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(B) The function is defined as $f(x) = \begin{cases} -x(2 - \sin(\frac{1}{x})), & x  0 \end{cases}

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not monotonic on $(-\infty, 0)$ and $(0, \infty)$

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(B) The function is defined as $f(x) = \begin{cases} -x(2 - \sin(\frac{1}{x})), & x  0 \end{cases}$.For $x > 0$,$f'(x) = (2 - \sin(\frac{1}{x})) + x(-\cos(\frac{1}{x}) \cdot (-\frac{1}{x^2})) = 2 - \sin(\frac{1}{x}) + \frac{1}{x}\cos(\frac{1}{x})$.As $x 	o 0^+$,the term $\frac{1}{x}\cos(\frac{1}{x})$ oscillates between arbitrarily large positive and negative values. Thus,$f'(x)$ changes sign infinitely often in any neighborhood of $0$,implying $f(x)$ is not monotonic on $(0, \infty)$.For $x Similarly,as $x 	o 0^-$,the term $-\frac{1}{x}\cos(\frac{1}{x})$ oscillates,causing $f'(x)$ to change sign infinitely often. Thus,$f(x)$ is not monotonic on $(-\infty, 0)$.Therefore,$f$ is not monotonic on $(-\infty, 0)$ and $(0, \infty)$.

Consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \begin{cases} (2 - \sin(\frac{1}{x}))|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f$ is

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