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

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(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{

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

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(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

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