Question

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Answer 1

Mean Value Theorem states that for a function f:[a,b]→R , if
(a) f is continuous on [a,b]
(b) f is differentiable on (a,b)

then, there exists some c∈(a,b) such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two 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
⇒ f(x) is not continuous in [5,9].
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,
limh→0- $f\frac{(n+h)-f(n)}{h}$=limh→0-$f\frac{(n+h)-f(n)}{h}$=limh→0-$\frac{n-1-n}{h}$=limh→0-$\frac{-1}{h}=∞$
The right hand limit of f at x=n is,
limh→0+$f\frac{(n+h)-f(n)}{h}$=limh→0+$\frac{[n+h]-[n]}{h}$=limh→0+ $\frac{n-n}{h}$=limh→0+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 Mean Value Theorem.
Hence, Mean Value 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].
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,
limh→0- $f\frac{(n+h)-f(n)}{h}$=limh→0-$\frac{[n+h]-[n]}{h}$=limh→0-n$\frac{-1-n}{h}$=limh→0-$\frac{-1}{h}=∞$
The right hand limit of f at x=n is,
limh→0+$f\frac{(n+h)-f(n)}{h}$=limh→0+$\frac{[n+h]-[n]}{h}$=limh→0+$\frac{ n-n}{h}$=limh→0+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 Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for f(x)=[x] for x∈[-2,2]

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(iii) f(x)=x2-1 for x∈[1,2]
It is evident that f, being a polynomial function, is continuous in [1,2] and is differentiable in (1,2).
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is applicable for f(x)=x2-1 for x∈[1,2]
It can be proved as follows.
$f(1)=(1)^2-1=1-1=0$
$f(2)=(2)^2-1=4-1=3$
$∴f\frac{(b)-f(a)}{b-a}=f\frac{(2)-f(1)}{2-1}=\frac{3-0}{1}=3$

f'(x)=2x
f'(c)=3
⇒2c=3
$⇒c=\frac{3}{2}$=1.5,where 1.5∈[1,2]