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;
the prior lesson covers this type of sentence.
For linear inequalities, the variable appears in the simplest possible wayall 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, and is covered in this lesson and the next two. In nonlinear sentences, the variable appears in a more complicated wayperhaps you have an $\,x^2\,$ (or higher power), or $\,|x|\,$, or $\,\sin x\,$.
For solving nonlinear inequalities, more advanced tools are needed.
To get started, review the essential concepts in these two lessons,
being sure to click-click-click to practice the concepts from each:
There are two basic methods for solving nonlinear inequalities in one variable.
Both are called ‘test point methods’,
because they involve identifying important intervals,
and then ‘testing’ a number from each of these intervals.
Below, the sentence ‘$\,x^2 \ge 3\,$’ is solved using both methods,
so you can get a sense of which appeals to you more.
The solutions below are written extremely compactlythis is pretty much the bare minimum that a teacher would want to see.
For all the underlying concepts and details, study the next two lessons:
YOU WRITE THIS DOWN | COMMENTS |
$x^2 \ge 3$ | original sentence |
$x^2 - 3\ge 0$ |
rewrite the inequality with zero on the right-hand 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 left-hand 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 $\,f\,$:
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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 |
When a graph is easy to obtain (as in this example),
then you may not need to ‘officially’ use the test point method.
Just graph $\,f(x) = x^2 - 3\,$ (see below);
the solutions of ‘$\,x^2 - 3 \ge 0\,$’
are the values of $\,x\,$ for which the graph of $\,f\,$ lies on or above the $x$-axis.
$x^2 \ge 3$ | original sentence |
$f(x) := x^2$, $g(x) := 3$ |
if desired,
define functions $\,f\,$ (the left-hand side) and $\,g\,$ (the right-hand 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 $\,g\,$ |
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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 |
When graphs are easy to obtain (as in this example),
then you may not need to ‘officially’ use the test point method.
Just graph $\,f(x) = x^2\,$ (in red below) and $\,g(x) = 3\,$ (in green below):
then, the solutions of ‘$\,x^2 \ge 3\,$’ are the values of $\,x\,$ for which
the graph of $\,f\,$ lies on or above the graph of $\,g\,$.
Just for fun, jump up to wolframalpha.com
and key in: x^2 >= 3
Voila!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
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