Using interval notation to express all real numbers less than or equal to
a Or greater than or equal to
b
Write the interval expressing all real numbers less than or equal to
$\text{\hspace{0.17em}}\mathrm{-1}\text{\hspace{0.17em}}$ or greater than or equal to
$\text{\hspace{0.17em}}1.$
We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at
$\text{\hspace{0.17em}}-\infty \text{\hspace{0.17em}}$ and ends at
$\text{\hspace{0.17em}}\mathrm{-1},$ which is written as
$\text{\hspace{0.17em}}\left(-\infty ,\mathrm{-1}\right].$
The second interval must show all real numbers greater than or equal to
$\text{\hspace{0.17em}}1,$ which is written as
$\text{\hspace{0.17em}}\left[1,\infty \right).\text{\hspace{0.17em}}$ However, we want to combine these two sets. We accomplish this by inserting the union symbol,
$\cup ,$ between the two intervals.
When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the
addition property and the
multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.
These properties also apply to
$\text{\hspace{0.17em}}a\le b,$$a>b,$ and
$\text{\hspace{0.17em}}a\ge b.$
Demonstrating the addition property
Illustrate the addition property for inequalities by solving each of the following:
(a)
$x-15<4$
(b)
$6\ge x-1$
(c)
$x+7>9$
The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.
Solving inequalities in one variable algebraically
As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.
Solving an inequality algebraically
Solve the inequality:
$\text{\hspace{0.17em}}13-7x\ge 10x-4.$
Solving this inequality is similar to solving an equation up until the last step.
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.