SOLUTION: FOR f(x)= 1 OVER x2-2x-8, FIND THE INTERVAL(S) WHERE f(x)<0.

Question 62922: FOR f(x)= 1 OVER x2-2x-8, FIND THE INTERVAL(S) WHERE f(x)<0.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
FOR f(x)= 1 OVER x^2-2x-8, FIND THE INTERVAL(S) WHERE f(x)<0.
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Find the zeroes of the denominator:
x^2-2x-8=0
(x-4)(x+2)=0
x=-2 or x=4
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Draw a number line and mark two points as x=-2 and x=4.
That breaks the number line into three intervals:
I: (-inf,-2)
II: (-2,4)
III: (4,inf)
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Check the inequality y=1/[(x+2)(x-4)] with a value from each interval
to determine where the solution set resides.
I: If x=-100, 1/(-102)(-96)<0 is false, so no solutions in interval I.
II: If x=0, 1/(-4)(+2)<0 is true, so interval II is part of the solution set.
III: If x=100, 1/(98)(102)<0 is false, so no solutions in interval III.
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Final Answer: -2 Cheers,
Stan H.
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