Question

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?
\((i)f(x)=[x]\,for\, x∈[5,9]\)
\((ii)f(x)=[x]\,for\, x∈[-2,2]\)
\((iii)f(x)=[x^2]\,for\, x∈[1,2]\)

Answer 1

By Rolle’s Theorem, for a function \(f:[a,b]→R\),if 
(a) \(f\) is continuous on \([a,b]\)
(b) \(f\) is differentiable on \((a,b)\)
(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. 

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\((i)f(x)=[x]\,for\, x∈[5,9]\)
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 \(x=9\)
\(⇒ f(x)\) is not continuous in \([5,9]\)
Also,\(f(5)=[5]=5\) and \(f(9)=[9]=9\)
\(∴f(5)≠f(9)\)
The differentiability of \(f\) in \((5,9)\) is checked as follows.
Let \(n\) be an integer such that \(n∈(5,9).\)
The left hand limit of \(f\) at \(x=n\) is 
\(\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}\)
\(=\underset{h→0}{lim}\frac{n-1-n}{h}=\underset{h→0}{lim}\frac{-1}{h}=∞\)
The right hand limit of \(f\) at \(x=n\) is 
\(\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}\)
\(=\underset{h→0}{lim}\frac{n-n}{h}=\underset{h→0}{lim}\,0=0\)
Since the left and right hand limits of f at \(x = n\) are not equal, \(f\) is not differentiable at \(x=n\)
\(∴f\) is not differentiable in \((5,9)\). 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 for \(f(x)=[x]\, for\, x∈[5,9]\)

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\((ii)f(x)=[x]\, for\, x∈[-2,2]\)
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 \(x=2\)
\(⇒ f(x)\) is not continuous in \([−2,2]\)
Also,\(f(-2)=[-2]=-2\) and \(f(2)=[2]=2\)
∴f(-2)≠f(2)
The differentiability of f in \((−2,2)\) is checked as follows.
Let \(n\) be an integer such that \(n ∈ (−2,2).\)
The left hand limit of \(f\) at \(x=n\) is 
\(\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}\)
\(=\underset{h→0}{lim}\frac{n-1-n}{h}=\underset{h→0}{lim}\frac{-1}{h}=∞\)
The right hand limit of \(f\) at \(x=n\) is 
\(\underset{h→0}{lim}\frac{f(n+h)-f(n)}{h}=\underset{h→0}{lim}\frac{[n+h]-[n]}{h}\)
\(=\underset{h→0}{lim}\frac{n-n}{h}=\underset{h→0}{lim}\,0=0\)
Since the left and right hand limits of \(f\) at \(x=n\) are not equal, \(f\) is not differentiable at \(x=n\) 
\(∴f\) is not differentiable in \((−2,2)\). 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 for \(f(x)=[x]\, for\, x∈[-2,2]\)

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