Question

\(f(x) = \) \(\frac{x^2+2x−15}{x^2−7x−18}\) is negative if and only if

• A. \(-5 < x < -2\) or \( 3 < x < 9\)
• B. \(-2 < x < 3\) or \(x > 9\)
• C. \(x < -5\) or \(3 < x < 9\)
• D. \(x < -5 \) or \(-2 < x < 3\)

Answer 1

The Correct Option is A

The objective is to identify the intervals where \( f(x) \) is negative. This requires determining the sign of \( f(x) \) within intervals defined by its zeros. The process begins with finding these zeros for both the numerator and the denominator.

Step 1: Finding the zeros of the numerator

The numerator is \( x^2 + 2x - 15 \). Factoring this expression yields: \[ (x + 5)(x - 3) = 0 \] The zeros of the numerator are therefore: \[ x = -5 \quad \text{and} \quad x = 3 \]

Step 2: Finding the zeros of the denominator

The denominator is \( x^2 - 7x - 18 \). Factoring this expression gives: \[ (x + 2)(x - 9) = 0 \] The zeros of the denominator are: \[ x = -2 \quad \text{and} \quad x = 9 \]

Step 3: Defining intervals

The zeros identified above define the following intervals: \[ (-\infty, -5), (-5, -2), (-2, 3), (3, 9), (9, \infty) \]

Step 4: Testing the sign of \( f(x) \) in each interval

The sign of \( f(x) \) will be tested in each interval using a representative test point:

• For \( (-\infty, -5) \), test point \( x = -6 \): \[ f(-6) = \frac{(-)}{(-)} = \text{positive} \]
• For \( (-5, -2) \), test point \( x = -3 \): \[ f(-3) = \frac{(+)}{(-)} = \text{negative} \]
• For \( (-2, 3) \), test point \( x = 0 \): \[ f(0) = \frac{(-)}{(-)} = \text{positive} \]
• For \( (3, 9) \), test point \( x = 6 \): \[ f(6) = \frac{(+)}{(-)} = \text{negative} \]
• For \( (9, \infty) \), test point \( x = 10 \): \[ f(10) = \frac{(+)}{(+)} = \text{positive} \]

Step 5: Conclusion

Based on the interval tests, \( f(x) \) is negative within the following intervals: \[ (-5, -2) \quad \text{and} \quad (3, 9) \]

Final Answer:

The correct solution is \( \boxed{(-5 < x < -2 \text{ or } 3 < x < 9)} \).