5,435 views
12 votes
12 votes

Consider the function $y=|x|$ in the interval $[-1, 1]$. In this interval, the function is

  1. continuous and differentiable
  2. continuous but not differentiable
  3. differentiable but not continuous
  4. neither continuous nor differentiable

6 Answers

2 votes
2 votes

$f(x) = \begin{cases} -x & \text{ if }  x < 0 \\ +x & \text{ if } x \geq 0  \end{cases}$

Continuity at $ x=0 $

$\begin{align*} \lim_{x \rightarrow 0^{-}}f(x)\ &= \lim_{x \rightarrow 0^{+}}f(x)\\ 0 &= 0\\ \end{align*}$

So $ f(x) = |x|$ is continous at x = 0


Differentiability at $x=0$

$\begin{align*} \lim_{h \rightarrow 0}\frac{f(0)-f(0-h)}{0-(0-h)} &= \lim_{h \rightarrow 0}\frac{f(0+h)-f(0)}{(0+h)-(0)}\\ \lim_{h \rightarrow 0}\frac{0-(-0+h)}{h)} &= \lim_{h \rightarrow 0}\frac{(0+h)-0}{h} \\ -1 &= 1\\ \end{align*}$

So $ f(x) = |x|$ is not differentiable at x = 0

Option $B$ is the answer

edited by
Answer:

Related questions

27 votes
27 votes
2 answers
2
0 votes
0 votes
2 answers
4
Arjun asked Sep 23, 2019
721 views
If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then$f(x)$ is continuous at $x=0$, but not differentiable at $x=0$$f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$...