If f(x) = begin{cases} x left( frac{e^{1/x} - | vedclass

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(C) First,check continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$. Since $\

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$f$ is differentiable at every point

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(C) First,check continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight)$. Since $\left| \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} ight| Now,check differentiability at $x = 0$: $LHD = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{-h \left( \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} ight)}{-h} = \lim_{h 	o 0^+} \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} = \lim_{h 	o 0^+} \frac{1 - e^{-2/h}}{1 + e^{-2/h}} = \frac{1-0}{1+0} = 1$. $RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h \left( \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} ight)}{h} = \lim_{h 	o 0^+} \frac{1 - e^{-2/h}}{1 + e^{-2/h}} = 1$. Since $LHD = RHD = 1$,the function is differentiable at $x = 0$. Thus,$f$ is differentiable at every point.

If $f(x) = \begin{cases} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then the correct statement is:

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