Key Concepts and Formulas

• Monotonicity and Derivatives: A function $f(x)$ is monotonic on an interval if its derivative $f'(x)$ is either always non-negative (increasing) or always non-positive (decreasing) on that interval. If $f'(x)$ changes sign, $f(x)$ is not monotonic.
• Product Rule: The derivative of a product of two functions $u(x)$ and $v(x)$ is given by $(uv)' = u'v + uv'$.
• Chain Rule: The derivative of a composite function $f(g(x))$ is given by $f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

Step 1: Rewrite the function to remove the absolute value

The given function is $f(x) = \left\{ \begin{array}{rcl} (2 - \sin(1/x))|x| &,& x \ne 0 \\ 0 &,& x = 0 \end{array} \right.$ We rewrite the function by considering the sign of $x$: $f(x) = \begin{cases} -x(2 - \sin(1/x)), & x < 0 \\ 0, & x = 0 \\ x(2 - \sin(1/x)), & x > 0 \end{cases}$ Reasoning: The absolute value function $|x|$ is defined as $x$ for $x \ge 0$ and $-x$ for $x < 0$. We use this definition to rewrite the piecewise function without the absolute value.

Step 2: Calculate the derivative for $x > 0$

For $x > 0$, $f(x) = x(2 - \sin(1/x))$. We apply the product rule with $u = x$ and $v = 2 - \sin(1/x)$. Then $u' = 1$ and $v' = -\cos(1/x) \cdot (-1/x^2) = \frac{1}{x^2} \cos(1/x)$. Thus, $f'(x) = u'v + uv' = (1)(2 - \sin(1/x)) + x\left(\frac{1}{x^2} \cos(1/x)\right) = 2 - \sin(1/x) + \frac{1}{x} \cos(1/x).$ Reasoning: We use the product and chain rules to find the derivative of $f(x)$ for $x > 0$.

Step 3: Calculate the derivative for $x < 0$

For $x < 0$, $f(x) = -x(2 - \sin(1/x))$. We apply the product rule with $u = -x$ and $v = 2 - \sin(1/x)$. Then $u' = -1$ and $v' = -\cos(1/x) \cdot (-1/x^2) = \frac{1}{x^2} \cos(1/x)$. Thus, $f'(x) = u'v + uv' = (-1)(2 - \sin(1/x)) + (-x)\left(\frac{1}{x^2} \cos(1/x)\right) = -2 + \sin(1/x) - \frac{1}{x} \cos(1/x).$ Reasoning: We use the product and chain rules to find the derivative of $f(x)$ for $x < 0$.

Step 4: Analyze the sign of $f'(x)$ for $x > 0$

We have $f'(x) = 2 - \sin(1/x) + \frac{1}{x} \cos(1/x)$. Since $-1 \le \sin(1/x) \le 1$, we have $1 \le 2 - \sin(1/x) \le 3$. Consider $x_n = \frac{1}{2n\pi}$. Then $f'(x_n) = 2 - \sin(2n\pi) + 2n\pi \cos(2n\pi) = 2 + 2n\pi > 0$. Consider $x_n = \frac{1}{(2n+1)\pi}$. Then $f'(x_n) = 2 - \sin((2n+1)\pi) + (2n+1)\pi \cos((2n+1)\pi) = 2 - 0 + (2n+1)\pi (-1) = 2 - (2n+1)\pi$. For large $n$, $2 - (2n+1)\pi < 0$. Thus, $f'(x)$ changes sign for $x > 0$, so $f(x)$ is not monotonic on $(0, \infty)$. Reasoning: We find values of $x$ for which the derivative is positive and negative to show that the function is not monotonic.

Step 5: Analyze the sign of $f'(x)$ for $x < 0$

We have $f'(x) = -2 + \sin(1/x) - \frac{1}{x} \cos(1/x)$. Since $-1 \le \sin(1/x) \le 1$, we have $-3 \le -2 + \sin(1/x) \le -1$. Consider $x_n = -\frac{1}{2n\pi}$. Then $f'(x_n) = -2 + \sin(-2n\pi) - (-2n\pi) \cos(-2n\pi) = -2 + 0 + 2n\pi = -2 + 2n\pi$. For large $n$, $-2 + 2n\pi > 0$. Consider $x_n = -\frac{1}{(2n+1)\pi}$. Then $f'(x_n) = -2 + \sin(-(2n+1)\pi) - (-(2n+1)\pi) \cos(-(2n+1)\pi) = -2 + 0 + (2n+1)\pi (-1) = -2 - (2n+1)\pi$. So $f'(x_n) = -2 - (2n+1)\pi < 0$. Thus, $f'(x)$ changes sign for $x < 0$, so $f(x)$ is not monotonic on $(-\infty, 0)$. Reasoning: We find values of $x$ for which the derivative is positive and negative to show that the function is not monotonic.

Step 6: Conclusion

Since $f(x)$ is not monotonic on either $(-\infty, 0)$ or $(0, \infty)$, option (A) is correct.

Common Mistakes & Tips

• Remember to use the chain rule correctly when differentiating $\sin(1/x)$.
• Be careful with the signs when dealing with $x < 0$.
• To show that a function is not monotonic, it is sufficient to find two points where the derivative has opposite signs.

Summary

We analyzed the monotonicity of the given function by first rewriting it to remove the absolute value. Then, we calculated the derivative for $x>0$ and $x<0$. Finally, we showed that the derivative changes sign in both intervals, implying the function is not monotonic on either interval. Therefore, the function is not monotonic on ($-\infty$, 0) and (0, $\infty$).

Final Answer

The final answer is \boxed{not monotonic on ($-\infty$, 0) and (0, $\infty$)}, which corresponds to option (A).