The range of the function f (x) ` = (x)/(1 + |x|) , x in R , is `

Options

A. R
B. (-1,1)
C. R - (0)
D. [-1,1]

Answer

To find the range of the function \( f(x) = \frac{x}{1 + |x|} \), we will analyze the function by breaking it into two cases based on the definition of the absolute value.

### Step 1: Define the function for \( x < 0 \)
For \( x < 0 \), we have \( |x| = -x \). Thus, the function becomes:
\[ f(x) = \frac{x}{1 - x} \]

### Step 2: Set up the equation for \( y \)
Let \( y = f(x) \):
\[ y = \frac{x}{1 - x} \]
Rearranging gives:
\[ y(1 - x) = x \implies y - yx = x \implies y = x + yx \implies x = \frac{y}{y + 1} \]

### Step 3: Determine the constraints for \( y \)
Since \( x < 0 \), we require:
\[ \frac{y}{y + 1} < 0 \]
This implies that \( y \) must be greater than \(-1\) (since \( y + 1 \) must be positive) and less than \( 0 \):
\[ -1 < y < 0 \]

### Step 4: Define the function for \( x \geq 0 \)
For \( x \geq 0 \), we have \( |x| = x \). Thus, the function becomes:
\[ f(x) = \frac{x}{1 + x} \]

### Step 5: Set up the equation for \( y \)
Let \( y = f(x) \):
\[ y = \frac{x}{1 + x} \]
Rearranging gives:
\[ y(1 + x) = x \implies y + yx = x \implies y = x - yx \implies x = \frac{y}{1 - y} \]

### Step 6: Determine the constraints for \( y \)
Since \( x \geq 0 \), we require:
\[ \frac{y}{1 - y} \geq 0 \]
This implies that \( y \) must be less than \( 1 \) (since \( 1 - y \) must be positive) and can take any value from \( 0 \) to \( 1 \):
\[ 0 \leq y < 1 \]

### Step 7: Combine the ranges from both cases
From the analysis of both cases, we find:
- For \( x < 0 \): \( -1 < y < 0 \)
- For \( x \geq 0 \): \( 0 \leq y < 1 \)

Thus, the overall range of the function \( f(x) \) is:
\[ (-1, 0) \cup [0, 1) \]

### Final Answer
The range of the function \( f(x) = \frac{x}{1 + |x|} \) is \( (-1, 1) \).

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