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

Options

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

Answer

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
We start by letting \( f(x) = y \). Thus, we have:
\[ y = \frac{x}{1 + x^2} \]
Rearranging this, we get:
\[ y(1 + x^2) = x \]
This can be rewritten as:
\[ yx^2 - x + y = 0 \]

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

### Step 3: Use the discriminant condition
For \( x \) to have real solutions, the discriminant of this 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: Solve the inequality
Rearranging the inequality gives:
\[ 4y^2 \leq 1 \]
Dividing by 4:
\[ y^2 \leq \frac{1}{4} \]
Taking the square root of both sides:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

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