The domain of the derivative of the function | vedclass

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(C) Given the function $f(x) = \frac{x}{1+|x|}$.We can write this as a piecewise function:If $x \ge 0$,then $|x| = x$,so $f(x) = \frac{x}{1+x}$.If $x

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$(-\infty, \infty)$

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(C) Given the function $f(x) = \frac{x}{1+|x|}$.We can write this as a piecewise function:If $x \ge 0$,then $|x| = x$,so $f(x) = \frac{x}{1+x}$.If $x Now,we find the derivative $f'(x)$:For $x > 0$,$f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$.For $x At $x = 0$,we check the left-hand derivative and right-hand derivative:$LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h 	o 0^-} \frac{1}{1-h} = 1$.$RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h 	o 0^+} \frac{1}{1+h} = 1$.Since $LHD = RHD = 1$,the function is differentiable at $x = 0$ and $f'(0) = 1$.Thus,the derivative exists for all real numbers $x \in (-\infty, \infty)$.

The domain of the derivative of the function $f(x) = \frac{x}{1+|x|}$ is

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