Cauchy's Mean Value Theorem and Its Proof

Cauchy's Mean Value Theorem

Suppose f and g are two functions such that:

• f and g are continuous on the closed interval [a, b].
• f and g are differentiable on the open interval (a, b).
• For all x in the open interval (a, b), g'(x) ≠ 0.

Then there exists a number z in the open interval (a, b) such that:

[f(b) - f(a)] / [g(b) - g(a)] = f'(z) / g'(z)

Proof of the Theorem

Proving that g(b) is not equal to g(a)

First, we show that g(b) ≠ g(a). We use a proof by contradiction; assume that g(b) = g(a). Since g satisfies the conditions of the Mean Value Theorem, there is some number c in (a, b) such that:

g'(c) = [g(b) - g(a)] / (b - a)

If we assume g(b) = g(a), then g(b) - g(a) = 0, which implies g'(c) = 0. This contradicts the third condition of our hypothesis, which states that g'(x) ≠ 0 in (a, b). Therefore, the assumption cannot be true, meaning g(b) ≠ g(a) and consequently g(b) - g(a) ≠ 0.

Defining the Auxiliary Function

Now consider the function h defined by:

h(x) = [f(x) - f(a)] - [[f(b) - f(a)] / [g(b) - g(a)]] * [g(x) - g(a)]

We shall show that h satisfies the conditions of Rolle's Theorem. The first two conditions are met because f(x) and g(x) are continuous and differentiable; since h(x) is the difference of continuous and differentiable functions, it is also a continuous and differentiable function.

Applying Rolle's Theorem

Now we compute h(a) and h(b):

• h(a) = [f(a) - f(a)] - [[f(b) - f(a)] / [g(b) - g(a)]] * [g(a) - g(a)] = 0
• h(b) = [f(b) - f(a)] - [[f(b) - f(a)] / [g(b) - g(a)]] * [g(b) - g(a)] = [f(b) - f(a)] - [f(b) - f(a)] = 0

Since h(x) satisfies the three conditions of Rolle's Theorem, there exists a number z in (a, b) such that h'(z) = 0.

Final Derivation

The derivative h'(x) is:

h'(x) = f'(x) - [[f(b) - f(a)] / [g(b) - g(a)]] * g'(x)

Replacing x with z, we have:

h'(z) = f'(z) - [[f(b) - f(a)] / [g(b) - g(a)]] * g'(z) = 0

This simplifies to:

f'(z) = [[f(b) - f(a)] / [g(b) - g(a)]] * g'(z)

Which results in the final form:

f'(z) / g'(z) = [f(b) - f(a)] / [g(b) - g(a)]

This is exactly what was claimed.

Special Case of the Theorem

The standard Mean Value Theorem is a special case of Cauchy's Theorem in the instance where g(x) = x.