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

We are tasked with finding when \( f(x) \) is negative. To do so, we need to determine the sign of \( f(x) \) in different intervals defined by its zeros. Let's first find the zeros of the numerator and denominator:

Step 1: Finding the zeros of the numerator

The numerator is \( x^2 + 2x - 15 \), which factors as: \[ (x + 5)(x - 3) = 0 \] This gives zeros at: \[ x = -5 \quad \text{and} \quad x = 3 \]

Step 2: Finding the zeros of the denominator

The denominator is \( x^2 - 7x - 18 \), which factors as: \[ (x + 2)(x - 9) = 0 \] This gives zeros at: \[ x = -2 \quad \text{and} \quad x = 9 \]

Step 3: Defining intervals

Using these zeros, we 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

Now, we will test the sign of \( f(x) \) in each of these intervals by picking a test point from each:

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

Step 5: Conclusion

From the above evaluations, \( f(x) \) is negative in the intervals: \[ (-5, -2) \quad \text{and} \quad (3, 9) \]

Final Answer:

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

Answer 2

We are given the rational function: 

\[ f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18} < 0 \] Which can be factored as: \[ \frac{(x + 5)(x - 3)}{(x - 9)(x + 2)} < 0 \]

Step 1: Identifying critical points

The critical points are the zeros of the numerator and denominator: \[ x = -5, \, -2, \, 3, \, 9 \]

Step 2: Testing the sign in different intervals

We now analyze the sign of the expression in the intervals defined by these critical points: \( (-\infty, -5), (-5, -2), (-2, 3), (3, 9), (9, \infty) \). - **For \( x < -5 \)**: All four terms \((x + 5), (x - 3), (x - 9), (x + 2)\) will be negative, resulting in a positive overall expression. - **For \( x > 9 \)**: All four terms will be positive, leading to a positive overall expression. - **For \( -2 < x < 3 \)**: Two terms are negative and two are positive, resulting in a positive overall expression. - **For \( -5 < x < -2 \) and \( 3 < x < 9 \)**: The expression will be negative as two terms are positive and two are negative.

Step 3: Conclusion

From the analysis above, the expression \( f(x) \) is negative in the intervals: \[ (-5 < x < -2) \quad \text{or} \quad (3 < x < 9) \]

Final Answer:

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