In Precalculus, it's essential that you can easily and efficiently solve sentences like ‘[beautiful math coming... please be patient]$\,\frac{3x}{2}1 \ge \frac 15  7x\,$’ and ‘[beautiful math coming... please be patient]$\,x^2 \ge 3\,$’ .
The first sentence is an example of a linear inequality in one variable; a prior lesson covers this type of sentence.
For linear inequalities, the variable appears in the simplest possible way—all you have are numbers, times $\,x\,$ to the first power
(i.e., terms of the form $\,kx\,$, where $\,k\,$ is a
real number).
The second sentence is an example of a nonlinear inequality in one variable.
In nonlinear sentences, the variable appears in a more complicated way—perhaps
you have an $\,x^2\,$ (or higher power), or $\,x\,$, or $\,\sin x\,$.
For solving nonlinear inequalities, more advanced tools are needed, which are discussed in detail in this lesson and the next.
There are two basic methods for solving nonlinear inequalities in one variable.
These two methods were introduced in the prior lesson,
Solving Nonlinear Inequalities in One Variable (Introduction).
Both are called ‘test point methods’, because they involve identifying important intervals, and then ‘testing’ a number from each of these intervals.
The two methods are:
There is a KEY IDEA before we get started:
For the ‘twofunction’ test point method, it is necessary to understand the
interaction between two graphs.
We need to know where one of the graphs lies above (or below) the other graph.
As illustrated below, there are only two types of places where a graph can change from being (say)
above another graph to below it—at an intersection point, or at a break in one of the graphs:
at an intersection point (where $\,f(x) = g(x)\,$) 
red is above purple to left of INT red is below purple to right of INT 

where there's a break in the graph of $\,f\,$ or the graph of $\,g\,$ 
red is above purple to the left of the break red is below purple to the right of the break 
It's important to note that there doesn't have to be an above/below change at an intersection point,
and there doesn't have to be an above/below change at a break.
These are just the candidates for places where an above/below change can occur.
That is, the following implications are true:
NO ABOVE/BELOW CHANGE AT THIS INTERSECTION POINT; NO ABOVE/BELOW CHANGE AT THIS BREAK 
In this first example, ‘$\,x^2 \ge 3\,$’ is solved using the ‘truth’ (or ‘twofunction’) method.
A full discussion
accompanies this first solution; after, an inanutshell version of the solution is given.
TRUTH OF ‘$\,x^2\ge 3\,$’ 
Here is the ‘inanutshell’ version of the previous solution.
This is probably the minimum amount of work that a teacher would want you to show.
In the exercises in this section, you will be expected to show all of these steps.
The comments are for your information,
and do not need to be included in your solutions.
$x^2 \ge 3$  original sentence 
$f(x) := x^2$, $g(x) := 3$  if desired, define functions $\,f\,$ (the lefthand side) and $\,g\,$ (the righthand side), so they can be easily referred to in later steps 
$x^2 = 3$ $x = \pm\sqrt 3$ (no breaks in either graph) 
find where $\,f(x)\,$ and $\,g(x)\,$ are equal (intersection points); find any breaks in the graphs of $\,f\,$ and $\,g\,$ 
TRUTH OF ‘$\,x^2 \ge 3\,$’ 

solution set: $(\infty,\sqrt 3] \cup [\sqrt 3,\infty)$ sentence form of solution: $x\le \sqrt 3\ \ \text{ or }\ \ x \ge\sqrt 3$ 
read off the solution set, using correct interval notation; or, give the sentence form of the solution 
Here is a second example, where there is both a break in one of the graphs, and an intersection point.
Only the inanutshell version of the solution is given.
YOU WRITE THIS DOWN  COMMENTS 
$\displaystyle \frac 1x \lt 2$  original sentence 
$\displaystyle f(x) := \frac 1x\,$, $g(x) := 2$  if desired, name the function on the lefthand side $\,f\,$ and the function on the righthand side $\,g\,$, so they can be easily referred to in later steps; recall that ‘$:=$’ means ‘equals, by definition’ 
When $\,x = 0\,$ there is a break, since division by zero is not allowed. For $\,x\ne 0\,$: $\displaystyle \frac 1x = 2$ $x = \frac 12$ (take reciprocals of both sides: if two numbers are equal, so are their reciprocals) 
identify the candidates for above/below relationship changes:

TRUTH OF ‘$\displaystyle\, \frac 1x \lt 2\,$’ 

solution set: $(\infty,0) \cup (\frac 12,\infty)$ sentence form of solution: $x < 0\ \ \text{ or }\ \ x > \frac 12$ 
Read off the solution set, using correct interval notation; or, give the sentence form of the solution. 
Just for fun, jump up to wolframalpha.com and key in the two examples explored in this lesson:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. For graphical insight into the solution set, a graph is sometimes displayed.
Click the “show/hide graph” button if you prefer not to see the graph. 
PROBLEM TYPES:
