#### Question

Examine if Rolle's theorem is applicable to any one of the following functions.

(i) f (x) = [x] for x ∈ [5, 9]

(ii) f (x) = [x] for x ∈ [−2, 2]

Can you say something about the converse of Rolle's Theorem from these functions?

#### Solution

By Rolle’s theorem, for a function f : [a , b] →R , if

(a) f is continuous on [a, b],

(b) f is differentiable on (a, b) and

(c) f (a) = f (b),

then there exists some *c* ∈ (*a*, *b*) such that f'(c) = 0 .

Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis

(i) f(x) = [x] for x ∈ [ 5 , 9 ]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9.

Thus, f (x) is not continuous on [5, 9].

Also , f (5) = [5] = 5 and f (9) = [9] = 9

∴ f (5) ≠ f (9)

The differentiability of f on (5, 9) is checked in the following way.

Let n be an integer such that n ∈ (5, 9).

The left hand limit of f at x = n is,

`lim_(h ->o)(f (n + h) - f (n)\)/h = lim_(h->o) ([n + h ] - [n])/h) = lim_(h->o)(n- 1- n)/h = lim_(h ->o)(-1)/h =oo`

The right hand limit of f at x = n is,

`lim_(h->o) (f (n +h )- f (n))/h = lim _(h->o)([n +h] - [n])/h = lim_(h->o)(n-n)/h = lim _(h->o) 0 = 0 `

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, f is not differentiable on (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on f (x) for x ∈ [5 , 9] .

(ii) f (x) = [x] for x ∈ [-2 , 2]

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2.

Thus, f (x) is not continuous on [−2, 2].

Also, f (-2) = [-2] = -2 and f (2) = [2] = 2

∴ f (-2) ≠ f (2)

The differentiability of f on (−2, 2) is checked in the following way.

Let n be an integer such that n ∈ (−2, 2).

The left hand limit of f at x = n is .

`lim_(h->o) (f (n + h) -f (n))/ h = lim_(h->o) ([n +h]- [n])/h = lim_(h->o)(n - 1- n)/h = lim_(h->o) (-1)/ h =oo`

The right hand limit of f at x = n is ,

`lim_(h->o) (f(n+h) -f (n))/h = lim_(h->o)([n+h] - [n])/h) lim_(h->o)(n-n)/h = lim_(h->o)(-1)/h = oo`

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, f is not differentiable on (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on f(x)= [x] for x ∈ [ -2 , 2 ].