Proving Cauchy's Mean Value Theorem
test The mean value theorem is one of the most useful results from calculus, as it allows us to study many of its mathematical consequences while also attaching an intuitive geometric interpretation.
This is a short post which guides you through the proof of a more general version of the mean value theorem, called Cauchy's mean value theorem (Cauchy's MVT) or the generalized mean value theorem.
The key to approaching these exercises is to turn your geometric understanding into mathematically precise statements. In order to do this, we need to take ideas that are currently understood geometrically and turn them into rigorous mathematics.
To get an idea of what this means in practice, take a look at the exercise below. This will be similar to the exercises you will encounter throughout the post. Try to complete it yourself, and open the hints if you need them.
Exercise 3: Show that if f defined on an open interval (a,b) has a local (or relative) extreme point x∗ in (a,b) and is differentiable at x∗, then f′(x∗)=0.
To complete the exercise we need a rigorous definition of "local extreme point." If you do not already know what a local extreme point is, then this is not a hint but rather context which is required to solve the problem.
Let g be a function defined on an open interval I. We say that x∗ in I is a local maximum point of g if there exists r>0 such that g(x∗)≥g(x) for all x in (x∗−r,x∗+r).
Exercise 3.1: Interpret this definition geometrically. What does r represent here?
Exercise 3.2: Define local minimum point.
Try to show that f′(x∗)≥0 and f′(x∗)≤0 simultaneously by using the definition of the derivative and the fact that x→c−limg(x)=x→c+limg(x) if limx→cg(x) exists.
We will assume first that x∗ is a local maximum point. Then we can choose r>0 so that f(x∗)≥f(x) for all x in I=(x∗−r,x∗+r). For all x− in I such that x−<x∗, we see that f(x−)≤f(x∗). Likewise, f(x+)≤f(x∗) for all x+ in I such that x+>x∗. Then
x→x∗−limx∗−xf(x∗)−f(x)≥0 and x→x∗+limx∗−xf(x∗)−f(x)≤0Since f is differentiable at x0, f′(x∗)=0 (since f′(x∗) can't be simultaneously positive and negative). If x∗ is instead a local minimum point of f, we see that f′(x∗)=−(−f)′(x∗)=0 since x∗ is then a local maximum point of −f.
You should spend a reasonable amount of time (maybe 1 hour max) on each exercise. It is not shameful to look at the solution for a hint if you are stuck; in fact, it is arguably more productive to do this than to agonize over a difficult exercise.
The proof of Cauchy's MVT relies on Rolle's theorem. To prove Rolle's theorem, you should be familiar with the extreme value theorem.
Rolle's Theorem: If a function f is continuous on [a,b] and differentiable on (a,b) and f(a)=f(b), then there exists at least one c in (a,b) such that f′(c)=0.
Exercise 4: Write the statement of the extreme value theorem.
The extreme value theorem states that if a function f defined on [a,b] is continuous (on [a,b]), then f must have at least one global minimum point c and at least one global maximum point d, both in [a,b].
That is, there exist c,d in [a,b] such that for all x in [a,b],
f(c)≤f(x)≤f(d).Exercise 5: Use the extreme value theorem and Exercise 3 to prove Rolle's theorem.
By the extreme value theorem, a function f defined on [a,b] satisfying the conditions for Rolle's theorem has a global minimum point c and a global maximum point d, both in [a,b].
If c or d is in (a,b), we see from Exercise 5 that f′(c)=0 or f′(d)=0. If both c and d are not in (a,b), they must both be in {a,b}. Hence f(x)=f(a)=f(b) for all x in [a,b], meaning f′ is zero on (a,b).
We can now prove Cauchy's mean value theorem using Rolle's theorem.
Let's look at what we want to prove:
Cauchy's Mean Value Theorem: Suppose functions f,g are both continuous on [a,b] and differentiable on (a,b).
If g(a)=g(b) and g′(x)=0 for all x in (a,b), there exists c in (a,b) so that g(a)−g(b)f(a)−f(b)=g′(c)f′(c).
Just like before, we'll prove this through a series of exercises.
Exercise 6: Let f, g, a, and b satisfy the hypothesis above. Define the function h by h(x)=f(x)−g(a)−g(b)f(a)−f(b)g(x). Prove that h(a)=h(b).
This problem is easier than the others, so no solution is provided. Instead, a rather generous hint is given below.
The proof is just algebraic manipulation. Observe that f(a)−f(b)=g(a)−g(b)f(a)−f(b)(g(a)−g(b))=g(a)−g(b)f(a)−f(b)g(a)−g(a)−g(b)f(a)−f(b)g(b).
Exercise 6.1: Complete the proof.
Exercise 7: Compute h′ for h in Exercise 6.
The important part is that g(a)−g(b)f(a)−f(b) is a constant. So, h′(x)=f′(x)−g(a)−g(b)f(a)−f(b)g′(x)
Exercise 8: Prove Cauchy's mean value theorem using Exercises 6, Exercise 7, and Rolle's Theorem.
Since products and sums of continuous (resp. differentiable) functions are continuous (resp. differentiable), h is continuous on [a,b] and differentiable on (a,b). Because h(a)=h(b) (Exercise 6), by Rolle's Theorem, there exists c in (a,b) such that h′(c)=f′(c)−g(a)−g(b)f(a)−f(b)g′(c)=0. Hence, g(a)−g(b)f(a)−f(b)=g′(c)f′(c) since g′(c)=0.
(Mini) Exercise 9: Show how the (regular) mean value theorem follows from Cauchy's mean value theorem.
Cauchy's mean value theorem is also called the extended mean value theorem. Defining g in Cauchy's mean value theorem by g(x)=x for all x in [a,b] gives the mean value theorem.
Exercise 9.1: Verify that g as defined above satisfies the conditions for Cauchy's mean value theorem.
Because of this, Cauchy's mean value theorem is said to be "more general" than the mean value theorem.
Last Updated: December 2022