Let f be defined on D = R - {-1, 1} by f(x) = | vedclass

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

Question

(B) Given: $f(x) = \frac{|x|}{1 - |x|}$.First,check continuity at $x = 0$:$	ext{LHL} = \lim_{x 	o 0^-} \frac{-x}{1 - (-x)} = \lim_{x 	o 0^-} \frac{

Vedclass · Accepted Answer

$f$ is differentiable on $D$ except at $x = 0$

Vedclass · Answer

(B) Given: $f(x) = \frac{|x|}{1 - |x|}$.First,check continuity at $x = 0$:$	ext{LHL} = \lim_{x 	o 0^-} \frac{-x}{1 - (-x)} = \lim_{x 	o 0^-} \frac{-x}{1 + x} = 0$.$	ext{RHL} = \lim_{x 	o 0^+} \frac{x}{1 - x} = 0$.Since $f(0) = 0$,$f(x)$ is continuous at $x = 0$.Now,check differentiability at $x = 0$:For $x For $x > 0$,$f(x) = \frac{x}{1 - x}$,so $f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$. Thus,$	ext{RHD} = f'(0^+) = 1$.Since $	ext{LHD} 
eq 	ext{RHD}$,$f(x)$ is not differentiable at $x = 0$.However,$f(x)$ is differentiable for all $x \in D \setminus \{0\}$.Therefore,$f$ is differentiable on $D$ except at $x = 0$.

Let $f$ be defined on $D = R - \{-1, 1\}$ by $f(x) = \frac{|x|}{1 - |x|}$,then

Similar Questions

If $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$ is differentiable at every point of the domain,then the values of $a$ and $b$ are respectively

If the function $g(x) = \begin{cases} ae^x, & x \le 0 \\ b\cos x + x, & x > 0 \end{cases}$ is differentiable,then the value of $a^2 + b^2$ is

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Let $f(x) = |x - \alpha| + |x - \beta|$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 2 = 0$. Then the number of points in $[\alpha, \beta]$ at which $f$ is not differentiable is

Vedclass Test Series

Exam Paper Generator

Online Exam Module