The range of the function `y=(x)/( 1+x^2)`is

Options

A. `[-1/2,1/2]`
B. `[-1/2,1/2]`
C. `[-1/2 ,1/2]`
D. `[-1/2,1/2]`

Answer

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

### Step 1: Rearranging the Equation
We start with the equation:
\[ y = \frac{x}{1 + x^2} \]
We can rearrange this to form a quadratic equation in \( x \):
\[ y(1 + x^2) = x \]
This simplifies to:
\[ yx^2 - x + y = 0 \]

### Step 2: Identifying the Quadratic Coefficients
The quadratic equation we have is:
\[ yx^2 - x + y = 0 \]
From this, we identify the coefficients:
- \( a = y \)
- \( b = -1 \)
- \( c = y \)

### Step 3: Discriminant Condition
For \( x \) to have real solutions, the discriminant \( D \) of the quadratic must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ (-1)^2 - 4(y)(y) \geq 0 \]
This simplifies to:
\[ 1 - 4y^2 \geq 0 \]

### Step 4: Solving 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 \( y = \frac{x}{1 + x^2} \) is:
\[ \boxed{[-\frac{1}{2}, \frac{1}{2}]} \]