The Lagrange's Theorem, also known as the Mean Value Theorem, is a fundamental result in mathematical analysis. This theorem states that, given a function that is continuous on a closed interval \( [a, b]\) and differentiable in \( (a, b) \), there exists at least one point where the derivative equals the difference quotient between the endpoints of the interval. The proof is based on the Rolle's Theorem and the construction of an auxiliary function.
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Lagrange's 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 allows 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:
\[ F'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} \]
Thus, \(F'(\xi) = 0\) implies the theorem's statement. This point \( \xi \) is not necessarily unique.
Corollaries of Lagrange's Theorem
Corollary 1. If a function has a zero derivative at every point of an interval, then the function is constant over that interval.
Proof. 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 \(x_2 - x_1 > 0\). Thus, \(f(x_2) \geq f(x_1)\). 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 satisfies \(m_1 \leq f'(x) \leq m_2\) 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:
\[ \frac{f(x) - f(a)}{x - a} = f'(\xi) \]
Since \(m_1 \leq f'(\xi) \leq m_2\), the statement follows immediately.