1. ## Inequality Help

Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: $x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0$ From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.

2. Originally Posted by OzzMan
Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: $x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0$ From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.
There are various and sundry ways to look at it but I think the most visual, and the easiest to explain is the graphing method.

Graph $(x + 1)(x - 2)(x - 3)$ (see below).

Your solution set is where the graph dips below the x axis, $x \in (-\infty, -1) \cup (2, 3)$

-Dan

3. Wow. Yea thanks man.

4. Oh is my solution set always the part of the graph that dips below the x axis?

5. Originally Posted by OzzMan
Oh is my solution set always the part of the graph that dips below the x axis?
Only because you have a "<" in the problem. You need to change how you get your answer according to whether you have a " $\leq$", " $\geq$", "<", or ">" in the problem.

-Dan

6. Could you help me with this one? Producing the graph and such and telling me what to look for in the graph. $\frac {x-8}{3-x}<0$ The graph is confusing me.

7. Originally Posted by OzzMan
Could you help me with this one? Producing the graph and such and telling me what to look for in the graph. $\frac {x-8}{3-x}<0$ The graph is confusing me.
Well, once again you have a "<" so you are looking for all points on the graph below the x-axis. (I never said it, but when setting up a problem like this you always want to get the RHS to be 0.)

So the only trick is to see where
$\frac{x-8}{3-x} = 0$

$x - 8 = 0$ for $x \neq 3$

$x = 8$.

So the solution set will be $x \in (-\infty, 3) \cup (8, \infty)$.
(The hyperbola moves from negative to positive over the vertical asymptote x = 3, if that's what's bothering you.)

-Dan

8. Alright. So the answer is $T>8\quad T<3$ right?

9. Originally Posted by OzzMan
Alright. So the answer is $T>8\quad T<3$ right?
Yes, though I'm not sure where you got the T from.

-Dan

10. Oh yeah $T$ was the variable used in my book instead of $X$. It is still a little confusing though, sadly. I don't know why but from the looks of the graph it looks like the opposite of what I listed the answers as above. But I'm going to listen to you anyways.

11. Originally Posted by OzzMan
Oh yeah $T$ was the variable used in my book instead of $X$. It is still a little confusing though, sadly. I don't know why but from the looks of the graph it looks like the opposite of what I listed the answers as above. But I'm going to listen to you anyways.
Make sure you understand it. You are looking for regions on the graph where the graph is below the x-axis. That is how you determine what intervals are part of the solution set.

-Dan

12. Yea i think i see what you are saying though about the hyperbola moving from negative to positive over the vertical asymptote of $x=3$. Thanks again.

13. ## number line method solution

Originally Posted by OzzMan
Can someone explain the basics of inequalities to me. I'm having trouble with the easiest of questions, just because the signs. For example: $x^3-4x^2+x+6<0\quad \Rightarrow (x+1)(x-2)(x-3)<0$ From here I have no idea what I'm doing. Can someone explain to me what to do after this point? I know it is something to do with negatives and positives. If someone could explain that to me I'd probably understand these a lot more. Thats my biggest issue.
(x+1)(x-2)(x-3)< 0
The solution range are -1, 2, 3 by using number line method.
________-____+_______-___
| | | | | | | | | | | | | | | |
-2-1 01 2 3
Since
(x +1)(x-2)(x-3)=0
Substitute -2 into x in quadrate equation.
(-2 +1)(-2-2)(-2-3) = -1* -4* -5 = -20
x< -1, -1< x< 3, x >3.
The inequality solution is x < -1 and x > 3

from clement
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