Lagrange's Mean Value Theorem
Theorem 1: Lagrange's Mean Value Theorem
Suppose a function \(f\) is a continuous on \([a, b]\) and differentiable on \((a, b)\). Then
there exists a number \(c \in (a, b)\) such that
\[
f'(c) = \frac{f(b) - f(a)}{b - a} \tag{1}
\]
Note: The special case of this theorem: if \(f(a) = f(b)\), then f'(c) = 0. is called Rolle's Theorem.
We ommit the proof but the outile of the proof can be:
- case 1: \(f(x)\) is constant, \(f'(x)\) = 0 for all \(c \in (a,b)\).
- case 2: \(f(x) > f(a)\), show that at \(c\), \(f\) has local maximum, then \(f'(c) = 0\) by the Extrem Value Theorem.
- case 3: \(f(x) < f(a)\), similarly, show that at \(c\), \(f\) has local minimum, then \(f'(c) = 0\).
Proof:
Suppose a function \(f\) is a continuous on \([a, b]\) and differentiable on \((a, b)\).
Consider the line connecting \(a, f(a)\) and \(b, f(b)\). The equation of the line can be written as \[ y - f(a) = \frac{f(b)-f(a)}{b -a} (x - a) \] Let g(x) be the vertical difference between \(x, f(x)\) and (x, y). Then \[ g(x) = f(x) - f(a) - \frac{f(b)-f(a)}{b -a} (x -a) \tag{2} \] \(g\) is continuous on \([a, b]\) and differentiable on \((a, b)\) because \(g\) is the sum of \(f\) and the first-degree polynomial, both of which are continuous and differentialble on the same interval.
From Equation (2), \[ g'(x) = f'(x) = \frac{f(b)-f(a)}{b -a} \] Also, clearly, \(g(a) = g(b) = 0\). Thus, by Rolle's Theorem \[ \exists c \in (a,b) \text{ such that } g'(c) = 0 \] This implies \[ g'(c) = f'(c) = \frac{f(b)-f(a)}{b -a} = 0 \] and so \[ f'(c) = \frac{f(b)-f(a)}{b -a} \]
Consider the line connecting \(a, f(a)\) and \(b, f(b)\). The equation of the line can be written as \[ y - f(a) = \frac{f(b)-f(a)}{b -a} (x - a) \] Let g(x) be the vertical difference between \(x, f(x)\) and (x, y). Then \[ g(x) = f(x) - f(a) - \frac{f(b)-f(a)}{b -a} (x -a) \tag{2} \] \(g\) is continuous on \([a, b]\) and differentiable on \((a, b)\) because \(g\) is the sum of \(f\) and the first-degree polynomial, both of which are continuous and differentialble on the same interval.
From Equation (2), \[ g'(x) = f'(x) = \frac{f(b)-f(a)}{b -a} \] Also, clearly, \(g(a) = g(b) = 0\). Thus, by Rolle's Theorem \[ \exists c \in (a,b) \text{ such that } g'(c) = 0 \] This implies \[ g'(c) = f'(c) = \frac{f(b)-f(a)}{b -a} = 0 \] and so \[ f'(c) = \frac{f(b)-f(a)}{b -a} \]