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 \(-2 < x < 3\)

Hint:

Given:
\(f(x)=\)\(\frac{x^2+2x−15}{x^2−7x−18}<0\)

\(\frac {(x+5)(x-3)}{(x-9)(x+2)}<0\)
We have identified four critical points: -5, -2, 3, and 9.
For x less than -5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative, resulting in a positive overall expression.
Likewise, for x greater than 9, all four terms will be positive.
In the interval (-2,3), two terms are negative and two are positive, leading to a positive overall expression.
That leaves us with the ranges (-5,-2) and (3,9), where the expression will be negative.

So, the correct option is (A): \(-5 < x < -2\) or \(3 < x < 9\)

Answer 1

The Correct Option is A

To find when \(f(x)\)  is negative, we need to determine the sign of \(f(x)\) in the different intervals given by its zeros. Let's first find the zeros of the numerator and denominator: 
For \(\left( x^2 + 2x - 15 \right): (x + 5)(x - 3) = 0\)
This gives us zeros at x = -5 and x = 3.

For \(\left( x^2 - 7x - 18 \right): (x + 2)(x - 9) = 0\)
This gives us zeros at x = -2 and x = 9.

Now, using these zeros, we get the following intervals:
(-∞, -5), (-5, -2), (-2, 3), (3, 9), and (9, ∞).

Let's test the sign of \(f(x)\) in each of these intervals by picking a test point from each:
1. Test \(( x = -6 ): ( f(-6) ) = \frac{(-)}{  (-)} =\)positive 
2. Test \(\left( x = -3 \right): \left( f(-3) \right) =\frac{ (+) }{ (-) }=\) negative 
3. Test \(\left( x = 0 \right): \left( f(0) \right) = \frac{(-) }{ (-) }=\) positive
4. Test\(\left( x = 6 \right): \left( f(6) \right) =\frac{ (+) }{ (-)} =\)  negative
5. Test \(( x = 10 ): ( f(10) ) =\frac{ (+) }{ (+)} =\) positive
From the above evaluations, \(f(x)\) is negative in the intervals: (-5, -2) and (3, 9).

So, the correct option is option(A): \(-5 < x < -2\) or \(3 < x < 9.\)