Cauchy’s Form of the Mean Value Theorem

If we apply the Mean value theorem to each of two functions f and g continuous on [a,b] and differentiable on (a, b), we can conclude that there are c₁, c₂ ∈ (a, b) such that

$$f^{\prime }(c_1)=\frac{f(b)-f(a)}{b-a}\text{and}{g}^{\prime }(c_2)=\frac{g(b)-g(a)}{b-a}$$

This gives

However it happens to exist one c ∈ (a,b) wich works for both function, as the next theorem precises.

Theorem 5.6.1 (Cauchy's Mean Value Theorem) Let f and g continuous on [a,b] and differentiable on (a, b). Then there exists c ∈ (a,b) such that

Proof. If g(a) = g(b) then the result holds by Rolle's Theorem applied to g. So we assume that g(a) ≠ g(b). In the proof of Lagrange's Mean value theorem we used the auxiliary function

We replace the function x (identity function) with g; We have

Then ω is continous on [a,b] and differentiable on (a,b). One can easily verify that ω(a) = ω (b), so by Rolle's Theorem there is a c ∈ (a,b) such that ω'(c) = 0. Finally ω'(c) = 0 gives

as desired.  □

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