Lagrange’s Mean Value Theorem or First Mean Value theorem states that,

If a function f is defined in the closed interval (a, b), agree with the following conditions

The function f is always continuous in the closed interval (a, b)

The function is always differentiable in the open interval (a, b)

And there includes a value x = c in such a manner that,

[f (b) – f (a)] / (b-a) = f’ (c)

Consider the following related function,

F (x) =f (x) +λ x

We selected a number λ such that the condition F(a) =F(b) agreed. Then,

F (a) +λ a =f (b) +λb,

= f(b) -f(a) = λ (a-b)

λ = - f(b) -f(a) /b-a

Therefore , we have

F(x) = f (x) - f(b) -f(a)/ b-a (x)

The function f is continuous in the closed interval (a, b), differentiable in the open interval (a, b) and holds equal values at the endpoints of the interval. Hence, it satisfies all the conditions of Rolle’s theorem, Then there includes a point c in the interval (a,b) such that

F’(c) = 0

It follows that,

f’c - f (b)-f (a) /b-a = 0

Or,

f(b) -f(a) =f’(c) (b-a)

Rolle’s Theorem Class 12 is one of the fundamental theorems in differential calculus. It is an exceptional case of mean value theorem which in turn is an important element in the proof of the fundamental theorem of calculus.

An exception case of Lagrange’s Mean Value Theorem is Rolle’s Theorem which states that,

A function f is defined in the closed interval (a, b), in such a manner that it agrees with the following conditions.

The function f is continuous in closed interval (a, b)

The function is always differentiable in the open interval (a, b)

Now if f (a) =f (b), then there includes at least one single value of x, let us consider this value as c which comes in between a and b i.e. (a < c <b) in such a manner that f ‘(c) = 0.

In short, if a function f is continuous on the closed interval and differentiable on the open interval then there includes a point x = c in (a, b) such that f ‘(c) =0

1. Discuss the conditions for the application of Rolle’s Theorem for the following function.

F(x) = x2/3 on [-1, 1]

f ’(x) = 2/3x1/3

f ’ (0) = 2 /3(0)1/3

f ’(0) = ∞

So f ‘(x) does not takes place at x=o € (-1, 1)

f ’(x) is not differentiable in X € (-1, 1)

So Rolle’s Theorem is not applicable on F(x) in [-1, 1]

2. Verify mean value theorem , if

f (x) = x2-4x -3 in the interval [a, b] where a =1 and b= 4

Solution: f (x) = x2-4x-3, the given variable is continuous in the interval [1,4] and derivable (1,4).

The given polynomial equation is holding all the conditions of Rolle’s Theorem.

The f ’(x) = 2x-4

f ’(x) 2c-4

f (4) = 16-16-3 = -3

f (1) = 1-4-3 =6

There included a value of C such that,

f ’(x) = f (b) –f(a)/ b-a = 2c-4 = -3-(6)/4-1 =3/3=1

Therefore, 2C = 4 +1 or C= 5/2€ (1, 4)

Rolle’s Theorem was initially proven in 1691.

The Rolle’s Theorem was proved just after the first paper including calculus was introduced.

Michel Rolle was the first famous Mathematician who was alive when the Calculus was first introduced by Newton and Leibnitz.

Initially, Michel Rolle was critical of calculus, but later he decided to prove this significant theorem.

1. In which of the following intervals is f(x) = -x satisfy Rolle’s Theorem?

a. (0,2)

b. (3,4)

c. (-3,-1)

d. None of these

2. Based on a Rolle's theorem for a continuous function f (x), if the starting point f(a) and the end-point f(b)equals to 0 then,

a. It can be somewhere between f (a) and f(b) and the instantaneous rate of change should be 0.

b. It can be somewhere between f (a) and f (b) and the function must be equal to 0.

c. The function is flat

d. Rolle’s Theorem cannot be applied.

3. Consider the function f (x) given below. Based on Rolle’s Theorem between x= o and x= pi, for how many values of x is the instantaneous rate of change equals to 0.

a. At least 1

b. Rolle’s theorem is not applicable

c. 2

d. 0

FAQ (Frequently Asked Questions)

1. Discuss the Geometrical Representation of Lagrange’s Mean Value Theorem.

Here, we will discuss the geometrical representation of Lagrange’s Mean Value Theorem:

In the graph given below the curve, y = f(x) is in the continuous form x-=a and x=b and also the differentiable within the closed interval (a, b), then based on Lagrange Mean Value Theorem, for any function that is continuous on (a,b) and also differentiable on (a,b) there includes some in the interval (a,b) such that the secant joining the endpoints of the interval ]a,b] is parallel to the tangent at point C.

f ’(c) is equivalent to f (b) –f (a)/b-a

Geometrical Representation of Lagrange’s Mean Value Theorem.

(image will be uploaded soon)

2. Discuss the Geometrical Representation of Rolle’s Theorem.

Here, we will discuss the geometrical representation of Rolle’s Theorem:

In the graph given above, the curve y =f(x) is continuous between x=a and x=b and at each point within the interval, it is possible to draw a tangent and ordinates corresponding to the abscissa and are equal then their includes at least one tangent to the curve which is parallel to the x-axis.

Algebraically, this theorem denotes that if (x) is characterizing a polynomial function in x and the two roots of the equation f(x) =0 are x= a and x =b, then there includes at least one equation f ’(x) = 0 lying between the values.

The opposite of Rolle's Theorem is not always true and it is also possible that there includes more than one value for x, for which the theorem holds but there should be the definite probability of the existence of one such value.

Geometrical Representation of Rolle’s Theorem. (image will be uploaded soon)