Consequences of the Mean Value Theorem Learning Goals

Consequences of the Mean Value Theorem

Learning goals: There are some nice results that come from the Mean Value Theorem and its variations. Let’s look at some.

There are a number of immediate consequences of the Mean Value Theorem that are bedrock for later work. Theorem. Let f(x) be continuous on [a, b] and differentiable on (a, b). Then: 1. if f 0(x) > 0 for all x ∈ (a, b) then f is strictly increasing. 2. if f 0(x) < 0 for all x ∈ (a, b) then f is strictly decreasing. 3. if f 0(x) = 0 for all x ∈ (a, b) then f is constant. 4. if g(x) is another such function and g0(x) = f 0(x) for all x ∈ (a, b) then f(x) − g(x) is constant. Proof. For the first item, let x < y be in the interval (a, b). Then y−x > 0 and f(y)−f(x) = f 0(c)(y − x) for some c ∈ [x, y] ⊂ (a, b). But f 0(c) > 0 for all c ∈ (a, b) so f(y) − f(x) > 0. Thus positive derivative everywhere implies strictly increasing. The second item is identical with f 0(c) being negative, and the third with f 0(c) = 0. The fourth follows from the third by differentiation. Of course, the converse of all these items is also true. For if f is increasing its difference quotients are always positive, so the derivative which is their limit must be nonnegative. The derivative of a strictly increasing function can be zero at single points; witness f(x) = x3 at x = 0. The proof of the following theorem will be in-class discussion or homework: Theorem. Let f(x) be continuous on (a, b] and differentiable on (a, b). Assume lim f 0(x) = x→b− L. Then f is (one-sided) differentiable at b and f 0(b) = L. Of course there is a corresponding version for f approaching a from the right. A very interesting consequence of the Mean Value Theorem is that derivatives satisfy the Intermediate Value Property. That is, if f 0(a) = s and f 0(b) = t then for each u between s and t there is a c between a and b such that f 0(c) = u. To be formal about it: Theorem (Darboux). Let f(x) be a function that is differentiable at all point in [a, b] (one- sided derivatives at the endpoints is OK). Let y be a value such that f 0(a) < y < f 0(b) or f 0(a) > y > f 0(b). Then there is some c ∈ (a, b) such that f 0(c) = y.

f(x)−f(a) 0 Proof. Consider the function g(x) = x−a when x 6= a and g(a) = f (a). Then g is continuous on [a, b] so by the Intermediate Value Theorem takes every value between 0 f(b)−f(a) g(a) = f (a) and g(b) = b−a . So the diﬀerence quotient of f takes each of these values, 0 f(b)−f(x) and, by MVT, f much take each of these values. Applying the same logic to h(x) = b−x 0 0 f(b)−f(a) 0 (setting h(b) = f (b), of course) forces f to take all values between b−a and f (b). So we’re done.

1 The basic idea of the proof is that the derivative has to take on all values that the difference quotient does, so take all the difference quotients between a and some point x, and between x and b, and these are all the numbers between f 0(a) and f 0(b) by the Intermediate Value Theorem applied to the continuous difference quotient function. This theorem has some interesting corollaries of its own:

• f 0 cannot change signs without going through zero.

• If f 0 6= 0 for any input, then f 0 never changes sign, so f is monotonic.

• If f 0 itself is monotonic then it must be continuous, for the only kinds of discontinuities of monotonic functions are jumps (we didn’t actually prove that but it isn’t incredibly hard) and derivatives can’t make jumps (we haven’t proven that either, and it is a little tougher).

Since derivatives satisfy the Intermediate Value condition, do they have to be continuous? It is not easy to ﬁnd a counterexample for this question, but the answer is no.

2 1 Example: Consider the function f(x) = x sin x (when x 6= 0, f(0) = 0). This function is easily seen to be continuous everywhere. Away from zero, standard differentiation rules 1 1 show the derivative to be 2x sin x − cos x . At zero, since the function has a different formula, you have to use the limit definition, but of course it is simple enough to show that 0 1 the derivative f (0) = 0. But the values of the derivative at x = πn do not converge, so the derivative itself is discontinuous at x = 0. We have seen crazy functions that are discontinuous everywhere, and also that are discontinuous at, say, the rationals but not the irrationals. Can a derivative be such a function? The answer to that is no; derivatives can be ugly but not too ugly. Examples will have to wait until we have more background, though. Functions that are differentiable but whose derivatives are not continuous are kind of a pain in the neck. So often theorems require functions that have continuous derivatives. For example, in MVC one can prove that a multivariable function is differentiable if each partial derivative is continuous. Functions where the derivative is actually continuous are called, obviously, continuously differentiable and are said to belong to class C1.