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# RolleŌĆÖs Theorem

Let f ( x) be continuous on a closed interval [a , b] and differentiable on the open interval (a , b), If f (a) = f (b) , then there is at least one point c Ōłł (a, b) where fŌĆÖ(c) =0.

RolleŌĆÖs Theorem

## Theorem 7.2 (RolleŌĆÖs Theorem)

Let f ( x) be continuous on a closed interval [a , b] and differentiable on the open interval (a , b)

If f (a) = f (b) , then there is at least one point c Ōłł (a, b) where fŌĆÖ(c) =0. Geometrically this means that if the tangent is moving along the curve starting at x = a towards as in FigŌĆē7.2 x = b then there exists a c Ōłł ( a, b) at which the tangent is parallel to the x -axis.

Example 7.19

Compute the value of 'c ' satisfied by the RolleŌĆÖs theorem for the function

f(x) = x 2 (1 ŌłÆ x)2 , x Ōłł[0,1] .

Solution

Observe that, f(0) = 0 = f (1), f (x) is continuous in the interval [0,1] and is differentiable in (0,1) . Now,

f ŌĆ▓( x) = 2x (1ŌłÆ x)(1ŌłÆ 2x) .

Therefore, f ŌĆ▓(c) = 0 gives c = 0,1 , and 1/2

which ŌćÆ c = 1/2 Ōłł(0,1) .

Example 7.20

Find the value in the interval (1/2, 2) satisfied by the Rolle's theorem for the function f(x) = x + 1/x , x Ōłł [1/2 , 2] .

Solution

We have, f (x) is continuous in [1/2 , 2] and differentiable in (1/2 , 2) with f(1/2) = 5/2 = f (2) . By the RolleŌĆÖs theorem there must exist a value c Ōłł (1/2 , 2) such that,

f (c) = 1 ŌĆō 1/c2 = 0 ŌćÆ c2 = 1 gives c = ┬▒1. As 1Ōłł (1/2 , 2) , we choose c = 1.

Example 7.21

Compute the value of 'c ' satisfied by RolleŌĆÖs theorem for the function in the interval [2, 3].

Solution

Observe that, f(2) = 0 = f (3) and f ( x) is continuous in the interval [2, 3] and differentiable in (2, 3) . Now, Observe that ŌłÆ ŌłÜ6 Ōłē(2, 3) and hence c = + ŌłÜ6 satisfies the RolleŌĆÖs theorem.

RolleŌĆÖs theorem can also be used to compute the number of roots of an algebraic equation in an interval without actually solving the equation.

### Example 7.22

Without actually solving show that the equation x4 + 2x3 ŌłÆ 2 = 0 has only one real root in the interval (0,1) .

Solution

Let f(x)= x4 + 2x3 ŌłÆ 2. Then f (x) is continuous in [0,1] and differentiable in (0,1) . Now,

f ŌĆ▓( x) = 4x3 + 6x2 . If f ŌĆ▓(x) = 0 then,

2x2 (2x + 3) = 0 .

Therefore, x = 0,ŌłÆ 3/2 but 0, ŌłÆ 3/2 Ōłē(0,1) .

Thus, fŌĆÖ(x) > 0, ŌłĆx Ōłł(0,1) .

Hence by the RolleŌĆÖs theorem there do not exist a , b Ōłł(0,1) such that, f (a) = 0 = f (b) . Therefore the equation f (x) = 0 cannot have two roots in the interval (0,1) . But, f (0) = ŌłÆ2 < 0 and f (1) = 1 > 0 tells us the curve y = f ( x) crosses the x -axis between 0 and 1 only once by the Intermediate value theorem. Therefore the equation x4 + 2x3 ŌłÆ 2 = 0 has only one real root in the interval (0,1) .

As an application of the RolleŌĆÖs theorem we have the following,

### Example 7.23

Prove using the RolleŌĆÖs theorem that between any two distinct real zeros of the polynomial

a0xn + an-1xnŌłÆ1 + ŌĆ”.. + a1x + a0

there is a zero of the polynomial

nanxn ŌłÆ1 + ( n ŌłÆ1)anŌłÆ1xnŌłÆ2 + ŌĆ” + a1.

Solution

Let P ( x) = a0xn + an-1xnŌłÆ1 + ŌĆ”.. + a1x + a0. Let ╬▒ < ╬▓ be two real zeros of P ( x) . Therefore, P (╬▒ ) = P(╬▓ ) = 0 . Since P ( x) is continuous in [╬▒ , ╬▓ ] and differentiable in (╬▒ , ╬▓ ) by an application of RolleŌĆÖs theorem there exists ╬│ Ōłł (╬▒, ╬▓ ) such that P (╬│ ) = 0. Since,

PŌĆ▓(x) = nanxn ŌłÆ1 + ( n ŌłÆ1)anŌłÆ1xnŌłÆ2 + ŌĆ” + a1.

which completes the proof.

### Example 7.24

Prove that there is a zero of the polynomial, 2x3 ŌłÆ 9x2 ŌłÆ11x +12 in the interval (2, 7) given that and 7 are the zeros of the polynomial x4 ŌłÆ 6x3 ŌłÆ11x2 + 24x + 28 .

Solution

Applying the above example 7.23 with

P(x) = x 4 ŌłÆ 6x3 ŌłÆ 11x2 + 24x + 28, ╬▒ = 2, ╬▓ = 7

and observing

PŌĆÖ( x) / 2 = 2x3 ŌłÆ 9x2 ŌłÆ 11x +12 = Q(x) , (say).

This implies that there is a zero of the polynomial Q ( x) in the interval (2, 7) .

For verification,

Q(2) = 16ŌłÆ36ŌłÆ22+12 = 28ŌłÆ58 =ŌłÆ30<0

Q(7) = 686 ŌłÆ 441ŌłÆ 77 +12 = 698 ŌłÆ 518 =180 > 0

From this we may see that there is a zero of the polynomial Q ( x) in the interval (2, 7) .

## Remark

There are functions for which RolleŌĆÖs theorem may not be applicable.

(1) For the function f ( x) = | x |, x Ōłł[ŌłÆ1,1] RolleŌĆÖs theorem is not applicable, even though f (ŌłÆ1) = 1 = f (1) because f ( x) is not differentiable at x = 0 .

(2) For the function, even though f (0) = f (1) = 1 , Rolle's theorem is not applicable because the function f ( x) is not continuous at x = 0 .

(3) For the function f (x) = sin x , x Ōłł [0, ŽĆ/2] RolleŌĆÖs theorem is not applicable, even though f (x) is continuous in the closed interval [0, ŽĆ/2] and differentiable in the open interval (0, ŽĆ/2) because, 0 = f (0) ŌēĀ f(ŽĆ/2) = 1.

If f ( x) is continuous in the closed interval [a , b] and differentiable in the open interval (a , b) and even if f ( a) ŌēĀ f (b) then the RolleŌĆÖs theorem can be generalised as follows.

## Theorem 7.1 (Intermediate value theorem)

If f is continuous on a closed interval [a , b] , and c is any number between f (a) and f (b) inclusive, then there is at least one number x in the closed interval [a , b] , such that f ( x) = c .

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12th Maths : UNIT 7 : Applications of Differential Calculus : RolleŌĆÖs Theorem | Mean Value Theorem | Mathematics