Question
>\(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\) is negative if and only if
>\(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\) is negative if and only if
- 5 < x < -2 or 3 < x < 9
- x < -5 or -2 < x < 3
- 2 < x < 3 or x > 9
- x < -5 or 3 < x < 9
Option: 1
Solution:
\(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\)<0
\(\frac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{\left( {x - 9} \right)\left( {x + 2} \right)}} < 0\)
We have four inflection points -5, -2, 3, and 9.
For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.
When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.
We are left with the range (-5,-2) and (3,9) where the expression will be negative.
Solution:
\(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\)<0
\(\frac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{\left( {x - 9} \right)\left( {x + 2} \right)}} < 0\)
We have four inflection points -5, -2, 3, and 9.
For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.
When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.
We are left with the range (-5,-2) and (3,9) where the expression will be negative.
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