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(A) The function is defined as $f(x) = \frac{x}{1+|x|}$.We can express this piecewise as:$f(x) = \begin{cases} \frac{x}{1-x}, & x < 0 \ \frac{x}{1+x}

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$(-\infty, \infty)$

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(A) The function is defined as $f(x) = \frac{x}{1+|x|}$.We can express this piecewise as:$f(x) = \begin{cases} \frac{x}{1-x}, & x Now,we check the differentiability at $x = 0$:Left-hand derivative $(LHD)$ at $x = 0$:$LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h 	o 0^-} \frac{1}{1-h} = 1$Right-hand derivative $(RHD)$ at $x = 0$:$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h 	o 0^+} \frac{1}{1+h} = 1$Since $LHD = RHD = 1$,the function is differentiable at $x = 0$.For $x 
eq 0$,the function is a rational function with a non-zero denominator,so it is differentiable everywhere.Thus,the derivative exists for all $x \in (-\infty, \infty)$.

The set of all points,where the derivative of the function $f(x) = \frac{x}{1+|x|}$ exists,is

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