for f(x)=1 divided by x2-2x-8,find the interval where f(x)<0.?

Answer 1

f(x) = 1/(x^2-2x - 8)
f(x) = 1/[(x-4)(x+2)]
f(x) is negative for all x less than 4
This is because the term x-4 will be negative for all x less than 4 and so f(x) will be negative (less than 0) for all x less than 4.

Note that f(x) is undefined for x = 4..

Answer 2

That equals 1/((x+2)(x-4)).
SO, look at intervals of x<-2, -24.

AT x<-2, its greater then 0, at -24, its greater then 0.

So, f(x)<0 at -2

Answer 3

given
f(x)=1/(x^2-2x-8)
for f(x)<0
i.e.1/(x^2-2x-8) <0

for the f(x) to be less than zero, the denominatore has to be less than zero..from this condition we can imply that..to satisfy f(x)<0 the following condition has to be fulfilled....

x^2-2x-8<0

=> (x-2)^2-12<0
=>(x-2)^2 - (2sqrt3)^2

=> (x-2-2sqrt3)(x-2+2sqrt3)<0

therefore,
2(1-sqrt3)< x <2(1+sqrt3)