Continuity and Differentiability

117. Solution :
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  $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f\left(x\right)=\left[x\right]$ for $x\in \left[5,9\right]$

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).

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-1-n}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{-1}{h}=\infty$

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-n}{h}=\underset{h\to 0}{{\displaystyle lim}}0=0$

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 Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for  $f\left(x\right)=\left[x\right]$ for $x\in \left[5,9\right]$ .

(ii)  $f\left(x\right)=\left[x\right]$ $inx\in \left[-2,2\right]$

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).

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-1-n}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{-1}{h}=\infty$

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-n}{h}=\underset{h\to 0}{{\displaystyle lim}}0=0$

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 Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for  $f\left(x\right)=\left[x\right]$ for $x\in \left[-2,2\right]$

(iii)  $f\left(x\right)=x^2-1$ for $x\in \left[1,2\right]$

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  $f\left(x\right)=x^2-1$ for $x\in \left[1,2\right]$

It can be proved as follows.

$\begin{array}{l}f\left(1\right)=1^2-1=0,f\left(2\right)=2^2-1=3\\ \therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{3-0}{1}=3\\ f\text{'}\left(x\right)=2x\\ \therefore f\text{'}\left(c\right)=3\\ \Rightarrow 2c=3\\ \Rightarrow c=\frac{3}{2}=1.5,where,1.5\in \left(1,2\right)\end{array}$

113. Solution:

By Rolle's Theorem, for a function $f\left[a,b\right]\to R,$ if

f is continuous on $\left[a,b\right]$

f is differentiable on $\left(a,b\right)$

f(a)= f(b)

then, there exists some $c\in \left(a,b\right)$ such that $f^{\text{'}}\left(c\right)=0$

therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f\left(x\right)=\left[x\right]$ for $x\in \left[5,9\right]$

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

$\Rightarrow$ f(x) is not continuous in $\left[5,9\right]$

Also, $\begin{array}{l}f\left(5\right)=\left[5\right]=5,and,f\left(9\right)=\left[9\right]=9\\ \therefore f\left(5\right)\ne f\left(9\right)\end{array}$

The differentiability of f in $\left(5,9\right)$ is checked as follows.

Let n be an integer such that $n\in \left(5,9\right)$ .

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-1-n}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{-1}{h}=\infty$

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-n}{h}=\underset{h\to 0}{{\displaystyle lim}}0=0$

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

$\therefore$ 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 $f\left(x\right)=\left[x\right]$ for $x\in \left[5,9\right]$

$f\left(x\right)=\left[x\right]$ for $x\in \left[-2,2\right]$

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

$\Rightarrow$ f(x) is not continuous in $\left[-2,2\right]$ .

Also, $\begin{array}{l}f\left(-2\right)=\left[-2\right]=-2,and,f\left(2\right)=\left[2\right]=2\\ \therefore f\left(-2\right)\ne f\left(2\right)\end{array}$

The differentiability of f in $\left(-2,2\right)$ is checked as follows.

Let n be an integer such that $n\in \left(-2,2\right)$ .

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-1-n}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{-1}{h}=\infty$

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

$\underset{h\to 0}{{\displaystyle lim}}\frac{f\left(n+h\right)-f\left(n\right)}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{\left[n+h\right]-\left[n\right]}{h}=\underset{h\to 0}{{\displaystyle lim}}\frac{n-n}{h}=\underset{h\to 0}{{\displaystyle lim}}0=0$

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

$\therefore$ 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 $f\left(x\right)=\left[x\right]$ for $x\in \left[-2,2\right]$

$f\left(x\right)=x^2-1$ for $x\in \left[1,2\right]$

It is evident that f, being a polynomial function, is continuous in $\left[1,2\right]$ and is differentiable on (1,2).

$\begin{array}{l}f\left(1\right)=1^2=0\\ f\left(2\right)=2^2-1=3\\ \therefore f\left(1\right)\ne f\left(2\right)\end{array}$

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 $f\left(x\right)=x^2-1$ for $x\in \left[1,2\right]$ .