Question

Solve the inequality for x given that (x-6)/(x-3)(x-9)<0
a) (-inf,-6)U(6,9)
b) (-inf,3)U(6,inf)
c) (-inf,6)U(6,inf)
d) (-inf,-3)U(6,9)
e) (-inf,3)U(6,9)

Answer 1

this is not as hard as it looks
changes sign at 3,6,9

Answer 2

divide number line up in to four intervals
pick some interval and see if it is positive or negative there there, then you will know all

Answer 3

3,6,9 would all be open circles right?

Answer 4

yeah but you don't need no circles, you need intervals

Answer 5

\[(-\infty, 3),(3,6),(6,9),(9\infty)\] are your intervals
check on one then alternate

Answer 6

somehow i got that (-inf,3)U(3,6)U(9,inf)

Answer 7

I have no idea what im doing wrong

Answer 8

oh wait is it e?

Answer 9

you have two adjacent intervals \((-\infty,3)\) and \((3,6)\) that can't be right

Answer 10

it changes sign at 3 so if it is positive on one interval it must be negative on the other
did you check on any one interval?
with some practice you won't have too, but it doesn't hurt

Answer 11

i did a little math wrong and forgot that i'm looking for negatives, not positives

Answer 12

what interval did you check?

Answer 13

i got that (-inf,3) was negative, (3,6) was positive, (6,9) was negative, and (9,inf) was positive

Answer 14

i would also point out that one only of the above answers is possible, even if you didn't check anything

Answer 15

yeah at the risk of repeating myself you only have to check one
if you know it is negative on \((-\infty,3)\) then it must be
\[\huge -,+,-,+\]

Answer 16

so the correct answer is (-inf,3)(6,9), right?

Answer 17

and yeah it is E

Answer 18

The graph confirms the answer you got

https://www.desmos.com/calculator/ipckz8kxhy

the portion on the interval (-inf, 3) is below the x axis
the portion on the interval (6,9) is below the x axis
everything else is either on the x axis or above it