Differentiation (Part-38) | S N De

Differentiation (Part-36) | S N De

 

$~3.~$ Show that Rolle's theorem is not applicable to the following functions  in the specified intervals :

$~(i)~~f(x)=1-\sqrt[3]{x^2}~~$ in $~~-1 \leq x \leq 1.$

Solution. 

Clearly, $~f(x)~$ has a definite and unique value at each point in $~[-1,1].$

$~\therefore~f(x)~$ is continuous in $~[-1,1].$

Now, $~f'(x)=\frac{d}{dx}(1-\sqrt[3]{x^2}) \\ \text{or,}~~f'(x)\\=0-\frac{d}{dx}(x^{2/3})\\=-\frac 23x^{\frac 23-1}\\=-\frac{2}{3x^{1/3}} \rightarrow(1)$

From $~(1),~$ we notice that $~f'(x)~$ does not exist at $~x=0\in(-1,1).$

So, Rolle's theorem is not applicable to the function $\,f(x)\,$ in the given interval. 

$~(ii)~f(x)=\tan x~$ in $~~0 \leq x \leq \pi.$

Solution.

Give : $~f(x)=\tan x,~\text{in}~~x \in[0,\pi]$

But, $~f(\pi/2+) \neq f(\pi/2-)~$ and so $~f(x)~$ is not continuous  at $~x=\pi/2~$ and not differentiable. 

Hence, first two conditions of Rolle's theorem are not satisfied and hence, Rolle's theorem is not applicable.

$~(iii)~~f(x)=|x-1|~~$ in $~~ 0\leq x \leq 2.$

Solution. 

Here, $~~f(x)=x-1,~~\text{for}~~x \geq 1,\\~~f(x)=-(x-1)~~\text{for}~x \leq 1.$

Since a polynomial function is continuous and differentiable everywhere in a finite interval, hence $~f(x)~$ is continuous and differentiable for values of $~x<1~$ or, $~x>1~$ of the given interval. Therefore, we are to examine its continuity and differentiability of $~f(x)~$ at $~x=1.$

Let us first examine the differentiability of $~f(x)~$ at $~x=1.$

Clearly, $~~Rf'(1)\\=\lim_{ h \to 0+}~\frac{f(1+h)-f(1)}{h}\\=\lim_{ h \to 0+}~ \frac{(1+h-1)-0}{h}\\=\lim_{ h \to 0+}~ \frac hh\\=1$

and $~~Lf'(1)\\=\lim_{ h \to 0-}~\frac{f(1+h)-f(1)}{h}\\=\lim_{ h \to 0-}~ \frac{-(1+h-1)-0}{h}\\=\lim_{ h \to 0-}~ \frac {-h}{h}\\=-1$

$~~\therefore~~Rf'(1) \neq Lf'(1),~$ hence $~f(x)~$ is not differentiable at $~x=1~; ~$ again the point $~x=1 \in (0,2).$

Therefore, the condition of differentiability of $~f(x)~$ in $~0<x<2~$ is not satisfied.

Hence, Rolle's theorem is not applicable to the function $~f(x)~$ in the given interval.

$~(iv)~~f(x)=2+(x-2)^{2/3}~~$ in $~~1\leq x \leq 3.$

Solution. 

Clearly, $~f(x)~$ has a definite and unique value at each point in $~[1,3].$

$~\therefore~f(x)~$ is continuous in $~[1,3].$

Now, $~~f'(x)\\=\frac{d}{dx}[2+(x-2)^{2/3}]\\=0+\frac{d}{dx}[(x-2)^{\frac 23}]\\=\frac 23(x-2)^{\frac 23-1}\\=\frac23(x-2)^{-\frac 13}\\=\frac{2}{3(x-2)^{\frac 13}} \rightarrow(1)$

From $~(1),~$ we notice that $~f'(x)~$ does not exist at $~x=2\in(1,3).$

So, Rolle's theorem is not applicable to the function $\,f(x)\,$ in the given interval.

Chhaya Mathematics by S.N.De for Class 11 Paperback

$~(v)~~f(x) = \begin{cases} \displaystyle x^2+1 & \text{when}~~~0 \leq x \leq 1 \\ 3-x & \text{when}~~ 1<x \leq 2 \end{cases}$

Solution. 

Since a polynomial function is continuous and differentiable everywhere in a finite interval, hence $~f(x)~$ is continuous and differentiable for values of $~x<1~$ or, $~x>1~$ of the given interval. Therefore, we are to examine its continuity and differentiability of $~f(x)~$ at $~x=1.$

Let us first examine the differentiability of $~f(x)~$ at $~x=1.$

Clearly, $~~Rf'(1)\\=\lim_{ h \to 0+}~\frac{f(1+h)-f(1)}{h}\\=\lim_{ h \to 0+}~ \frac{[3-(1+h)]-(1+1)}{h}\\=\lim_{ h \to 0+}~ \frac {2-h-2}{h}\\=\lim_{ h \to 0+}~ \frac{-h}{h}\\=-1$

and $~~Lf'(1)\\=\lim_{ h \to 0-}~\frac{f(1+h)-f(1)}{h}\\=\lim_{ h \to 0-}~ \frac{[(1+h)^2+1]-(1+1)}{h}\\=\lim_{ h \to 0-}~ \frac {1+2h+h^2+1-2}{h}\\=\lim_{ h \to 0-}~ \frac{2h+h}{h}\\=\lim_{ h \to 0-}~ \frac{h(2+h)}{h}\\=2 $

$~~\therefore~~Rf'(1) \neq Lf'(1),~$ hence $~f(x)~$ is not differentiable at $~x=1~; ~$ again the point $~x=1 \in (0,2).$

Therefore, the condition of differentiability of $~f(x)~$ in $~0<x<2~$ is not satisfied.

Hence, Rolle's theorem is not applicable to the function $~f(x)~$ in the given interval.