Why does f'(x) = 0 for all x in (a,b) imply f is constant?:

Theorem 1 part c of section 4.1 says: If f'(x) = 0 for each value of x in an interval (a,b) then f is constant on (a,b). This is an extremely important theorem, as will become obvious when we get to Chapter 6.

It should look rather likely to be true: after all, f'(x)=0 says that the slope of the graph of f is zero everywhere, and that ought to mean that f is constant. But it is a bit subtle -- it is very important that f be differentiable at every single point in the interval.

Let f be the function with value 0 on the interval (0,1/3), value 1/2 on the interval [1/3,2/3], and value 1 on the interval (2/3,1); here the interval (a,b) is (0,1).

This function f is NOT constant on the whole interval (0,1) (there is no one single number C such that f(x) = C for each x in (0,1)). But f certainly has f'(x) = 0 for all x in the three intervals (0,1/3), (1/3,2/3), and (2/3,1). So why doesn't it contradict the theorem? Because it is not differentiable (in fact, not even continuous) at x=1/3 and at x=2/3.

Things can get worse. A fascinating example is Cantor's "Devil's Staircase." This is the CONTINUOUS function with graph

This function is defined as follows: for all x in the interval [1/3,2/3], f(x) =1/2; for x in [1/9,2/9], f(x) = 1/4; for x in [7/9,8/9], f(x)=3/4, etc. etc. Again f'(x)=0 for each x in the interior of each of the intervals in its definition, and the sum of all the lengths of the intervals is 1. (It can be written as the sum of a geometric series, for those of you who know about geometric series.)

Yet clearly this f is not constant. The important fact is that although f is continuous, it is not differentiable at ALL points in (0,1)-- in advanced calculus courses, students can prove that f is not differentiable at the endpoints of the intervals in its definition, as well as that f is continuous. So this f is not a counterexample to Theorem 1c. And in fact, there are no counterexamples, since it is a true theorem.

Theorem 1c is proved as a special case of a more general theorem:

The Mean Value Thereom: Suppose a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then there is a number c in (a,b) such that the slope of the graph of f at the point (c,f(c)), namely f'(c), is equal to the slope of the line connecting the endpoints (a,f(a)) and (b,f(a)) of the graph, namely (f(b)-f(a))/(b-a).

In the example above, there are in fact not just one but three numbers c for which the conclusion holds. Two of them are shown, as dots on the x-axis, and the tangent lines for those values of c can indeed be seen to be parallel to the line joining the endpoints.

Assignment: Find the third c (approximately), by finding a third parallel tangent line and the point on the x-axis below its point of tangency.

The Mean Value Theorem says that there is at least one specific point at which the instantaneous rate of change is the same as the average rate of change over the interval. In other words, if you drive 120 miles in 2 hours, then at least once during that time you were driving exactly 60 miles per hour ( = difference in position f(b)-f(a) (120 miles)/difference in time b-a (2 hours).

To prove Theorem 1c, you apply the Mean Value Theorem, using an arbitrary interval [a,b]. The Mean Value Theorem then says that there is some c between a and b such that f'(c) = (f(b)-f(a)/(b-a). But f'(x) is always 0, by assumption! Therefore f(b) must equal f(a). Since b is completely arbitrary, it says f(b) = f(a) for every single number b. That means f is constant.

To prove the Mean Value Theorem is indeed beyond the scope of this course! It is proved, however, in Math 151. Ask your Lecturer for a proof if you are interested.