`f(x)={((2-Sin(1/x))absx,,xne0),(0,,x=0):}`

Options

A. Monotonic in `(-oo,0)uu(0,oo)`
B. Not Monotonic in `(-oo,0)uu(0,oo)`
C. Monotonic in `(-oo,oo)`
D. Not Monotonic in `(-oo,oo)`

Answer

To analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} (2 - \sin(1/x)) |x|, & x \neq 0 \\ 0, & x = 0 \end{cases} \]

we need to determine the monotonicity of the function across its domain.

### Step 1: Analyze the function for \( x < 0 \)

For \( x < 0 \), the absolute value \( |x| \) becomes \( -x \). Thus, we can rewrite the function as:

\[ f(x) = (2 - \sin(1/x)) (-x) = -x(2 - \sin(1/x)) \]

### Step 2: Analyze the function for \( x > 0 \)

For \( x > 0 \), the absolute value \( |x| \) remains \( x \). Therefore, the function can be expressed as:

\[ f(x) = (2 - \sin(1/x)) x \]

### Step 3: Find the derivative \( f'(x) \)

To determine the monotonicity, we need to find the derivative \( f'(x) \).

1. For \( x < 0 \):
\[ f'(x) = -\left(2 - \sin(1/x)\right) + x \cos(1/x) \cdot \frac{1}{x^2} = - (2 - \sin(1/x)) + \frac{\cos(1/x)}{x} \]

2. For \( x > 0 \):
\[ f'(x) = (2 - \sin(1/x)) + x \cos(1/x) \cdot \frac{1}{x^2} = (2 - \sin(1/x)) + \frac{\cos(1/x)}{x} \]

### Step 4: Analyze the behavior of \( f'(x) \)

- For both \( x < 0 \) and \( x > 0 \), the terms \( \sin(1/x) \) and \( \cos(1/x) \) oscillate between -1 and 1 as \( x \) approaches 0. This means that \( f'(x) \) can take both positive and negative values depending on the oscillation of the sine and cosine functions.

### Step 5: Conclusion about monotonicity

Since \( f'(x) \) can be both positive and negative in any interval around zero, we conclude that the function \( f(x) \) is neither monotonic nor non-decreasing over the entire real line.

Thus, the function is non-monotonic over the interval \( (-\infty, \infty) \).

### Final Answer

The function \( f(x) \) is non-monotonic over the entire real line \( (-\infty, \infty) \).

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