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 will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \frac{x - 1}{x^2 - 3x + 3} \]
To analyze the transformation of the real line, we can rearrange this into a standard form.

### Step 2: Cross-multiply
Cross-multiplying gives:
\[ y(x^2 - 3x + 3) = x - 1 \]
This can be rewritten as:
\[ yx^2 - 3yx + 3y = x - 1 \]
Rearranging terms, we have:
\[ yx^2 - (3y + 1)x + (3y + 1) = 0 \]

### Step 3: Identify the quadratic form
This is a quadratic equation in \( x \):
\[ yx^2 - (3y + 1)x + (3y + 1) = 0 \]
For this quadratic to have real solutions, the discriminant must be non-negative.

### Step 4: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = y \)
- \( b = -(3y + 1) \)
- \( c = 3y + 1 \)

Thus, the discriminant is:
\[ D = (-(3y + 1))^2 - 4(y)(3y + 1) \]
Calculating this gives:
\[ D = (3y + 1)^2 - 4y(3y + 1) \]
\[ D = 9y^2 + 6y + 1 - (12y^2 + 4y) \]
\[ D = 9y^2 + 6y + 1 - 12y^2 - 4y \]
\[ D = -3y^2 + 2y + 1 \]

### Step 5: Set the discriminant greater than or equal to zero
To find the interval for \( y \), we set the discriminant \( D \) to be greater than or equal to zero:
\[ -3y^2 + 2y + 1 \geq 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 3y^2 - 2y - 1 \leq 0 \]

### Step 6: Factor the quadratic
Next, we factor \( 3y^2 - 2y - 1 \):
\[ 3y^2 - 3y + y - 1 = 0 \]
Factoring gives:
\[ (3y + 1)(y - 1) \leq 0 \]

### Step 7: Find the critical points
Setting each factor to zero gives:
1. \( 3y + 1 = 0 \) → \( y = -\frac{1}{3} \)
2. \( y - 1 = 0 \) → \( y = 1 \)

### Step 8: Test intervals
We test the intervals determined by the critical points \( y = -\frac{1}{3} \) and \( y = 1 \):
- For \( y < -\frac{1}{3} \): Choose \( y = -1 \) → \( (3(-1) + 1)(-1 - 1) = (-2)(-2) > 0 \)
- For \( -\frac{1}{3} < y < 1 \): Choose \( y = 0 \) → \( (3(0) + 1)(0 - 1) = (1)(-1) < 0 \)
- For \( y > 1 \): Choose \( y = 2 \) → \( (3(2) + 1)(2 - 1) = (7)(1) > 0 \)

### Step 9: Conclusion
The function \( y = f(x) \) transforms the real line in the interval:
\[ y \in \left[-\frac{1}{3}, 1\right] \]