((x-2)(x+3))/(x-1) is greater or equal to zero. Graph the solution set.?

Answer 1

It's greater than zero if the numerator and denominator have the same sign. This will happen if x > 2, as both will be positive, or if x is between -3 and 1, in which case both would be negative. If the numerator equals zero (but the denominator doesn't), the whole thing equals zero. This happens at 2 and -3. So the solution set is all x such that x>2 or -3 <= x < 1.

Make your number line, then put a filled-in dot over the 2 and draw a ray going to the right. Put a filled-in dot over the -3 and an open dot over the 1 and draw a segment between them.

Answer 2

Well: Let f(x) = ((x - 2)(x + 3))/(x - 1)

if x<3, then x-2<-5<0, x+3<0 and x-1<-4<0. thus f(x)<0 for x<-3
if -3≤x<1 then x-2≤-1<0, 0≤x+3, and x-1<0, thus f(x)≥0 for -3≤x<1
(Note that x cannot equal 1)
if 14>0 and x-1>0, thus f(x)<0 for 1 if x≥2, then x-2≥0, x+3≥5>0, and x-1≥1>0, thus f(x)≥0 for x≥2

Therefore f(x)≥0 when -3≤x<1 or x≥2


The graph would be on the real line (number line) and would be:

---------------------[3____ ______1)---- --[2____ ________ ______ __∞)

Where the -------- are not bold, and the ________ is. Please excuse that I had to break the number line up.

Answer 3

[(x-2)(x+3)]/(x-1)>=0

x -inf -3 1 2 inf
-----------------------------------------------------------
x-2 - - - - 0 + +
-------------------------------------------------------------
x+3 - - - - - 0 + + + + + +
--------------------------------------------------------------
x-1 - - - - - 0 + + + +
--------------------------------------------------------------
(x-2)(x+3) - 0 + / - 0 +
/(x-1)

So the solution set is x is in [-3 , 1) U [2 , Inf)
This is graphed on a line.

----------------[/////////////////)----[//////////////////////////////////////
-3 1 2 Inf]

I'm sorry this looks so bad here, see what values each factor
has on the real line and then multiply the signs. Because you have two multiplications and a division.

Answer 4

i used to know how to answer this question last year. but i don't any more. sorry! i remember something about x-int, y-int, and the vert. and horiz. asymptotes. one of them is where y is undefined, right? i just don't remember what the graph looks like...

Answer 5

eulercrosser's answer is correct.