Lagrange's Mean Value Theorem
If \(f(x) \) is
a) continuous in [a,b]
b) differentiable in (a,b)
then there exists at least ne value \(c \in (a,b) \) such that
\[ \frac{f(b) - f(a)}{b - a} = f'(c) \]
Sample Problem
Verify Lagrange's Mean Value Theorem for \[ f(x) = x^2 \text{ in } (1,5) \]
Solution
\(f(x) \) is continuous and differentiable in (1,5)
Here \(a = 1\) and \(b = 5\)
\[ \text{Now } f'(x) = 2x \]
So by Lagrange's Mean Value Theorem,
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
\[ 2c = \frac{5^2 - 1^2}{5 - 1} = \frac{25 - 1}{4} \]
\[ \Rightarrow \boxed{c = 3} \]
