The function `f(x) = (x )/( 1+ |x|) ` is

Options

A. strictly increasing
B. strictly decreasing
C. neither increasing nor decreasing
D. not differentiable at x=0

Answer

To determine the nature of the function \( f(x) = \frac{x}{1 + |x|} \), we need to analyze its behavior by differentiating it in two cases based on the definition of the absolute value function.

### Step 1: Define the function based on cases
The absolute value function \( |x| \) can be defined in two cases:
1. When \( x \geq 0 \), \( |x| = x \)
2. When \( x < 0 \), \( |x| = -x \)

Thus, we can rewrite the function \( f(x) \) as:
- For \( x \geq 0 \):
\[ f(x) = \frac{x}{1 + x} \]
- For \( x < 0 \):
\[ f(x) = \frac{x}{1 - x} \]

### Step 2: Differentiate \( f(x) \) for \( x \geq 0 \)
Now, we differentiate \( f(x) \) for the case \( x \geq 0 \):
\[ f'(x) = \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \]

### Step 3: Differentiate \( f(x) \) for \( x < 0 \)
Next, we differentiate \( f(x) \) for the case \( x < 0 \):
\[ f'(x) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2} \]

### Step 4: Analyze the sign of \( f'(x) \)
For both cases:
- For \( x \geq 0 \): \( f'(x) = \frac{1}{(1 + x)^2} > 0 \)
- For \( x < 0 \): \( f'(x) = \frac{1}{(1 - x)^2} > 0 \)

Since \( f'(x) > 0 \) for all \( x \) (both for \( x \geq 0 \) and \( x < 0 \)), it indicates that the function is strictly increasing in both intervals.

### Conclusion
The function \( f(x) = \frac{x}{1 + |x|} \) is strictly increasing for all \( x \).

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