The mean value theorem

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

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

$$ f \text{ is continuous on } \bigl[a,b \bigr] \text{ and derivable on } \bigl ]a,b \bigr[ \ \Longrightarrow \ \exists c \in \bigl ]a,b \bigr[, \ f'(c) = \frac{ f(b) - f(a)}{b-a} \qquad \bigl(\text{The mean value theorem} \bigr) $$

Proof

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

Let us also consider an affine function $g(x)$ secant to the curve of $f$ at the boundary points $A(a, \, f(a))$ and $B(b, \, f(b))$.

Points $A$ and $B$ represent the endpoints of the curve of $f$ over the studied interval.

Let us now consider an auxiliary function $\Phi$ also defined on $\bigl[a,b \bigr]$ such as:

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

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

$$ \forall x \in \bigl[a,b \bigr], $$

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

But,

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

So,

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

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

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

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

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

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 $\bigl[a,b \bigr]$, and derivable on $\bigl ]a,b \bigr[$:

$$ f(a) = f(b) \Longrightarrow \ \exists c \in \bigl ]a,b \bigr[, \ f'(c) = 0 $$

In our case,

$$ \Phi(a) = \Phi(b) \ \Longrightarrow \ \exists c \in \bigl ]a,b \bigr[, \ \Phi'(c) = 0 \qquad(1) $$

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,

$$\exists c \in \bigl ]a,b \bigr[, \ f'(c) = \frac{ f(b) - f(a)}{b-a} $$

And as a result,

$$ f \text{ is continuous on } \bigl[a,b \bigr] \text{ and derivable on } \bigl] \hspace{0.1em} a, b \hspace{0.1em} \bigr[ \ \Longrightarrow \ \exists c \in \bigl ]a,b \bigr[, \ f'(c) = \frac{ f(b) - f(a)}{b-a} \qquad \bigl(\text{The mean value theorem} \bigr) $$