Absolute Value

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Absolute value makes negative numbers positive and leaves nonnegative numbers alone. Of course, there is no algebra called ”make positive”. Instead, we need a piecewise defined function.

Absolute Value $|k| = \begin {cases} -k & \text {if $k<0$,}\\ k & \text {if $k\ge 0$}. \end {cases}$

The symbol for the absolute value function are two vertical bars surrounding the domain value.

Graph of $y = |k|$ .

Shifting

Graph of $y = P(t) = |t+4|$ .

The graph is positioned by its corner, which occurs when the inside of the absolute value formula equals $0$ . $t+4=0$ , when $t=-4$ .

$P$ has a global minimum of $\answer {0}$ , which occurs at $\answer {-4}$ . $P$ has no maximums.

$P$ decreases on $(-\infty , -4]$ and increases on $[-4, \infty )$

$P(t) = |t+4| = \begin {cases} -(t+4) & \text {if $t<\answer {-4}$,}\\ t+4 & \text {if $t\ge \answer {-4}$}. \end {cases}$

Solving Equations

Let $B(r) = |r-3| + 2$ .

$\blacktriangleright$ Solve $B(r) = 5$ .

We can see from the graph that there are two solutions to this equation. Our first goal in solving this equation is to get the absolute term by itself.

$|r-3| + 2 = 5$

$|r-3| = 3$

There are only two numbers whose absolute value is $3$ . They are $\answer {-3}$ or $\answer {3}$ .

We have two possibilities, either the inside of the absolute value, $r-3=-3$ or $r-3=3$ .

If $r-3=-3$ , then $r=0$ . If $r-3=3$ , then $r=6$ .

We can also turn to the definition of the absolute value function.

The function $|r-3| + 2$ can be written as

$|r-3| + 2 = \begin {cases} -(r-3) + 2 & \text {if $t<\answer {3}$,}\\ r-3 + 2 & \text {if $t\ge \answer {3}$}. \end {cases}$

We are looking for where $-(r-3) + 2 = 5$ and $t < 3$

where $(r-3) + 2 = 5$ and $t \ge 3$ .

These produce the solutions $r=0$ and $r=6$

Behavior

Formulas for absolute value functions look like

$abs(x) = A | B \, x + C | + D$

where $A \ne 0$ , $B \ne 0$ , $C$ , and $D$ are real numbers.

The critical number for an absolute value function is the domain number that makes the inside of the absolute value bars equal to $0$ .

$B \, x + C = 0$

The absolute value function increases on one side of the critical number and decreases on the other. The sign of the leading coefficient determines which way it goes.

If $A > 0$ , then decreasing switching to increasing.
If $A < 0$ , then increasing switching to decreasing.

End-Behavior

The end-behavior is the same on both sides for an absolute value function. The sign of the leading coefficient determines which way it goes.

If the leading coefficent is positive, then $\lim _{x \to -\infty } abs(x) = \infty$

$\lim _{x \to \infty } abs(x) = \infty$

If the leading coefficent is negative, then $\lim _{x \to -\infty } abs(x) = -\infty$

$\lim _{x \to \infty } abs(x) = -\infty$

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