Find the range of each of the following function: `f(x)=x/(1+x^(2))`

Explanation

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

### Step 1: Set the function equal to \( y \)
We start by rewriting the function:
\[ y = \frac{x}{1 + x^2} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(1 + x^2) = x \]
This simplifies to:
\[ yx^2 - x + y = 0 \]

### Step 3: Rearrange into a standard quadratic form
We can rearrange this into a standard quadratic equation in terms of \( x \):
\[ yx^2 - x + y = 0 \]

### Step 4: Identify coefficients for the quadratic formula
In this quadratic equation, the coefficients are:
- \( a = y \)
- \( b = -1 \)
- \( c = y \)

### Step 5: Apply the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute the coefficients:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(y)(y)}}{2(y)} \]
This simplifies to:
\[ x = \frac{1 \pm \sqrt{1 - 4y^2}}{2y} \]

### Step 6: Determine the condition for real \( x \)
For \( x \) to be real, the discriminant must be non-negative:
\[ 1 - 4y^2 \geq 0 \]

### Step 7: Solve the inequality
Rearranging gives:
\[ 4y^2 \leq 1 \]
Dividing by 4:
\[ y^2 \leq \frac{1}{4} \]

### Step 8: Take the square root
Taking the square root of both sides gives:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

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