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(Cauchy MVT). Let $f,g: [a, b] \longrightarrow \mathbb{R}$ continuous functions and differentiable em $(a, b)$, then there exist $c \in > (a,b)$ such that: $$[f(b) - f(a)]g'(c) = f'(c)[g(b) - g(a)]$$

$\textbf{My proof:}$ Define $h: [a,b] \longrightarrow \mathbb{R}$ by $h(x) = [f(b) - f(a)]g(x) - [g(b) - g(a)]f(x)$. Note that $h$ satisfies de Rolle's Theorem hypothesis and $$h(a) = f(b)g(a) - f(a)g(a) - g(b)f(a) + g(a)f(a) = f(b)g(b) - f(a)g(b) - g(b)f(b) + g(a)f(b) = h(b).$$

So, there exists $c \in (a, b)$ such that $h'(c) = 0.\blacksquare$

Does anyone know an alternative proof? (Without use auxiliary functions)

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