In Precalculus, it's essential that you can easily and efficiently solve sentences like:
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 are a few KEY IDEAS before we get started:
For example, subtracting $\,3\,$ from both sides of ‘$\,x^2 \ge 3\,$’ gives the equivalent inequality ‘$\,x^2  3\ge 0\,$’.
where $\,f(x) = 0$ (at a zero of the function) 
SIGN CAN CHANGE AT A ZERO 

where there's a break in the graph of $\,f$ 
SIGN CAN CHANGE AT A BREAK 
It's important to note that there doesn't have to be a sign change at a zero,
and there doesn't have to be a sign change at a break.
These are just the candidates for places where a sign change can occur.
That is, the following implications are true:
NO SIGN CHANGE AT THIS ZERO; NO SIGN CHANGE AT THIS BREAK 
In this first example, ‘$\,x^2 \ge 3\,$’ is solved using the ‘compare to zero’ (or ‘onefunction’) method.
A full discussion accompanies this first solution;
after, an inanutshell version of the solution is given.
SIGN OF $\,f(x)\,$ 
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.
YOU WRITE THIS DOWN  COMMENTS 
$x^2 \ge 3$  original sentence 
$x^2  3\ge 0$ 
rewrite the inequality with zero on the righthand side; since the inequality symbol is ‘$\,\ge\,$’, we need to determine where the graph of $\,x^2  3\,$ lies on ($\,=\,$) or above ($\,\gt\,$) the $x$axis 
$f(x) := x^2  3$ 
if desired, name the function on the lefthand side $\,f\,$, so it can be easily referred to in later steps; recall that ‘$:=$’ means ‘equals, by definition’ 
$x^2  3 = 0$ $x^2 = 3$ $x = \pm\sqrt 3$ (no breaks in graph of $\,f\,$) 
identify the candidates for sign changes for $\,f\,$:

SIGN OF $\,f(x)\,$ 

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. For this problem, since the inequality symbol is ‘$\,\ge\,$’, we need to read off all values corresponding to plus signs and zeroes. 
Here is a second example, where there is both a break in the graph and a zero.
Only the inanutshell version of the solution is given.
YOU WRITE THIS DOWN  COMMENTS 
$\displaystyle \frac 1x \lt 2$  original sentence 
$\displaystyle \frac 1x  2 \lt 0$ 
rewrite the inequality with zero on the righthand side; since the inequality symbol is ‘$\,\lt\ $’, we need to determine where the graph of $\displaystyle\,\frac 1x  2\,$ lies below the $x$axis 
$\displaystyle f(x) := \frac 1x  2$ 
if desired, name the function on the lefthand side $\,f\,$, so it 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 $\,x\ne 0\,$: $\displaystyle \frac 1x  2 = 0$ $\displaystyle \frac 1x = 2$ $x = \frac 12$ (take reciprocals of both sides: if two numbers are equal, so are their reciprocals) 
identify the candidates for sign changes for $\,f\,$:

SIGN OF $\,f(x)\,$ 

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. For this problem, since the inequality symbol is ‘$\,\lt\,$’, we need to read off all values corresponding to minus signs. 
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.
Equivalently, you are finding where the red graph lies on or above the $x$axis. Click the “show/hide graph” button if you prefer not to see the graph. 
PROBLEM TYPES:
