Rolle's Theorem

Rolle's Theorem Class 12 is a variant of the mean value theorem that meets specific requirements. Lagrange's mean value theorem is both the mean value theorem and the first mean value theorem at the same time. In general, the mean can be defined as the average of a set of values. The procedure of determining the mean value of two separate functions is different in the case of integrals. Let's look at Rolle's theorem, as well as the mean value and geometrical meaning of such functions.

State and Prove Lagrange’s Mean Value Theorem (LMVT) 

Lagrange’s Mean Value Theorem or First Mean Value theorem states that a function f is defined in the closed interval (a, b), it agrees 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)f(b)–f(a) / (b-a) = f’ (c)

LMVT Theorem Proof

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; 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)

Geometrical Representation of Lagrange’s Mean Value Theorem

The graph given below the curve y = f(x) is in the continuous form, and x-=a and x=b are also differentiable within the closed interval (a, b), then,

According to the Lagrange Mean Value Theorem, 

Any function that is continuous on (a,b) and also differentiable on (a,b) has some in the interval (a,b) such that the secant joining the interval's ends is parallel to the tangent at point C.

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

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Solved Example

1. Verify mean value theorem , if

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

Solution: f (x) = x2-4x-3, the given variable is continuous in the interval 1,41,4 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)

State and Prove Rolle’s Theorem 

Statement of Rolle's Theorem

Rolle's Theorem is a specific example of Lagrange's mean value theorem, which states:

If a function f is defined in the closed interval [a, b] in such a way that it meets the conditions below.

• On the closed interval [a, b], the function f is continuous.

• On the open interval, the function f is differentiable (a, b)

• If f (a) = f (b), then at least one value of x exists; let us assume that this value is c, which is between a and b, i.e. (a c b) in such a way that f'(c) = 0.

If a function is continuous on the closed interval [a, b] and differentiable on the open interval [a, b], there exists a point x = c in (a, b) where f'(c) = 0. (a, b).

Rolle's theorem can be expressed mathematically as:

Let f: [a, b] R be continuous on [a, b] and differentiable on (a, b), with f(a) = f(b). Then some c in (a, b) exists such that f′(c) = 0.

Rolle's Theorem Geometric Explanation

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If the curve y = f(x) is continuous between x = a and x = b, and it is possible to draw a tangent at any point within the interval, and the ordinates corresponding to the abscissa and are identical, then there is at least one tangent to the curve that is parallel to the x-axis.

This theorem states that if f (x) represents a polynomial function in x and the two roots of the equation f(x) = 0 are x = a and x = b, then at least one root of the equation f'(x) = 0 is between these two values.

The inverse of Rolle's theorem is not true; it is possible that more than one value of x exists for which the theorem is true, but there is a reasonable possibility that only one such value occurs.

Examples of Rolle's Theorem

Ans. The function y = x2 + 2 is continuous in [– 2, 2] and differentiable in (– 2, 2), according to Rolle's theorem.

From the given question,

f(x) = x(2) + 2

f(-2) = (-2)2 + 2 = 4 + 2 = 6

f(2) = (2)2 + 2 = 4 + 2= 6

Thus, f(– 2) = f( 2) 

Hence, the value of f(x) at –2 and 2 coincide.

Now, f'(x) = 2x 

Rolle’s theorem states that there is a point c ∈ (– 2, 2) such that f′(c) = 0. 

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

2. Discuss the conditions for the application of Rolle’s Theorem for the following function: F(x) = x2/3 on −1,1,−1,1

Ans. 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,−1,1  

Facts About Rolle’s Theorem

• Rolle’s Theorem was initially proven in 1691.

• 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 Calculus was first introduced by Newton and Leibnitz.

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

Quiz Time

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

a. At least 1

b. Rolle’s theorem is not applicable

c. 2

d. 0