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

Options

A. `(1//3,3)`
B. `[1//3,3]`
C. `(0,3)`
D. None of these

Answer

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

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

### Step 2: Rearrange the equation
Multiply both sides by the denominator to eliminate the fraction:
\[ y(x^2 + x - 1) = x^2 + x + 1 \]
This simplifies to:
\[ yx^2 + yx - y = x^2 + x + 1 \]

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

### Step 4: Identify coefficients
In this quadratic equation, the coefficients are:
- \( A = y - 1 \)
- \( B = y - 1 \)
- \( C = 1 + y \)

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

### Step 6: Expand the discriminant
Expanding this gives:
\[ (y - 1)^2 - 4(y^2 - 1) \geq 0 \]
This simplifies to:
\[ y^2 - 2y + 1 - 4y^2 + 4 \geq 0 \]
Combining like terms:
\[ -3y^2 - 2y + 5 \geq 0 \]

### Step 7: Multiply by -1 and reverse the inequality
Multiplying through by -1 (which reverses the inequality):
\[ 3y^2 + 2y - 5 \leq 0 \]

### Step 8: Find the roots of the quadratic
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
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:
\[ y_1 = 1 \quad \text{and} \quad y_2 = -\frac{5}{3} \]

### Step 9: Determine the intervals
The quadratic \( 3y^2 + 2y - 5 \) opens upwards (since the coefficient of \( y^2 \) is positive). Thus, it is less than or equal to zero between its roots:
\[ -\frac{5}{3} \leq y \leq 1 \]

### Step 10: Conclusion
Therefore, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{5}{3}, 1\right] \]