Applications Of Derivatives Question 384
Question: Rolle’s theorem is not applicable to the function $ f(x)=|x| $ defined on
[-1, 1] because [AISSE 1986; MP PET 1994, 95]
Options:
A) f is not continuous on [ -1, 1]
B) f is not differentiable on (-1,1)
C) $ f(-1)\ne f(1) $
D) $ f(-1)=f(1)\ne 0 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)= \begin{cases} & -x,,\text{when –1}\le x<0 \\ & \text{ }x,\ when\ 0\le x\le 1 \\ \end{cases} . $
Clearly $ f(-1)=|-1|=1=f(1) $
But $ Rf’(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{|h|}{h} $
$ =\underset{h\to 0}{\mathop{\lim }},\frac{h}{h}=1 $
$ Lf’(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(0-h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{|-h|}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }},\frac{h}{-h}=-1 $
$ \therefore Rf’(0)\ne Lf’(0) $
Hence it is not differentiable on $ (-1,1) $ .
BETA
