Rolle's Theorem
If \(f(x) \) is
a) continuous in [a,b]
b) differentiable in (a,b)
c) \(f(a) = f(b) \)
then there exists at least one value \(c \in (a,b) \) such that \( f'(c) = 0 \)
Sample Problem
Verify Rolle's Theorem for \[ f(x) = x(x - 2)e^{\large\frac{3x}{4}} \text{ in } (0,2) \]
Solution
The function is continuous and differentiable in (0,2)
Here \(a = 0 \text{ and } b = 2 \)
\[ f(a) = f(0) = 0 \]
\[ f(b) = f(2) = 2(2 - 2)e^{\large\frac{6}{4}} = 0 \]
\[ \text{As } f(a) = f(b) \]
\(\Rightarrow \) Rolle's Theorem is applicable. Now,
\[ f'(x) = (x^2 - 2x)e^{\large\frac{3x}{4}}\times \frac{3}{4} + e^{\frac{3x}{4}}(2x - 2) \]
According to Rolle's Theorem,
\[ f'(c) = 0 \]
\[ \Rightarrow (x^2 - 2x)e^{\large\frac{3x}{4}}\times \frac{3}{4} + e^{\frac{3x}{4}}(2x - 2) \]
\[ e^{\large\frac{3x}{4}} \neq 0 \]
\[ \frac{3}{4}(x^2 - 2x) + 2x - 2 = 0 \]
\[ 3x^2 - 2x + 8x - 8 = 0 \]
\[ 3x^2 + 6x - 8 = 0 \]
\[ \Rightarrow x = -2, x = \frac{8}{6} \]
\[ \text{Or } c = -2 , c = \frac{8}{6} \]
\[\text{Since } c = -2 \notin (0,2) \]
\[ \Rightarrow \boxed{c = \frac{8}{6}} \]
