Rolle's Theorem and Lagrange's Mean Value Theorem

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.

Rolle's Theorem

Rolle's Theorem is a special case of Lagrange's Mean Value Theorem. It is also used to find the mean value of any function in a defined interval. It is applied when the initial value of the function f(a) equals the final value of the function f(b). For any function f(x) that is defined on the closed interval [a, b] Rolle's Theorem is applied if,

• f(x) is continuous in the closed interval [a, b]
• f(x) is differentiable in the open interval (a, b)
• f(b) = f(a)

Then, according to Rolle's Theorem, there exists at least one number c ∈ (a, b) such that 

f'(c) = 0

where, f'(c) is the differentiation of f(x) at point c.

Geometric Interpretation of Rolle’s Theorem

Rolle’s Theorem states that if any function satisfies all the conditions of Rolle's Theorem, i.e.

• f(x) is continuous in the closed interval [a, b]
• f(x) is differentiable in the open interval (a, b)
• f(b) = f(a)

Then there exists f'(c) = 0, which means that the tangent at point c is parallel to the x-axis.

Note: It is not necessary that the converse of this theorem is also true. That is we cannot say if at some point f'(c) = 0 then all the conditions of Rolle's theorem are satisfied. 

Proof of Rolle's Theorem

Statement: 

According to, Rolle’s theorem,

Let a function f is taken such that, f : [a, b] → R be continuous on [a, b], differentiable on (a, b), and f(a) = f(b), for some real number a and b. Then there exists some c in the interval (a, b) such that f′(c) = 0.

Proof: 

Taking the function f(x) which is,

• Continuous in the interval [a, b],
• Differentiable in the interval (a, b),
• And f(a) = f(b)

Now, for any function f(x) f′(x)=0 occurs at extreme values i.e. either at maxima or at minima.

Then, according to the Extreme Value Theorem theorem for any continuous function there exist both the maxima and minima in any finite interval.

Now, there exist two possibilities for our function,

Case 1: Function is constant.

Case 2: Function is not constant.

For, Case 1: Function is constant.

A constant function has a horizontal graph i.e. parallel to the x-axis which satisfies Rolle's Theorem at all the points in the interval since the derivative is zero everywhere.

For, Case 2: Function is not constant.

As, the function is not constant and its initial value and final value are equal, i.e. f(a) = f(b) (given), it must change its direction at least once inside the interval. Thus, it is safe to assume that the function will either have a minimum, a maximum, or both.

Now we need to show that at any of the interior points, the derivative is equal to zero. The derivative of the function is zero either at maxima or minima. As, we have shown that the function will have either a minimum, a maximum, or both. It is safe to say that,

f'(c) = 0 for atleast some c in the interval (a, b).

Thus, Rolle's Theorem is Verified.

Example of Rolle's Theorem

For the function f(x) = 2x² + 1 defined in the interval [-2, 2] verify Rolle's Theorem.

Solution: 

Given,

a = -2 and b = 2

f(x) = 2x² + 1

f'(x) = 4x

f(a) = f(-2) = 2(-2)² + 1 = 9

f(b) = f(2) = 2(2)² + 1 = 9

Now,

f(b) = f(a) thus condition for Rolle's Theorem is verified.

we know that, f'(c) = 0

4c = 0

c = 0

c = 0 ∈ (-2, 2)

Hence, Rolle's Theorem is verified.

Lagrange’s Mean Value Theorem

Lagrange’s Mean Value Theorem is used to find the mean value of any function in a defined interval. For any function f(x) that is defined on the closed interval [a, b] mean value theorem is applied if,

• f(x) is continuous in the closed interval [a, b]
• f(x) is differentiable in the open interval (a, b)

Then, according to Lagrange’s Mean Value Theorem, there exists at least one number c ∈ (a, b) such that 

f'(c) = [f(b) - f(a)] / (b-a)

where, f'(c) is the differentiation of f(x) at point c.

This theorem is also called the "Mean Value Theorem or First Mean Value Theorem"

Geometrical Interpretation of Lagrange’s Mean Value Theorem

Geometrically speaking, the derivative at c denotes the slope of the tangent of at x = c for f(x). It says, there must exist a point in between that interval where the slope of the tangent is equal to the slope of the line joining points x = a and x = b. 

Note: Lagrange's Mean Value Theorem is also called Finite Mean Value Theorem as there can be any number of c-points in the interval (a, b)

Proof of Lagrange's Mean Value Theorem

Statement:

According to Lagrange mean value theorem, "For a function f which is continuous over the closed interval [a,b], and differentiable over the open interval (a,b), there exists at least one point c in the interval (a,b) such that the slope of the tangent at the point c equals the slope of the secant through the endpoints of the curve which can be stated as,

f'(c) = f(b)−f(a ) / (b−a)

where, c ∈ (a, b)

Proof:

For any secant line g(x) to f(x) which passes through the points {a, f(a)} and {b, f(b)}. Now the slope of the secant line is given as,

m = {f(b)−f(a)} / (b−a) 

Also the formula for the secant line is y-y₁ = m (x- x₁). Now, the equation of the secant line is given as,

y - f(a) = {f(b)−f(a)} / (b−a) × (x-a)

y = {f(b)−f(a)} / (b−a) × (x-a) + f(a)

As the equation of secant line is given as g(x) = y, now

g(x) = {f(b)−f(a)} / (b−a) × (x-a) + f(a)....(1)

Now, let's define a function h(x) such that

h(x) = f(x) - g(x)

from eq (1)

h(x) = f(x) - [{f(b)−f(a)} / (b−a) × (x-a) + f(a)}

h(x) =- {f(b)−f(a)} / (b−a) × (x-a)

Now if the function h(x) is continuous on [a,b] and differentiable on (a,b). By applying the Rolles theorem, we get

h'(c) = 0 where, c ∈ (a, b)

h'(x) = f'(x) - [{f(b)−f(a)} / (b−a)]

Now, for some c in (a,b), 

h'(c) = 0

f'(c) - {f(b)−f(a)} / (b−a) = 0

f'(c) = {f(b)−f(a)} / (b−a)

Thus the Lagrange mean value theorem has been proved.

Physical Interpretation of Lagrange's Mean Value Theorem

Lagrange's Mean Value Theorem, tells us the average change in function over the closed interval [a, b], whereas f'(c) can be interpreted as the instantaneous change at ‘c’. Thus, Lagrange's Mean Value Theorem states that the average change in the function f(x) over the interval, [a, b] is equal to the instantaneous change f'(x) at point c for some c belonging to the interval (a, b).

Application of Lagrange's Mean Value Theorem

Mean Value Theorem is known to be one of the most essential theorems in analysis and therefore, all its applications have major significance. Some of the applications are listed below:

• Leibniz's Rule.
• L' Hospital's Rule.
• Strictly Increasing and Strictly decreasing functions.
• Symmetry of Second derivatives.
• Function is constant if f: (a, b) \rightarrow  R is Differentiable and f'(x)=0 for all x∈(a, b).

Example of Lagrange's Mean Value Theorem

For function f(x) = 2x² – 4x + 5 defined in the interval [0, 2], verify Lagrange's Mean Value Theorem

Solution:

Given,

a = 0 and b = 2

f(x) = 2x² – 4x + 5

f'(x) = 4x – 4

f(a) = f(0) = 2(0)² – 4(0) + 5 = 5

f(b) = f(2) = 2(2)² – 4(2) + 5 = 5

Now,

[f(b) – f(a)]/ (b – a) = 5 - 5 / 2 - 0 = 0

we know that, f'(c) = [f(b) – f(a)]/ (b – a)

4c – 4 = 0

4c = 4

c = 4/4

c = 1 ∈ (0, 2)

Hence, Lagrange's Mean Value Theorem is verified.

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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. 

12c^{2} - 16c + 7 = \frac{333 - 12}{5 - 2} = 107 \hspace{0.5cm} \to \hspace{0.5cm} 12c^{2} - 16c - 100 = 0 \\ c = \frac{2 + \sqrt{79}}{3}, \frac{2 - \sqrt{79}}{3} 

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 f(t) = 8t + e^{-3t}         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, 

8 - 3e^{-3c} = \frac{24 + e^{-9} -(-16 + e^{6}}{3-(-2)} = -72.6857 \\ 3e^{-3c} = 80.6857 \\ e^{-3c} = 26.8952 \\ -3c = ln(26.8952) \\ c = -1.0973

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.