Exercise 1953Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

Cauchy's extended mean value theorem

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Statement

Let $f,g$ be continuous on $[a,b]$ , differentiable on $]a,b[$ , with $g'(x)\ne0$ on $]a,b[$ . Show that there is $c\in]a,b[$ with
$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}.$

Hint

Apply Rolle to $h(x)=(f(b)-f(a))g(x)-(g(b)-g(a))f(x)$ .

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Solution

Setup. For $h(x)=(f(b)-f(a))g(x)-(g(b)-g(a))f(x)$ , we have
$h(a)=(f(b)-f(a))g(a)-(g(b)-g(a))f(a)$ and $h(b)=(f(b)-f(a))g(b)-(g(b)-g(a))f(b).$ Hence $h(b)-h(a)=(f(b)-f(a))(g(b)-g(a))-(g(b)-g(a))(f(b)-f(a))=0$ .

Rolle. $h$ satisfies $h(a)=h(b)$ ; there exists $c\in]a,b[$ with $h'(c)=0$ :
$0=(f(b)-f(a))g'(c)-(g(b)-g(a))f'(c).$
Rearranging gives the result (note $g(b)-g(a)\ne0$ since $g'\ne0$ ).

Cauchy's extended mean value theorem

Statement

Solution

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