The mean value theorem

This theorem is a direct consequence of Rolle's theorem.

Let be $f(x)$ a continuous function on an interval $[a,b]$, and derivable on $]a,b[$.

$$ f \ continuous \ on \ [a,b] \ and \ derivable \ on \ ]a,b[ \ \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = \frac{ f(b) - f(a)}{b-a}$$

Demonstration

Let be $f(x)$ a continuous function on an interval $[a,b]$, and derivable on $]a,b[$.

Let it be as well $ c \in [a,b] $ a real number, the tangent $T_c(x)$ to the curve of $f$ at point $c$, and also an affine function $g(x)$ connecting two points $A$ and $B$.

Points $A, B, C$ are the points of the curve of $f$ corresponding to the abscissae $a, b, c$.

Let us consider a function $\Phi$ also defined on $[a,b]$ such as:

$$\Phi(x) = f(x) - g(x) \qquad (\Phi) $$

Since the slope between $a$ and any point in $x \in [a, b]$ is worth:

$$ \forall x \in [a,b], \ \frac{g(x) - g(a)}{x-a} = \frac{ g(b) - g(a)}{b-a} $$

But,

$$ \Biggl \{ \begin{align*} f(a) = g(a) \\ f(b) = g(b) \end{align*} $$

So,

$$ \frac{g(x) - f(a)}{x-a} = \frac{ f(b) - f(a)}{b-a} $$

$$ g(x) = f(a) + \frac{ f(b) - f(a)}{b-a}(x-a) \qquad (g) $$

Thus, injecting $(g)$ in $(\Phi)$,

$$\Phi(x) = f(x) - f(a) - \frac{ f(b) - f(a)}{b-a}(x-a) \qquad (\Phi^*) $$

Functions $ f, g $ being derivable on $ ]a, b[ $, they are derivable on this same interval, it will be the same for $\Phi$.

Et comme :

$$\Phi(a) = \Phi(b) = 0 $$

Rolle's theorem may therefore apply.

Rolle's theorem tells us that:

For any a continuous function $f(x)$ on an interval $[a,b]$, and derivable on $]a,b[$:

$$ f(a) = f(b) \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = 0 $$

In our case,

$$ \Phi(a) = \Phi(b) \ \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ \Phi'(c) = 0 \qquad(1) $$

And, by applying the derivative of $(\Phi^*)$ we obtain $\Phi'$:

$$ \Phi'(x) = f'(x) - \frac{ f(b) - f(a)}{b-a} \qquad(\Phi ') $$

And thanks to both results $(1)$ and $(\Phi ')$, we obtain this:

$$ \Phi'(c) = 0 \Longrightarrow f'(c) - \frac{ f(b) - f(a)}{b-a} = 0 $$

Thus,

$$\exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = \frac{ f(b) - f(a)}{b-a} $$

And as a result,

$$ f \ continuous \ on \ [a,b] \ and \ derivable \ on \ ]a,b[ \ \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = \frac{ f(b) - f(a)}{b-a}$$