These statements also apply to
$\text{\hspace{0.17em}}\left|X\right|\le k\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\left|X\right|\ge k.$
Determining a number within a prescribed distance
Describe all values
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ within a distance of 4 from the number 5.
We want the distance between
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 5 to be less than or equal to 4. We can draw a number line, such as in
[link], to represent the condition to be satisfied.
The distance from
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to 5 can be represented using an absolute value symbol,
$\text{\hspace{0.17em}}\left|x-5\right|.\text{\hspace{0.17em}}$ Write the values of
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that satisfy the condition as an absolute value inequality.
$\left|x-5\right|\le 4$
We need to write two inequalities as there are always two solutions to an absolute value equation.
If the solution set is
$\text{\hspace{0.17em}}x\le 9\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}x\ge 1,$ then the solution set is an interval including all real numbers between and including 1 and 9.
So
$\text{\hspace{0.17em}}\left|x-5\right|\le 4\text{\hspace{0.17em}}$ is equivalent to
$\text{\hspace{0.17em}}\left[1,9\right]\text{\hspace{0.17em}}$ in interval notation.
Using a graphical approach to solve absolute value inequalities
Given the equation
$y=-\frac{1}{2}|4x-5|+3,$ determine the
x -values for which the
y -values are negative.
We are trying to determine where
$\text{\hspace{0.17em}}y<0,$ which is when
$\text{\hspace{0.17em}}-\frac{1}{2}|4x-5|+3<0.\text{\hspace{0.17em}}$ We begin by isolating the absolute value.
Now, we can examine the graph to observe where the
y- values are negative. We observe where the branches are below the
x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at
$\text{\hspace{0.17em}}x=-\frac{1}{4}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}x=\frac{11}{4},$ and that the graph opens downward. See
[link].
$k\le 1\text{\hspace{0.17em}}$ or
$\text{\hspace{0.17em}}k\ge 7;$ in interval notation, this would be
$\text{\hspace{0.17em}}(-\infty ,1]\cup [7,\infty ).$
Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See
[link] and
[link].
Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See
[link],[link] ,
[link] , and
[link].
Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See
[link] and
[link] .
Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See
[link] and
[link].
Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See
[link] .
show that the set of natural numberdoes not from agroup with addition or multiplication butit forms aseni group with respect toaaddition as well as multiplication