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 will analyze it step by step.

### Step 1: Analyze the function based on the absolute value
The function involves an absolute value, so we need to consider two cases based on the value of \( x \):

1. **Case 1**: \( x \geq 0 \)
- In this case, \( |x| = x \).
- Thus, the function simplifies to:
\[ f(x) = \frac{x}{1 + x} \]

2. **Case 2**: \( x < 0 \)
- Here, \( |x| = -x \).
- The function simplifies to:
\[ f(x) = \frac{x}{1 - x} \]

### Step 2: Find the derivative \( f'(x) \) for both cases
Next, we will find the derivative of \( f(x) \) in both cases using the quotient rule.

**For Case 1**: \( f(x) = \frac{x}{1 + x} \)

Using the quotient rule:
\[ f'(x) = \frac{(1 + x)(1) - x(1)}{(1 + x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \]

**For Case 2**: \( f(x) = \frac{x}{1 - x} \)

Again, using the quotient rule:
\[ f'(x) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1 - x + x}{(1 - x)^2} = \frac{1}{(1 - x)^2} \]

### Step 3: Analyze the sign of the derivative
Now we will analyze the sign of \( f'(x) \) in both cases.

1. **For \( x \geq 0 \)**:
- \( f'(x) = \frac{1}{(1 + x)^2} \)
- Since \( (1 + x)^2 > 0 \) for all \( x \geq 0 \), we have \( f'(x) > 0 \).

2. **For \( x < 0 \)**:
- \( f'(x) = \frac{1}{(1 - x)^2} \)
- Since \( (1 - x)^2 > 0 \) for all \( x < 0 \), we have \( f'(x) > 0 \).

### Step 4: Conclusion
Since \( f'(x) > 0 \) for all \( x \) in both cases, we conclude that the function \( f(x) \) is strictly increasing for all \( x \in \mathbb{R} \).

### Final Answer
The function \( f(x) = \frac{x}{1 + |x|} \) is **strictly increasing**.

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