Solve this

Question:

Let $f(x)=\left\{\begin{array}{ll}1, & x \leq-1 \\ |x|, & -1

(a) continuous at x = − 1
(b) differentiable at x = − 1
(c) everywhere continuous
(d) everywhere differentiable

Solution:

(b) differentiable at $x=-1$

$f(x)= \begin{cases}1, & x \leq-1 \\ |x|, & -1

Differentiabilty at $x=-1$

$(\mathrm{LHD} x=-1)$

$\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x+1}$

$=\lim _{x \rightarrow-1} \frac{f(x)-f(-1)}{x+1}$

$=\lim _{x \rightarrow-1} \frac{1-1}{-1+1}$

 

$=0$

$(\mathrm{RHD} x=-1)$

$=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x+1}$

$=\lim _{x \rightarrow-1} \frac{f(x)-f(-1)}{x+1}$

$=\lim _{x \rightarrow-1} \frac{f(x)-f(-1)}{x+1}$

$=\lim _{x \rightarrow-1} \frac{|x|-|-1|}{x+1}$

$=\frac{1-1 \mid}{-1+1}$

 

$=0$

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