Proof of Cauchy's Mean Value Theorem 📂Calculus

Proof of Cauchy's Mean Value Theorem

Theorem¹

Let’s say $a < b$ . If the function $f,g : \mathbb{R} \to \mathbb{R}$ is continuous at all points of $[a,b]$ , differentiable at all $x \in (a,b)$ , and if $g ' (x) \ne 0$ , then there exists at least one $c \in (a,b)$ that satisfies the following: ${{f ' (c)}\over{g ' (c)}}={{f(b)-f(a)}\over{g(b)-g(a)}}$

Explanation

If there’s any difference from the mean value theorem, it’s just that there’s one more function added. Seen as $g(x) = x$ , it means that $g$ becomes more flexible, which can be considered a generalization of the mean value theorem.

Proof

Contrapositive of Rolle’s theorem: If there does not exist $c$ that satisfies $g ' (c)=0$ in $(a,b)$ , then $g(a) \ne g(b)$

Given the premise of the theorem $g ' (x) \ne 0$ , we can guarantee $g(a) \ne g(b)$ through the contrapositive of Rolle’s theorem. Thus, let us define a new function $h : \mathbb{R} \to \mathbb{R}$ as follows: $h(x) := \left[ g(b)-g(a) \right] \left[ f(b)-f(x) \right] - \left[ f(b)-f(a) \right] \left[ g(b)-g(x) \right]$ According to the definition of $h$ , we have $h(a)= 0 = h(b)$ , and by Rolle’s theorem, there exists at least one $c$ that satisfies $h ' (c)=0$ in $(a,b)$ . Differentiating both sides with respect to $x$ yields: $h ' (x)=- \left[ g(b)-g(a) \right] f ' (x)+ \left[ f(b)-f(a) \right] g ' (x)$ Substituting $x=c$ into $h(x)$ gives: $h ' (c)=- \left[ g(b)-g(a) \right] f ' (c)+ \left[ f(b)-f(a) \right] g ' (c)=0$ Therefore, we can conclude that there exists at least one $c$ that satisfies $\displaystyle {{f ' (c)}\over{g ' (c)}}={{f(b)-f(a)}\over{g(b)-g(a)}}$ in $(a,b)$ .

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