The previous four lessons give a thorough discussion of solving inequalities:
Linear inequalities, like ‘$\,\frac{3x}{2}1 \ge \frac 15  7x\,$’,
can be solved by using familiar tools to get $\,x\,$ all by itself on one side, and a number on the other side.
The only thing different from working with equations is that
if you multiply or divide an inequality by a negative number,
then you must change the direction of the inequality symbol.
Nonlinear inequalities, like
‘$\,x^2 \ge 3\,$’, have the variable
appearing in a more complicated way.
They require more advanced tools—and that's where the ‘test point method’ comes into the picture.
The ‘test point method’ can also be used for linear inequalities, but it's like killing a mosquito with
a sledgehammer—that is,
you'd be working much harder than needed to accomplish the task.
The ‘test point method’ involves identifying important intervals, and then ‘testing’ a number from each interval—so the name is appropriate.
There are two slightly different ‘flavors’ of the test point method,
but only one is discussed here.
Every inequality can be written in a form with zero on the right hand side.
The solution sets (the values of $\,x\,$ that make each sentence true) are determined by
which inequality symbol is used:
INEQUALITY  SOLUTION SET 
$f(x) > 0$  value(s) of $\,x\,$ for which the graph of $\,f\,$ is ABOVE the $x$axis 
$f(x) \ge 0$  value(s) of $\,x\,$ for which the graph of $\,f\,$ is ON or ABOVE the $x$axis 
$f(x) < 0$  value(s) of $\,x\,$ for which the graph of $\,f\,$ is BELOW the $x$axis 
$f(x) \le 0$  value(s) of $\,x\,$ for which the graph of $\,f\,$ is ON or BELOW the $x$axis 
There are only two types of places where the graph of a function $\,f\,$ can go from above to
below the $x$axis (or vice versa).
That is, there are only two types of places where a
function $\,f\,$ can change its sign (from positive to negative, or vice versa):
For example, suppose a function has a zero at $\,x = 1\,$ and a break at $\,x = 3\,$. That's it—no other zeros or breaks. You've marked these on a number line, like the one shown at right. Notice that these two numbers break the number line into three subintervals. So, what could the graph look like on (say) the subinterval from $\,1\,$ to $\,3\,$? Could the graph be partly above the $\,x\,$axis, and partly below, on this subinterval? NO!! Because if it were, then there would have to be another zero or break where it switches—and there aren't any other zeros or breaks. So, you can pick any number between $\,1\,$ and $\,3\,$ (called the ‘test point’) and check whether you're above or below there—and then you know that this same behavior holds on the entire subinterval. That's the beauty of the test point method! 
The steps outlined here are for the ‘one function (compare with zero)’ flavor of the test point method.
STEPS IN TEST POINT METHOD  EXAMPLE: Solve ‘$\,x^2 + 5x > 6\,$’  COMMENTS 
STEP 1:

$\,\overbrace{x^2 + 5x + 6}^{:= f(x)} > 0$ 

STEP 2:

$$ \begin{gather} x^2 + 5x + 6 = 0\cr (x+2)(x+3) = 0\cr x = 2\ \ \text{ or } x = 3 \end{gather} $$ 

STEP 3:

(These calculations, mostly done mentally, are used to construct the green part of the test point diagram shown below.) $$f(x) = x^2 + 5x + 6 = (x+2)(x+3)$$ $T = 4$: $f(4) = (4+2)(4+3) = ()() = +$ $T = 2.5$: $f(2.5) = (2.5+2)(2.5+3) = ()(+) = $ $T = 0$: $f(0) = (0+2)(0+3) = (+)(+) = +$ 

STEP 4:

solution set of ‘$\,f(x) > 0\,$’: $(\infty,3)\cup (2,\infty)$ sentence form of solution set: $x < 3\ \ \text{ or }\ \ x > 2$ 

STEP 5 (optional):

$4$ is in the solution set:
$$\begin{gather}
(4)^2 + 5(4) \overset{?}{\ >\ } 6\cr
16  20 \overset{?}{\ >\ } 6\cr
4 > 6\ \ \text{TRUE!}
\end{gather}
$$
$2.5$ is not in the solution set: $$\begin{gather} (2.5)^2 + 5(2.5) \overset{?}{\ >\ } 6\cr 6.25 \overset{?}{\ >\ } 6\ \ \text{FALSE!} \end{gather} $$ 

THE TEST POINT DIAGRAM 

Here's a second example, where the function has both zeros and a break.
Only the bare minimum is shown here—when you get good at this method,
this is probably all you'll need to write down.
In this example, the choice was made to take a few steps and rename the function $\,f\,$ as a single fraction. Then:
STEPS IN TEST POINT METHOD  EXAMPLE: Solve ‘$\displaystyle\,x \le \frac{2}{x1}\,$’ 
STEP 1:

$\displaystyle\,\overbrace{x  \frac{2}{x1}}^{:= f(x)} \le 0$ 
STEP 2:

$$f(x) = x  \frac{2}{x1}
= \frac{x(x1)2}{x1}
= \frac{x^2  x  2}{x1}
= \frac{(x2)(x+1)}{x1}
$$
zeros: $\,2, 1\,$ break: $\,1$ 
STEP 3:


STEP 4:

solution set of ‘$\,f(x) \le 0\,$’: $(\infty,1]\cup (1,2]$ sentence form of solution set: $x\le 1\ \ \text{ or }\ \ 1 \lt x \le 2$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
