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 \(]a,b[\).

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

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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.

In our case,

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

And, by applying the derivative of \((\Phi^*)\):

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

Thus,

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

And as a result,

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