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

Options

A. `[1//3,2]`
B. `[-1//3,1]`
C. `[-1//3,2]`
D. none of these

Answer

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

### Step 1: Set up the equation
We start with the equation:
\[ y = \frac{x - 1}{x^2 - 3x + 3} \]
To eliminate the fraction, we can multiply both sides by the denominator:
\[ y(x^2 - 3x + 3) = x - 1 \]
This simplifies to:
\[ yx^2 - 3yx + 3y = x - 1 \]

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

### Step 3: Apply the discriminant condition
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = y \), \( b = -(3y + 1) \), and \( c = 3y + 1 \). Thus, the discriminant is:
\[ D = (-(3y + 1))^2 - 4(y)(3y + 1) \geq 0 \]
Calculating this gives:
\[ D = (3y + 1)^2 - 4y(3y + 1) \geq 0 \]

### Step 4: Expand and simplify the discriminant
Expanding the discriminant:
\[ (3y + 1)^2 = 9y^2 + 6y + 1 \]
\[ 4y(3y + 1) = 12y^2 + 4y \]
Thus:
\[ D = 9y^2 + 6y + 1 - 12y^2 - 4y \geq 0 \]
This simplifies to:
\[ -3y^2 + 2y + 1 \geq 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 3y^2 - 2y - 1 \leq 0 \]

### Step 5: Factor the quadratic
Next, we factor the quadratic:
\[ 3y^2 - 2y - 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ \sqrt{4 + 12} = \sqrt{16} = 4 \]
Thus:
\[ y = \frac{2 \pm 4}{6} \]
This gives us the roots:
\[ y = 1 \quad \text{and} \quad y = -\frac{1}{3} \]

### Step 6: Determine the intervals
Now we need to test the intervals determined by these roots:
1. \( y < -\frac{1}{3} \)
2. \( -\frac{1}{3} < y < 1 \)
3. \( y > 1 \)

Using test points in these intervals, we find:
- For \( y < -\frac{1}{3} \), the quadratic is positive.
- For \( -\frac{1}{3} < y < 1 \), the quadratic is negative.
- For \( y > 1 \), the quadratic is positive.

### Conclusion
The solution to the inequality \( 3y^2 - 2y - 1 \leq 0 \) is:
\[ y \in \left[-\frac{1}{3}, 1\right] \]
Thus, the interval into which the function transforms the entire real line is:
\[ \boxed{\left[-\frac{1}{3}, 1\right]} \]