The Lagrange Theorem, also known as the Mean Value Theorem, is a fundamental result in mathematical analysis. This theorem states that, given a function continuous on a closed interval \( [a, b]\) and differentiable on \( (a, b) \), there exists at least one point where the derivative coincides with the difference quotient between the endpoints of the interval. The proof is based on Rolle's Theorem and the construction of an auxiliary function.
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Lagrange Theorem (or Mean Value Theorem)
Let \( f : [a,b] \to \mathbb{R} \) be continuous on \([a,b]\) and differentiable at every point of \( (a,b) \). Then there exists at least one point \( \xi \in (a,b) \) such that:
\[ f'(\xi) = \frac{f(b) - f(a)}{b - a} \]
Proof. To prove this theorem, we construct an auxiliary function \(F(x)\) that will allow us to apply Rolle's Theorem. We define:
\[ F(x) = f(x) - \left[f(a) + (x-a)\frac{f(b)-f(a)}{b-a}\right] \]
This function \(F(x)\) is the difference between \(f(x)\) and the line passing through the points \((a,f(a))\) and \((b,f(b))\). It is easy to verify that \(F(a) = F(b) = 0\). Moreover, \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\), inheriting these properties from \(f\).
Applying Rolle's Theorem to \(F\), there exists at least one point \(\xi \in (a,b)\) such that \(F'(\xi) = 0\). Computing the derivative of \(F\) we obtain:
\[ F'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} \]
Therefore, \(F'(\xi) = 0\) implies the thesis of the theorem. This point \( \xi \) is not necessarily unique.
Corollaries of Lagrange Theorem
Corollary 1. If a function has zero derivative at every point of an interval, then the function is constant on that interval.
Proof. We fix a point \(x_0\) in the interval. For any other point \(x\), applying Lagrange's theorem to the interval \([x_0,x]\), we obtain:
\[ f(x) - f(x_0) = f'(\xi)(x - x_0) = 0 \]
Therefore \(f(x) = f(x_0)\) for every \(x\) in the interval.
Corollary 2. If \(f\) is differentiable on an interval \(I\) and \(f'(x) \geq 0\) for every \(x \in I\), then \(f\) is non-decreasing on \(I\). Similarly, if \(f'(x) \leq 0\), then \(f\) is non-increasing. If \(f'(x) > 0\) for every \(x \in I\), then \(f\) is strictly increasing; if \(f'(x) < 0\), then \(f\) is strictly decreasing.
Proof. Taking any two points \(x_1 < x_2\) in \(I\), Lagrange's theorem tells us that:
\[ f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1) \geq 0 \]
since \(f'(\xi) \geq 0\) and \(x_2 - x_1 > 0\). Therefore \(f(x_2) \geq f(x_1)\). If \(f'(\xi) > 0\), then \(f(x_2) > f(x_1)\).
Corollary 3. If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\) and \(m_1 \leq f'(x) \leq m_2\) for every \(x \in (a,b)\), then:
\[ m_1(x - a) \leq f(x) - f(a) \leq m_2(x - a) \]
Proof. Applying Lagrange's theorem, we know that there exists \(\xi\) between \(a\) and \(x\) such that:
\[ \frac{f(x) - f(a)}{x - a} = f'(\xi) \]
And since \(m_1 \leq f'(\xi) \leq m_2\), the thesis follows immediately.