# Thread: [SOLVED] Issues finding the domain of this inequality

1. ## [SOLVED] Issues finding the domain of this inequality

I need to find the domain of
$\displaystyle \frac{(x-1)(3-x)}{(x-2)^2}<=0$

my choices are:
$\displaystyle A) (-\infty, 1) or (3, \infty)$
$\displaystyle B) (-\infty, -3) or(-1,\infty)$
$\displaystyle C) (-\infty, -3] or (-2,-1)or[1,\infty)$
$\displaystyle D) (-\infty, 1] or [3, \infty)$

Im coming up with
$\displaystyle 1<=x<2 or 2<x<=3$
What am I doing wrong?

2. Originally Posted by ninobrn99
I need to find the domain of $\displaystyle \frac{(x-1)(3-x)}{(x-2)^2}<=0$
The "domain" is all allowable x-values, and relates to functions, not inequalities. Were you maybe actually supposed to find the solution intervals for the rational inequality...?

Originally Posted by ninobrn99
What am I doing wrong?
Until you show your work, I'm afraid there is no way to answer this question. Sorry!

You have three different factors, one of which occurs twice. Find the zeroes of each factor. These zeroes will divide the number line into intervals.

Check the sign of the rational expression on each interval (by picking a random x-value within the interval and plugging it into the expression). The solution intervals will be those for which you got a negative value.

3. Originally Posted by ninobrn99
I need to find the domain of
$\displaystyle \frac{(x-1)(3-x)}{(x-2)^2}<=0$

my choices are:
$\displaystyle A) (-\infty, 1) or (3, \infty)$
$\displaystyle B) (-\infty, -3) or(-1,\infty)$
$\displaystyle C) (-\infty, -3] or (-2,-1)or[1,\infty)$
$\displaystyle D) (-\infty, 1] or [3, \infty)$

Im coming up with
$\displaystyle 1<=x<2 or 2<x<=3$
What am I doing wrong?
Since you don't show what you are doing, it's hard to say what you are doing wrong!
Did you miss the fact that the denominator is squared? It doesn't matter whether x-2 is positive or negative, the square is positive.

4. well i know it cant be 2 for the denom. so {x|x=/2}

for the numerator:
I have:
$\displaystyle (-\infty,1)U(3,\infty)$
because the zeros of the numerator are 1 and 3.
where Im confused on is since x cant be 2, that does exclude it from my solution set, so im left with 1 and 3 only?
im trying to follow the example in my book.

5. Originally Posted by ninobrn99
well i know it cant be 2 for the denom. so {x|x=/2}
I think you mean that the zero of the denominator is x = 2, so that this is one of your interval endpoints. But since you cannot divide by zero, the endpoint will not actually be included in any interval.

Originally Posted by ninobrn99
for the numerator: I have:
$\displaystyle (-\infty,1)U(3,\infty)$
because the zeros of the numerator are 1 and 3.
where Im confused on is since x cant be 2, that does exclude it from my solution set, so im left with 1 and 3 only?
I'm not sure what you mean by being "left with 1 and 3 only"...?

If you mean that x = 1 and x = 3 are the zeroes of the numerator, and thus endpoints to intervals, then you are correct. Obviously, it's okay to have zero in the numerator of a fraction, so x = 1 and x = 3 will be acceptable inside the expression, and (possibly) as part of the solution to the inequality.

Now that you have the endpoints (and thus know what the four intervals are), try doing the testing explained earlier (or use the "table" method explained in the link provided earlier) to figure out which of the intervals gives the correct sign for the inequality.

6. thank you both for your assistance. I appreciate it.
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