Rolle’s Theorem and Lagrange’s Mean Value Theorem Solved Examples

Example 1: Find the value of “c” which satisfy the conclusion of the Mean Value Theorem for h(z) = 4z³ – 8z² + 7z – 2 on [2,5]. 

Solution:

Since the function is a polynomial, so it is continuous and differentiable both in this interval. Thus mean value theorem can be applied here. 

Given,

h(z) = 4z³ – 8z² + 7z – 2

h(2) = 12

h(5) = 333

h'(z) = 12z² -16z + 7

Now plug this into the formula of mean value theorem and solve for c. 

 

Example 2: Find the value of “c” which satisfy the conclusion of the Mean Value Theorem in the interval [-2,2] for f(x) = |x – 2| + |x| 

Solution:

Given function, f(x) = |x – 2| + |x| is not differentiable. Thus mean value theorem cannot be applied on this function in the interval [-2, 2]. 

Example 3: Find the value of “c” which satisfy the conclusion of Mean Value Theorem in the interval for  on [-2,3].

Solution:

This function is a sum of both exponential and polynomial functions. Since both the functions are continuous and differentiable everywhere, the function f(t) is continuous and differentiable everywhere. Thus mean value theorem is applicable. 

f(-2) = -16 + e^{(-3 .-2)} = -16 + e⁶
f(3) = 24 + e^-9
f'(t) = 8 -3e^-3t

Let’s plug it into the formula, 

The value of c = -1.0973 ∈ (-2, 3) thus, Lagrange’s Mean value theorem is satisfied.

Example 4: Find the value of ‘c’ which satisfy Rolle’s theorem for f(x) = x² – 2x – 8 on [-1, 3].

Solution:

Before finding out the points, we need to make sure that all the conditions of rolle’s theorem are applied on this interval. 

This function is a polynomial, so it’s both differentiable and continuous on the interval. 

Now let’s evaluate the function at the ends of the interval. f(-1) = -5 and f(3) = -5. 

Function values are equal at both the ends. So, now all the conditions for rolle’s theorem are satisfied.

f'(x) = 2x – 2

f'(c) = 2c – 2 = 0

c = 1

The value of c = 1 ∈ (-1, 3) thus, Rolle’s theorem is satisfied.

Example 5: Examine if Rolle’s theorem is applicable for the functions f(x) = [x] for x ∈ [5, 9].

Solution:

This function is a step function, and is not continuous. So, Rolle’s Theorem is not applicable in the following interval. 

Rolle’s Theorem and Lagrange’s Mean Value Theorem: Mean Value Theorems (MVT) are the basic theorems used in mathematics. They are used to solve various types of problems in Mathematics. Mean Value Theorem is also called Lagrenges’s Mean Value Theorem. Rolle’s Theorem is a subcase of the mean value theorem and they are both widely used. These theorems are used to find the mean values of different functions.

Rolle’s theorem, a special case of the mean-value theorem in differential calculus, asserts that under certain conditions, if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), and if f(a) equals f(b), then there exists at least one point x in the interval (a, b) where the derivative of f, denoted as f^‘(x), equals zero.