ROLLE'S THEOREM

If f(x) be a real valued function that satisfies the following three conditions.

1) f(x) is defined and continuous on [a, b] 

2) f(x) is differentiable on (a, b)

3) f (a) = f (b)

then there exists at least one point c ∊ (a, b).

Such that f'(c)  =  0.

Example 1 :

Using Rolle's theorem find the values of c.

f(x)  =  x²-4x+3 ,  1 ≤ x ≤ 3

Solution :

(i) f (x) is continuous on [1, 3].

(ii) f(x) is differentiable (1, 3).

(iii) f(1)  =  f(3).

All conditions are satisfied.

f'(c)  =  2x-4

To find that particular point we need to set f'(c)  =  0.

2c-4  =  0

c  =  2

So, c is 2.

Example 2 :

Using Rolle's theorem find the values of c.

f(x)  =  sin² x ,   0 ≤ x ≤ Π

Solution :

(i) f (x) is continuous on [0 ,Π].

(ii) f(x) is differentiable (0,Π).

f(0)  =  sin²(0)  ==> 0

f(Π)  =  sin² (Π)  ==>  0

f(0)  =   f(Π). All conditions are satisfied.

f'(x)  =  2sinx cos x

=  sin 2x 

f'(c)  =  sin 2c 

To find that particular point we need to set f'(c)  =  0

sin 2c  =  0

2c  =  sin^-1 (0)

2c  =  0, Π, 2 Π, 3 Π , .............

c  =  Π/2∊  0 ≤ x ≤ Π

Example 3 :

Using Rolle's theorem find the points on the curve

y  =  x²+1,   - 2 ≤ x ≤ 2

where the tangent is parallel to x - axis.

Solution :

If f(x) be a real valued function that satisfies the following three conditions.

1)  f(x) is defined and continuous on [-2, 2] 

2)  f(x) is differentiable on interval (-2,2).

y  =  x²+1 

f(-2)  =  (-2)² + 1

f(-2)  = 5

f(2)  =  2²+1

f(2)  =  5

f(-2)  =  f(2)

from this we come to know that the given function satisfies all the conditions of Rolle's theorem.

c ∈ (-2, 2)

we can find the value of c by using the condition

f'(c)  =  0

f(x)  =  x² + 1

f'(x)  =  2x

f'(c)  =  2c

f'(c)  =  0

2c  =  0  

 c  =  0

f (0) = 0² + 1  ==>  1

Therefore the required point on the curve is (0, 1).

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