SOLUTION: Um..for f(x)=1/x^2-2x-8 I need to find the interval/intervals where f(x)<0 and I cant seem to figure it out.

Question 65556: Um..for f(x)=1/x^2-2x-8 I need to find the interval/intervals where f(x)<0 and I cant seem to figure it out.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
for f(x)=1/x^2-2x-8 I need to find the interval/intervals where f(x)<0
f(x)=1/(x-4)(x+2)
You have vertical asymptotes at x=-2 and at x=4
Draw a number line.
Mark x=-2 and x=4 on the line
That breaks the line into 3 intervals;
Determine if f(x)<0 in each of those intervals, as follows.
If x=-10 f(-10)=1/(-)(-) which is positive; so no solutions in (-inf,-2)
If x=0 f(0)=1/(-)(+) which is negative; so (-2,4) is part of the solution
If x=10 f(10)=1/(+)(+) which is positive; so no solutions in (4,inf)
SOLUTION: -2 Cheers,
Stan H.
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