Let `f (x) =(x^(2) +x-1)/(x ^(2) - x+1)`, In above problem, the range of `f (x) AA x in [-1, 1]` is:

Options

A. `[-1, (3)/(5)]`
B. ` [-1, (5)/(3)]`
C. ` [-(1)/(3), 1]`
D. `[-1,1]`

Answer

To find the range of the function \( f(x) = \frac{x^2 + x - 1}{x^2 - x + 1} \) for \( x \in [-1, 1] \), we will follow these steps:

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

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

### Step 3: Rearrange the equation
Rearranging the equation leads to:
\[ x^2(1 - y) + x(1 + y) + (-1 - y) = 0 \]

### Step 4: Identify coefficients
This is a quadratic equation in \( x \):
\[ a = 1 - y, \quad b = 1 + y, \quad c = -1 - y \]

### Step 5: Use the discriminant
For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Calculating the discriminant:
\[ D = (1 + y)^2 - 4(1 - y)(-1 - y) \]
Expanding this:
\[ D = (1 + 2y + y^2) + 4(1 - y)(1 + y) \]
\[ = 1 + 2y + y^2 + 4(1 - y^2) \]
\[ = 1 + 2y + y^2 + 4 - 4y^2 \]
\[ = 5 - 3y^2 + 2y \]

### Step 6: Set the discriminant greater than or equal to zero
Now, we need:
\[ 5 - 3y^2 + 2y \geq 0 \]
Rearranging gives:
\[ -3y^2 + 2y + 5 \geq 0 \]
Multiplying through by -1 (reversing the inequality):
\[ 3y^2 - 2y - 5 \leq 0 \]

### Step 7: Find the roots of the quadratic
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6} \]
This gives us the roots:
\[ y_1 = \frac{10}{6} = \frac{5}{3}, \quad y_2 = \frac{-6}{6} = -1 \]

### Step 8: Analyze the intervals
The quadratic \( 3y^2 - 2y - 5 \) opens upwards (as the coefficient of \( y^2 \) is positive). The function is less than or equal to zero between the roots:
\[ y \in [-1, \frac{5}{3}] \]

### Step 9: Conclusion
Thus, the range of \( f(x) \) for \( x \in [-1, 1] \) is:
\[ \text{Range of } f(x) = [-1, \frac{5}{3}] \]