Find the rang of `f(x)=(x^2+34 x-71)/(x^2+2x-7)` `f(x)=(x^2-x+1)/(x^2+x+1)`

Explanation

To find the range of the given functions, we will analyze each function step by step.

### Part 1: Finding the Range of \( f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \)

1. **Set the function equal to \( y \)**:
\[ y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \]

2. **Cross-multiply**:
\[ y(x^2 + 2x - 7) = x^2 + 34x - 71 \]

3. **Rearranging the equation**:
\[ yx^2 + 2yx - 7y = x^2 + 34x - 71 \]
\[ (y - 1)x^2 + (2y - 34)x + (7y - 71) = 0 \]

4. **For the quadratic equation to have real solutions, the discriminant must be non-negative**:
\[ D = (2y - 34)^2 - 4(y - 1)(7y - 71) \geq 0 \]

5. **Expanding the discriminant**:
\[ D = (2y - 34)^2 - 4[(y - 1)(7y - 71)] \]
\[ = 4y^2 - 136y + 1156 - 4(7y^2 - 71y - 7y + 71) \]
\[ = 4y^2 - 136y + 1156 - (28y^2 - 312y + 284) \]
\[ = 4y^2 - 136y + 1156 - 28y^2 + 312y - 284 \]
\[ = -24y^2 + 176y + 872 \geq 0 \]

6. **Dividing the entire inequality by -8** (and reversing the inequality):
\[ 3y^2 - 22y - 109 \leq 0 \]

7. **Finding the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot (-109)}}{2 \cdot 3} \]
\[ = \frac{22 \pm \sqrt{484 + 1308}}{6} = \frac{22 \pm \sqrt{1792}}{6} \]
\[ = \frac{22 \pm 42.426}{6} \]
\[ y_1 = \frac{64.426}{6} \approx 10.738 \quad \text{and} \quad y_2 = \frac{-20.426}{6} \approx -3.404 \]

8. **Finding the range**:
The quadratic \( 3y^2 - 22y - 109 \leq 0 \) is satisfied between the roots:
\[ y \in \left[-3.404, 10.738\right] \]

### Part 2: Finding the Range of \( f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \)

1. **Set the function equal to \( y \)**:
\[ y = \frac{x^2 - x + 1}{x^2 + x + 1} \]

2. **Cross-multiply**:
\[ y(x^2 + x + 1) = x^2 - x + 1 \]

3. **Rearranging the equation**:
\[ yx^2 + yx + y = x^2 - x + 1 \]
\[ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 \]

4. **For the quadratic equation to have real solutions, the discriminant must be non-negative**:
\[ D = (y + 1)^2 - 4(y - 1)(y - 1) \geq 0 \]

5. **Expanding the discriminant**:
\[ D = (y + 1)^2 - 4(y - 1)^2 \]
\[ = y^2 + 2y + 1 - 4(y^2 - 2y + 1) \]
\[ = y^2 + 2y + 1 - 4y^2 + 8y - 4 \]
\[ = -3y^2 + 10y - 3 \geq 0 \]

6. **Dividing the entire inequality by -3** (and reversing the inequality):
\[ y^2 - \frac{10}{3}y + 1 \leq 0 \]

7. **Finding the roots using the quadratic formula**:
\[ y = \frac{\frac{10}{3} \pm \sqrt{(\frac{10}{3})^2 - 4 \cdot 1}}{2} \]
\[ = \frac{\frac{10}{3} \pm \sqrt{\frac{100}{9} - 4}}{2} = \frac{\frac{10}{3} \pm \sqrt{\frac{100}{9} - \frac{36}{9}}}{2} \]
\[ = \frac{\frac{10}{3} \pm \sqrt{\frac{64}{9}}}{2} = \frac{\frac{10}{3} \pm \frac{8}{3}}{2} \]
\[ y_1 = \frac{18/3}{2} = 3 \quad \text{and} \quad y_2 = \frac{2/3}{2} = \frac{1}{3} \]

8. **Finding the range**:
The quadratic \( y^2 - \frac{10}{3}y + 1 \leq 0 \) is satisfied between the roots:
\[ y \in \left[\frac{1}{3}, 3\right] \]

### Final Results:
- The range of the first function \( f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \) is \( \left[-3.404, 10.738\right] \).
- The range of the second function \( f(x) = \frac{x^2 - x + 1}{x^2 + x + 1} \) is \( \left[\frac{1}{3}, 3\right] \).