Let `f: R to R ` be defined by `f(x)=(x)/(1+x^(2)), x in R.` Then, the range of f is

Options

A. `[-(1)/(2),(1)/(2)]`
B. `(-1,1)-{0}`
C. `R-[-(1)/(2),(1)/(2)]`
D. `R-[-1,1]`

Explanation

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

### Step 1: Set up the equation
Let \( y = f(x) = \frac{x}{1 + x^2} \). We can rearrange this equation to express \( x \) in terms of \( y \):
\[ y(1 + x^2) = x \]
This simplifies to:
\[ yx^2 - x + y = 0 \]

### Step 2: Identify the quadratic equation
The equation \( yx^2 - x + y = 0 \) is a quadratic equation in \( x \). The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where:
- \( a = y \)
- \( b = -1 \)
- \( c = y \)

### Step 3: Determine the discriminant
For the quadratic equation to have real solutions for \( x \), the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac = (-1)^2 - 4(y)(y) = 1 - 4y^2 \]
We require:
\[ 1 - 4y^2 \geq 0 \]

### Step 4: Solve the inequality
Rearranging the inequality gives:
\[ 4y^2 \leq 1 \]
Dividing both sides by 4:
\[ y^2 \leq \frac{1}{4} \]
Taking the square root of both sides, we find:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

### Step 5: Conclusion
Thus, the range of the function \( f(x) = \frac{x}{1 + x^2} \) is:
\[ \text{Range of } f = \left[-\frac{1}{2}, \frac{1}{2}\right] \]