Posted by ted on Tuesday, July 24, 2012 at 8:07am.
You must mean:
3x-2 < (x+4)/(x-2) , so clearly x ≠ 2
(x-2)(3x-2) < x+4
3x^2 - 8x + 4 - x - 4 < 0
3x^2 - 9x < 0
3x(x-3) < 0
So the "critical values" are x=0 , x = 2, and x=3
test for a value of x<0
let x=-5 in original
-17 < -1/-7
-17 < 1/7 ? YES
test for a value between 0 and 2
let x=1
1 < 5/-1
1 < -5 ? NO
test for a value between 2 and 3
let x = 2.5
5.5 < 6.5/.5
5.5 < 13 ? , YES
test for a value x>3
let x=5
13 < 9/3
13 < 3 ? , NO
so we have x ≤ 0 OR 2 < x ≤ 3 , notice that x=2 in not included
To see that this is correct
go to my favourite graphing program
http://rechneronline.de/function-graphs/
and enter
3x-2 - (x+4)/(x-2)
in the window for "first graph"
change the "Range y-axis from" entries to -200 to 200
you will see the curve below the x-axis from -infinitity to 0
above the x-axis from 0 to the asymptote of 2
below the x-axis from 2 to 3, and
above the x-axis for x>3
thanks so much :)
what if all the values you put in do not work. how would you write the intervals.?
If the inequality can not be satisfied for any real value of x, then you just have to say that.