Lagrange Mean Value Theorem and Rolle’s Theorem

Rolle’s Theorem is a special case of the mean value theorem that is true if and only if specific conditions are met. At the same time, Lagrange’s mean value theorem is the mean value theorem itself, or the first mean value theorem, as the term is used in the literature. In general, mean can be thought of as the average of the values that have been provided. It is a different method in the case of integrals, however, when determining the mean value of two separate functions. 

Lagrange’s Mean Value Theorem: 

It is possible to define a function f on the closed interval [a,b] that meets all of the following requirements:

1. The function f is continuous on the closed interval [a, b] and on the closed interval [a, b]. 

2. In addition, the function f differencing on the open interval is differentiable (a, b) 

3. Then there is a value x = c that is defined in such a way that 

f’(c) = [f(b) – f(a)]/(b-a)

This theorem is also referred to as the first mean value theorem or Lagrange’s mean value theorem, depending on who is talking about it. 

Geometrical Interpretation of Lagrange’s Mean Value Theorem: 

The curve y = f(x) in the given graph is continuous between the points x = a and x = b and differentiable within the closed interval [a,b].According to Lagrange’s mean value theorem, for any function that is continuous on [a, b] and differentiable on (a, b), there exists some c in the interval (a, b) such that the secant linking the endpoints of the interval [a,b] exists.

f’(c) = [f(b) – f(a)]/(b-a)

With the help of the following example, you will be able to comprehend this concept better. 

Example: 

Check the Mean Value Theorem for the function f(x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4. 

Solution: 

Given,

f(x) = x² – 4x – 3

f’(x) = 2x – 4

A = 1 and b = 4 (given)

f(a) = f(1) = (1)² – 4(1) – 3 = 1 – 4 – 3 = -6

f(b) = f(4) = (4)² – 4(4) – 3 = -3

Now,

[f(b) – f(a)]/ (b – a) = (-3 + 6)/(4 – 1) = 3/3 = 1 

Following the mean value theorem, there exists a point at c ∈ (1,4) such that f’(c) = [f(b) – f(a)]/(b-a), which is to say that f’(c) = 1. 

2c – 4 = 1

2c = 5

c = 5/2 ∈ (1, 4)

Verification: f’(c) = 2(5/2) – 4 = 5 – 4 = 1 

As a result, the mean value theorem was verified. 

Rolle’s Theorem: 

Rolle’s Theorem, which is a specific case of Lagrange’s mean value theorem, states that the following:

If a function f is defined in the closed interval [a, b] in such a way that it satisfies the following conditions, then the function f is said to satisfy the conditions.

1. The function f is continuous on the closed interval [a, b] and on the closed interval [a, b].

2. In addition, the function f differencing on the open interval is differentiable (a, b)

3. After all of this, let’s assume that there is at least one value of x that is between the two values of a and b, i.e. (a<c<b), and that this value is c; this value is assumed to be zero.

To put it another way, if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point x = c in the closed interval [a, b) where f’(c) = 0. 

Geometric interpretation of Rolle’s Theorem: 

y = f(x) is continuous between x = a and x = b in the above graph, and at every point inside the interval, it is possible to draw a tangent to the curve, and ordinates that correspond to the abscissa and are equal, then there exists at least one tangent to the curve that is parallel to the x-axis.

When we look at this theorem from an algebraic perspective, it tells us that, given a polynomial function f (x) as its representation in x and two roots of the equation f(x)=0 as the values a and b, there exists at least one root of the equation f’(x) = 0 that lies between these values.

However, the converse of Rolle’s theorem is not true. Furthermore, there is a possibility that there is more than one value of x for which the theorem holds true; however, there is an extremely high probability of the existence of only one such value. 

Rolle’s Theorem statement: 

Rolle’s theorem can be expressed mathematically as follows:

Suppose f: [a, b] R is a continuous function on [a, b] and differentiable on (a, b), such that f(a) = f(b), where the real numbers a and b are used as examples. In that case, there exists at least one c in (a, b) such that f′(c) = 0. 

Example: 

Check the validity of Rolle’s theorem for the function y = x² + 2 for which a = –2 and b = 2. 

Solution: 

According to the statement of Rolle’s theorem, the function y = x² + 2 is continuous in the interval [–2, 2] and differentiable in the interval (–2, 2).

Taking into consideration the given, 

f(x) = x² + 2

f(-2) = (-2)² + 2 = 4 + 2 = 6

f(2) = (2)² + 2 = 4 + 2= 6

Thus, f(– 2) = f( 2) = 6 

As a result, the values of f(x) at –2 and 2 are the same. 

Now, f’(x) = 2x 

In general, Rolle’s theorem states that there is at least one point c ∈ (– 2, 2) at which f′(c) = 0.

At c = 0, f′(c) = 2(0) = 0, where c = 0 ∈ (– 2, 2).

As a result, it was confirmed.  

Conclusion: 

Rolle’s theorem contains three hypotheses (or a three-part hypothesis), whereas the Mean Values Theorem contains only two hypotheses. The outcomes appear to be different. Rolle’s Theorem states that if the third hypothesis is true (f(a)=f(b)), then both theorems state that there is a c in the open interval (a,b) where f’(c)=0.