Generalized Mean Value Theorem

Theorem

If $f, g$ are continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ then $\exists c \in (a, b)$ where:

[f (b) - f (a)] g^{'} (c) = [g (b) - g (a)] f^{'} (c)

If $g^{'} \neq 0$ over $(a, b)$ then the conclusion can be stated as:

\frac{f^{'} (c)}{g^{'} (c)} = \frac{f (b) - f (a)}{g (b)} - g (a)

Proof

Apply Rolle's Theorem (Baby MVT) (or just Mean Value Theorem (MVT)) to the function $h (x) = [f (b) - f (a)] g (x) - [g (b) - g (a)] f (x)$ . This is just the linear combination of $f (x), g (x)$ . Notice that $h (a) = h (b)$ so then by Rolle's Theorem (Baby MVT) then $\exists c \in (a, b)$ such that $h^{'} (c) = 0$ . Taking the derivative gives:

h^{'} (x) = [f (b) - f (a)] g^{'} (x) - [g (b) - g (a)] f^{'} (x)

Thus:

h^{'} (c) = [f (b) - f (a)] g^{'} (c) - [g (b) - g (a)] f^{'} (x)

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Application

Remember in Calculus the following limit problem:

lim_{x \to 1} \frac{\ln (x)}{x - 1}

(we know we haven't formally constructed $\ln$ but just roll with it). Normally we'd use L'Hopital's Rule(s). We'd say:

\overset{L^{'} H}{=} lim_{x \to 1} \frac{\frac{1}{x}}{1} = 1

See the L'Hopital's Rule(s) for more on where this comes into play.