`f(x)=(x-1)/(x^(2)-2x+3)` Find the range of f(x).

Answer

To find the range of the function \( f(x) = \frac{x-1}{x^2 - 2x + 3} \), we can follow these steps:

### Step 1: Set \( y = f(x) \)
Let \( y = \frac{x-1}{x^2 - 2x + 3} \).

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(x^2 - 2x + 3) = x - 1 \]
This simplifies to:
\[ yx^2 - 2yx + 3y = x - 1 \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ yx^2 - (2y + 1)x + (3y + 1) = 0 \]

### Step 4: Identify the coefficients
This is a quadratic equation in \( x \):
- Coefficient of \( x^2 \) is \( y \)
- Coefficient of \( x \) is \( -(2y + 1) \)
- Constant term is \( 3y + 1 \)

### Step 5: Apply the discriminant condition
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ D = (-(2y + 1))^2 - 4(y)(3y + 1) \geq 0 \]
This simplifies to:
\[ (2y + 1)^2 - 4y(3y + 1) \geq 0 \]

### Step 6: Expand and simplify the discriminant
Expanding gives:
\[ 4y^2 + 4y + 1 - (12y^2 + 4y) \geq 0 \]
Combining like terms results in:
\[ -8y^2 + 1 \geq 0 \]
Rearranging gives:
\[ 8y^2 \leq 1 \]

### Step 7: Solve the inequality
Dividing by 8:
\[ y^2 \leq \frac{1}{8} \]
Taking the square root:
\[ -\frac{1}{2\sqrt{2}} \leq y \leq \frac{1}{2\sqrt{2}} \]

### Step 8: Write the final range
Thus, the range of the function \( f(x) \) is:
\[ \boxed{\left[-\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right]} \]