists at least one point, c, in (a,b) at which f(b) – f(a) = f'(c)(b-a). Geometrically, the mean value theroem implies that there is a point in the specified interval at which the slope of the tangent at that point is equal to the slope of the chord between the endpoints of the interval. In terms of velocity, one might say that if a trip over some time interval averaged 60mph, then there must exist at least one point during the trip at which the instantaneous velocity was exactly 60mph.
For the problem at hand, f(b)-f(a)/b-a = 0/16 = 0, so the slope of the chord between the endpoints is zero. The derivative, f'(x) is (2/3)x-1/3, which exists everywhere in (-8,8) except zero. Theoretically, we could just stop here. The derivative is negative in [-8,0) and positive in (0,8]. It is never exactly zero in the interval (-8,8). Like the spoon in The Matrix, there is no point satisfying the mean value theorem in this problem.
Now, it is possible that a function may not be differentiable in an interval (a,b) and still have one or more points satisfying the conditions of the MVT. However, one can not use the MVT to guarantee the existence of such a point(s). In this example, it was game over as soon as you showed the function was not differentiable in the interval.
The Algorithmist 