The interval in which the function `y = f(x) = (x-1)/(x^2-3x+3)` transforms the real line is

Options

A. `(0,oo)`
B. `(-oo,oo)`
C. `[0,1]`
D. `[-1/3,1]

Answer

To find the interval in which the function \( y = f(x) = \frac{x-1}{x^2 - 3x + 3} \) transforms the real line, we need to determine the range of the function. Here is a step-by-step solution:

### Step 1: Set up the equation
We start with the equation:
\[ y = \frac{x-1}{x^2 - 3x + 3} \]

### Step 2: Cross-multiply
To eliminate the fraction, we cross-multiply:
\[ y(x^2 - 3x + 3) = x - 1 \]
This simplifies to:
\[ yx^2 - 3yx + 3y = x - 1 \]

### Step 3: Rearrange the equation
Rearranging gives us a quadratic equation in \( x \):
\[ yx^2 - (3y + 1)x + (3y + 1) = 0 \]

### Step 4: Identify coefficients
From the quadratic equation \( ax^2 + bx + c = 0 \), we identify:
- \( a = y \)
- \( b = -(3y + 1) \)
- \( c = 3y + 1 \)

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

### Step 6: Expand and simplify
Expanding gives:
\[ (3y + 1)^2 - (12y^2 + 4y) \geq 0 \]
\[ 9y^2 + 6y + 1 - 12y^2 - 4y \geq 0 \]
Combining like terms:
\[ -3y^2 + 2y + 1 \geq 0 \]

### Step 7: Multiply through by -1
To simplify the inequality, we multiply through by -1 (which reverses the inequality):
\[ 3y^2 - 2y - 1 \leq 0 \]

### Step 8: Factor the quadratic
Next, we factor the quadratic:
\[ (3y + 1)(y - 1) \leq 0 \]

### Step 9: Find the critical points
Setting each factor to zero gives the critical points:
- \( 3y + 1 = 0 \) → \( y = -\frac{1}{3} \)
- \( y - 1 = 0 \) → \( y = 1 \)

### Step 10: Test intervals
We test the intervals determined by the critical points:
1. \( y < -\frac{1}{3} \)
2. \( -\frac{1}{3} < y < 1 \)
3. \( y > 1 \)

Testing points in these intervals:
- For \( y = -1 \): \( (3(-1) + 1)(-1 - 1) = (-2)(-2) > 0 \) (not in range)
- For \( y = 0 \): \( (3(0) + 1)(0 - 1) = (1)(-1) < 0 \) (in range)
- For \( y = 2 \): \( (3(2) + 1)(2 - 1) = (7)(1) > 0 \) (not in range)

### Step 11: Conclusion
The function \( y \) is less than or equal to zero in the interval:
\[ y \in \left[-\frac{1}{3}, 1\right] \]

Thus, the interval in which the function transforms the real line is:
\[ \text{Range of } f(x) = \left[-\frac{1}{3}, 1\right] \]