# Non-Linear Inequalities, Sign Charts?

• Sep 22nd 2012, 03:56 PM
samiileigh94
Non-Linear Inequalities, Sign Charts?
I don't really understand how to solve non-linear inequalities after the factoring step? I don't understand how to use sign charts!
For example,
(x-5)(x+4)>=0
or
x^2>=9
???

• Sep 22nd 2012, 05:41 PM
Prove It
Re: Non-Linear Inequalities, Sign Charts?
Quote:

Originally Posted by samiileigh94
I don't really understand how to solve non-linear inequalities after the factoring step? I don't understand how to use sign charts!
For example,
(x-5)(x+4)>=0
or
x^2>=9
???

Well, if you think about where the function actually equals 0, it has to be at the x intercepts. Surely you can figure those out.

Then, we make note of the fact that a function can only ever change from positive to negative, or vice versa, AT the x-intercepts.

Therefore, the x intercepts break up the domain into a number of regions where the function could be positive or could be negative.

Test each of those regions by substituting a point from each of those regions into the function. See if you get positive or negative.

The solution to each inequality should then easily follow.
• Sep 22nd 2012, 05:48 PM
samiileigh94
Re: Non-Linear Inequalities, Sign Charts?
Well I understand the part you've explained, so let me be clearer. I don't understand how the positive and negative changes/leads to the solution. Like, what does it lend to the solution if its positive or negative? If that isn't clear enough, I'd appreciate explanation with an example.
• Sep 22nd 2012, 05:53 PM
Prove It
Re: Non-Linear Inequalities, Sign Charts?
Quote:

Originally Posted by samiileigh94
Well I understand the part you've explained, so let me be clearer. I don't understand how the positive and negative changes/leads to the solution. Like, what does it lend to the solution if its positive or negative? If that isn't clear enough, I'd appreciate explanation with an example.

In the first example you are asked to solve \displaystyle \displaystyle \begin{align*} (x - 5)( x + 4) \geq 0 \end{align*}. So you are asked to find all the points where this function is either 0 or positive. Finding where the function is 0 should be easy. Finding where the function is positive is done by testing points in each region determined by the intercepts (where the function is 0).
• Sep 22nd 2012, 06:01 PM
samiileigh94
Re: Non-Linear Inequalities, Sign Charts?
So the solution would be x>=-4, x>=5?
• Sep 22nd 2012, 06:06 PM
Prove It
Re: Non-Linear Inequalities, Sign Charts?
Quote:

Originally Posted by samiileigh94
So the solution would be x>=-4, x>=5?

No. Like I said, test points in each region. The first region is \displaystyle \displaystyle \begin{align*} (-\infty, -4) \end{align*} so you could test say \displaystyle \displaystyle \begin{align*} x = -5 \end{align*}, the second region is \displaystyle \displaystyle \begin{align*} (-4, 5) \end{align*} so you could test say \displaystyle \displaystyle \begin{align*} x = 0 \end{align*}, and the final region is \displaystyle \displaystyle \begin{align*} (5, \infty) \end{align*} so you could test say \displaystyle \displaystyle \begin{align*} x = 6 \end{align*}.
• Sep 22nd 2012, 06:18 PM
samiileigh94
Re: Non-Linear Inequalities, Sign Charts?
Okay, I get that, but the whole time I've been saying I don't understand how to interpret the results!

• Sep 22nd 2012, 06:23 PM
Prove It
Re: Non-Linear Inequalities, Sign Charts?
Quote:

Originally Posted by samiileigh94
Okay, I get that, but the whole time I've been saying I don't understand how to interpret the results!

OK, you have just shown that when \displaystyle \displaystyle \begin{align*} x = -5, (x - 5)(x + 4) > 0 \end{align*}. That means that ALL POINTS in the region \displaystyle \displaystyle \begin{align*} (-\infty, -4) \end{align*} will be positive. You have that when \displaystyle \displaystyle \begin{align*} x = 0, (x - 5)(x + 4) < 0 \end{align*}, so ALL POINTS in the region \displaystyle \displaystyle \begin{align*} (-4, 5) \end{align*} will be negative. Finally you have that when \displaystyle \displaystyle \begin{align*} x = 6, (x - 5)(x + 4) > 0 \end{align*}, so ALL POINTS in the region \displaystyle \displaystyle \begin{align*} (5, \infty) \end{align*} will be positive.
Your questions asks "at what points is \displaystyle \displaystyle \begin{align*} (x - 5)(x + 4) \geq 0 \end{align*}?"