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).
f(1) = 12-4(1)+3 = 1-4+3 = 4-4 = 0 |
f(3) = 32-4(3)+3 = 9-12+3 = 12-12 = 0 |
(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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