`f(x)=x/(1+x^(2))`. Find range of f(x).

Answer

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

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

### Step 2: Cross-multiply
Next, we cross-multiply to eliminate the fraction:
\[ y(1 + x^2) = x \]
This simplifies to:
\[ yx^2 - x + y = 0 \]

### Step 3: Identify the quadratic equation
Now we have a quadratic equation in terms of \( x \):
\[ yx^2 - x + y = 0 \]
This is in the standard form \( ax^2 + bx + c = 0 \) where \( a = y \), \( b = -1 \), and \( c = y \).

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

### Step 5: Solve the inequality
Rearranging gives:
\[ 4y^2 \leq 1 \]
Dividing 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 6: Conclusion
Thus, the range of \( f(x) \) is:
\[ \boxed{[-\frac{1}{2}, \frac{1}{2}]} \]

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