The critical values are -2, 1 ,and 3 , with of course x ? 1
I then pick an arbitrary value in each of the regions and determine the sign of the answer, the actual answer does not matter.
e.g let x = -5
(-)(-)/(-1) < 0 , so x <-2 works
let x = 0 , a value between -2 and 1
(-3)(+).(-) > 0 , no good
let x = 2 , a value between 1 and 3
(-)(+)/(+) < 0 , good
let x = 5 , a value > 3
(+)(+)/(+) > 0 , no good
so
x < -2 OR 1 < x < 3
How to solve this:
((x-3)(x+2))/((x-1)) < 0
I don't understand since theres x-1 in the denominator
2 answers
so for this: (-)(-)/(-1) < 0 , so x <-2 works
you replace the 0 with -2 ? .
Also for this part let x = 5 , a value > 3
(+)(+)/(+) > 0 , no good
you had to change the sign because it was all positive?
you replace the 0 with -2 ? .
Also for this part let x = 5 , a value > 3
(+)(+)/(+) > 0 , no good
you had to change the sign because it was all positive?