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Mathematics Question on Continuity and differentiability

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?
(i)f(x)=[x] for x∈[5,9](i)f(x)=[x]\,for\, x∈[5,9]
(ii)f(x)=[x] for x∈[−2,2](ii)f(x)=[x]\,for\, x∈[-2,2]
(iii)f(x)=[x2] for x∈[1,2](iii)f(x)=[x^2]\,for\, x∈[1,2]

Answer

By Rolle’s Theorem, for a function f:[a,b]→Rf:[a,b]→R,if
(a) ff is continuous on [a,b][a,b]
(b) ff is differentiable on (a,b)(a,b)
(c) f(a)=f(b)f(a)=f(b)
then, there exists some c∈(a,b)c∈(a, b) such that f′(c)=0f'(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](i)f(x)=[x]\,for\, x∈[5,9]
It is evident that the given function f(x)f (x) is not continuous at every integral point.
In particular, f(x)f(x) is not continuous at x=5x=5 and x=9x=9
⇒f(x)⇒ f(x) is not continuous in [5,9][5,9]
Also,f(5)=[5]=5f(5)=[5]=5 and f(9)=[9]=9f(9)=[9]=9
∴f(5)≠f(9)∴f(5)≠f(9)
The differentiability of ff in (5,9)(5,9) is checked as follows.
Let nn be an integer such that n∈(5,9).n∈(5,9).
The left hand limit of ff at x=nx=n is
limh→0f(n+h)−f(n)h=limh→0[n+h]−[n]h\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}
=limh→0n−1−nh=limh→0−1h=∞=\underset{h→0}{lim}\frac{n-1-n}{h}=\underset{h→0}{lim}\frac{-1}{h}=∞
The right hand limit of ff at x=nx=n is
limh→0f(n+h)−f(n)h=limh→0[n+h]−[n]h\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}
=limh→0n−nh=limh→0 0=0=\underset{h→0}{lim}\frac{n-n}{h}=\underset{h→0}{lim}\,0=0
Since the left and right hand limits of f at x=nx = n are not equal, ff is not differentiable at x=nx=n
∴f∴f is not differentiable in (5,9)(5,9). It is observed that ff does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for f(x)=[x] for x∈[5,9]f(x)=[x]\, for\, x∈[5,9]

(ii)f(x)=[x] for x∈[−2,2](ii)f(x)=[x]\, for\, x∈[-2,2]
It is evident that the given function f(x)f (x) is not continuous at every integral point.
In particular, f(x)f(x) is not continuous at x=−2x=−2 and x=2x=2
⇒f(x)⇒ f(x) is not continuous in [−2,2][−2,2]
Also,f(−2)=[−2]=−2f(-2)=[-2]=-2 and f(2)=[2]=2f(2)=[2]=2
∴f(-2)≠f(2)
The differentiability of f in (−2,2)(−2,2) is checked as follows.
Let nn be an integer such that n∈(−2,2).n ∈ (−2,2).
The left hand limit of ff at x=nx=n is
limh→0f(n+h)−f(n)h=limh→0[n+h]−[n]h\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}
=limh→0n−1−nh=limh→0−1h=∞=\underset{h→0}{lim}\frac{n-1-n}{h}=\underset{h→0}{lim}\frac{-1}{h}=∞
The right hand limit of ff at x=nx=n is
limh→0f(n+h)−f(n)h=limh→0[n+h]−[n]h\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}
=limh→0n−nh=limh→0 0=0=\underset{h→0}{lim}\frac{n-n}{h}=\underset{h→0}{lim}\,0=0
Since the left and right hand limits of ff at x=nx=n are not equal, ff is not differentiable at x=nx=n
∴f∴f is not differentiable in (−2,2)(−2,2). It is observed that ff does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for f(x)=[x] for x∈[−2,2]f(x)=[x]\, for\, x∈[-2,2]

(iii)f(x)=x2−1 for x∈[1,2](iii) f(x)=x^2-1 \,for\, x∈[1,2]
It is evident that ff, being a polynomial function, is continuous in [1,2][1, 2] and is differentiable in (1,2).(1, 2).
f(1)=(1)2−1=0f(1)=(1)2-1=0
f(2)=(2)2−1=3f(2)=(2)2-1=3
∴f(1)≠f(2)∴f(1)≠f(2)
It is observed that ff does not satisfy a condition of the hypothesis of Rolle’s Theorem
Hence, Rolle’s Theorem is not applicable for f(x)=x2−1 for x∈[1,2]f(x)=x^2-1\, for\, x∈[1,2]