Consider f(x) = begin{cases} frac{x^2}{|x|}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We have $f(x) = \begin{cases} \frac{x^2}{|x|}, & x 
e 0 \ 0, & x = 0 \end{cases} = \begin{cases} x, & x > 0 \ 0, & x = 0 \ -x, & x < 0 \end{ca

Vedclass · Accepted Answer

$f(x)$ is continuous everywhere

Vedclass · Answer

(B) We have $f(x) = \begin{cases} \frac{x^2}{|x|}, & x 
e 0 \ 0, & x = 0 \end{cases} = \begin{cases} x, & x > 0 \ 0, & x = 0 \ -x, & x We have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0} (-x) = 0$ and $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0} (x) = 0$.Since $f(0) = 0$,we have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 0$.Thus,$f(x)$ is continuous at $x = 0$.Since $f(x)$ is continuous for all $x 
e 0$ as well,$f(x)$ is continuous everywhere.Now,checking differentiability at $x = 0$:$Lf'(0) = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{-h - 0}{h} = -1$.$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h - 0}{h} = 1$.Since $Lf'(0) 
e Rf'(0)$,$f(x)$ is not differentiable at $x = 0$.

Consider $f(x) = \begin{cases} \frac{x^2}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$

Similar Questions

The function $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$ is

If $f(x) = \begin{cases} \frac{x^2 - (a+2)x + a}{x-2} & x \neq 2 \\ 2 & x = 2 \end{cases}$ is continuous at $x = 2$,then the value of $a$ is

If $[.]$ denotes the greatest integer function,then $f(x) = [x]^2 - [x^2]$ is discontinuous at

If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

The function $f(x) = \frac{\log(1+ax) - \log(1-bx)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module