The set of points where f(x) = frac{x}{4+|x|} | vedclass

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(A) Given $f(x) = \frac{x}{4+|x|}$.We can write $f(x)$ as a piecewise function:$f(x) = \begin{cases} \frac{x}{4+x}, & x \geq 0 \ \frac{x}{4-x}, & x <

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$(-\infty, \infty)$

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(A) Given $f(x) = \frac{x}{4+|x|}$.We can write $f(x)$ as a piecewise function:$f(x) = \begin{cases} \frac{x}{4+x}, & x \geq 0 \ \frac{x}{4-x}, & x To check for differentiability at $x = 0$,we find the Left Hand Derivative $(LHD)$ and Right Hand Derivative $(RHD)$.$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{\frac{h}{4+h} - 0}{h} = \lim_{h 	o 0^+} \frac{1}{4+h} = \frac{1}{4}$.$LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{\frac{h}{4-h} - 0}{h} = \lim_{h 	o 0^-} \frac{1}{4-h} = \frac{1}{4}$.Since $LHD = RHD = \frac{1}{4}$,the function is differentiable at $x = 0$.For $x > 0$,$f(x) = \frac{x}{4+x}$ is a rational function with a non-zero denominator,so it is differentiable.For $x Thus,$f(x)$ is differentiable for all $x \in (-\infty, \infty)$.

The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is

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