Let `f(x) = (x^(2))/((1+x^(2))` .Then range (f ) =?

Options

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

Explanation

To find the range of the function \( f(x) = \frac{x^2}{1 + x^2} \), we will follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) = \frac{x^2}{1 + x^2} \).

### Step 2: Rearrange the equation
We can rearrange the equation to express \( x^2 \) in terms of \( y \):
\[ y(1 + x^2) = x^2 \]
This simplifies to:
\[ y + yx^2 = x^2 \]
Rearranging gives:
\[ yx^2 - x^2 + y = 0 \]
Factoring out \( x^2 \):
\[ x^2(y - 1) = -y \]

### Step 3: Solve for \( x^2 \)
From the equation \( x^2(y - 1) = -y \), we can isolate \( x^2 \):
\[ x^2 = \frac{-y}{y - 1} \]

### Step 4: Determine the conditions for \( x^2 \)
Since \( x^2 \) must be non-negative (i.e., \( x^2 \geq 0 \)), we need:
\[ \frac{-y}{y - 1} \geq 0 \]
This inequality holds when both the numerator and denominator are either positive or negative.

### Step 5: Analyze the inequality
1. **Numerator**: \( -y \geq 0 \) implies \( y \leq 0 \).
2. **Denominator**: \( y - 1 < 0 \) implies \( y < 1 \).

Combining these conditions, we find that:
- \( y \leq 0 \) and \( y < 1 \) are contradictory since \( y \) cannot be negative in the context of the function \( f(x) \).

### Step 6: Find the valid range
Next, we analyze the behavior of \( f(x) \):
- As \( x \) approaches \( 0 \), \( f(0) = \frac{0^2}{1 + 0^2} = 0 \).
- As \( x \) approaches \( \infty \), \( f(x) \) approaches \( 1 \) since:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2}{1 + x^2} = \lim_{x \to \infty} \frac{1}{\frac{1}{x^2} + 1} = 1 \]

### Step 7: Conclusion
Thus, the function \( f(x) \) takes values from \( 0 \) to \( 1 \) (not including \( 1 \)). Therefore, the range of \( f \) is:
\[ \text{Range}(f) = [0, 1) \]