WHY ROLLES THEOREM IS NOT APPLICABLE FOR THE GIVEN FUNCTION

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 ∈ (a,b) where f '(c) = 0.

Geometrically this means that if the tangent is moving along the curve starting at x = a towards x = b then there exists a c ∈ (a, b) at which the tangent is parallel to the x -axis.

Problem 1 :

Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.

(i)  f(x)  =  |(1/x)|, x ∊ [-1, 1]

Solution :

The function is not defined at the interval [-1, 1], because when we apply x  =  0, we get

f(0)  =  |1/0|

f(0)  =  

So, rolle's theorem is not applicable for the given function.

(ii) f (x) = tan x, x ∊ [0, π]

Solution :

The function is not defined at the interval [0, π], because when we apply x  =  π/2, we get

f(π/2)  =  tan (π/2)

f(π/2)  =  ∞

So, rolle's theorem is not applicable for the given function.

(iii)  f(x)  =  x - 2 log x, ∊ [2, 7]

Solution :

f(x)  =  x - 2 log x

f(x) is defined and continuous on the interval [2, 7] and differentiable on (2, 7).

f(x)  =  x - 2 log x

f(2)  =  2 - 2 log 2

f(2)  =  2 - log 22

f(2)  =  2 - log 4 ----(1)

f(7)  =  7 - 2 log 7

f(7)  =  2 - log 72

f(7)  =  2 - log 49 ----(2)

f(2)  ≠ f(7)

(1)  ≠ (2)

So, rolle's theorem is not applicable for the given function.

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