Absolute Value

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distance

The core absolute value function, $|x|$, gives the distance on the number line between a number, $x$, and $0$. Since distance cannot be negative, it appears that the function just “makes numbers positive”.

To do this, the core absolute value function just returns nonnegative numbers unharmed, and makes negative numbers turn positive. Algebraically, this is accomplished through a piecewise defined function.

\[ |x| = \begin{cases} -x &\text {if $x<0$,}\\ x & \text {if $x\ge 0$}. \end{cases} \]

Technically speaking, $|-3| = -(-3)$ and then $-(-3) = 3$. The core absolute value function negates negative numbers.

Note: Just because you see a negative sign doesn’t mean you have a negative number.

$|4| = 4$
$|0| = 0$
$|-\pi | = -(-\pi ) = \pi $
$|\cos (\pi )| = |-1| = -(-1) = 1$
$|\sin (\tfrac {\pi }{4})| = \tfrac {1}{\sqrt {2}}$
$-(-8) = 8$

$|-\sqrt {5}| = \answer {\sqrt {5}}$
$|4-4| = \answer {0}$
$\left |\frac {-4}{-3}\right | = \answer {\frac {4}{3}}$
$|\cos (\tfrac {\pi }{2})| = \answer {0}$
$|\sin (\tfrac {3\pi }{2})| = \answer {1}$
$|\tan (\tfrac {3\pi }{4})| = \answer {1}$

There is no arithmetic operation called “make positive”. If a number is negative, then you make it positive by negating it.

Negating is arithmetic. “Make positive” is not arithmetic.

Graph of $y = A(t) = |t|$.

Solve $|m| = 18$.

Either $m = 18$ or $m = -18$

Continuity

The absolute value function is a continuous function.

On $(-\infty , 0)$ we have $| x | = -x$, which is a linear function, which is continuous.
On $[0, \infty )$ we have $| x | = x$, which is a linear function, which is continuous.

Both sides match up at $0$, since

\[ \lim \limits _{x \to 0^-} |x| = 0 \]

That graph suggests that $|x|$ decreases on $(-\infty , 0]$ and increases on $[0, \infty )$.

This is confirmed by the formula.

On $(-\infty , 0]$, $|x| = -x$, which is a decreasing linear functions.

On $[0, \infty )$, $|x| = x$, which is an increasing linear functions.

Behavior

Using the $iRoC$, or derivative, to say where $A(x) = | x |$ increasing and decreasing.

explanation

First, get rid of the absolute value bars.

$x < 0$ when $|x| = \answer {-x}$
$x > 0$ when $|x| = \answer {x}$

\[ |x| = \begin{cases} -x & \text { on } (-\infty , 0) \\ x & \text { on } [0, \infty ) \end{cases} \]

Each piece is a restricted linear function. We can get their $iRoC$.

\[ iRoC_{|x|} = \begin{cases} -1 & \text { on } (-\infty , 0) \\ 1 & \text { on } (0, \infty ). \end{cases} \]

$iRoC_{|x|}$ is negative on $(-\infty , 0)$ and positive on $(0, \infty )$.

$| x |$ is decreasing on $(-\infty , 0)$ and increasing on $(0, \infty )$.

Absolute value is a continuous function, which is decreasing on $(-\infty , 0)$ and increasing on $(0, \infty )$. That makes $| 0 | = 0$ the global minimum.

Since the absolute value function is an unbounded increasing linear function on $(0, \infty )$, we know that

\[ \lim \limits _{x \to \infty }| x | = \infty \]

Therefore, the range is $[0, \infty )$.

Critical Numbers

What are the cricital numbers for $A(x) = | x |$ ?

explanation

First, $0$ is in the domain.

The previous examples shows that the $iRoC_{|x|}$ is defined and nonzero everywhere except $0$.

The only possible critical number is $0$. We will show that there is no tangent line at $(0,0)$.

Suppose there is a tangent line to the graph of $y = | x |$ at $(0,0)$. Then, it has to match the slope on the left, which is $-1$ and it has to match the slope on the right, which is $1$. A line cannot have two slopes.

There is no tangent line at $(0,0)$, which means there is no slope.

\[ iRoC_{|x|}(0) \, \text { does not exist } \]

$0$ is the only critical number.

Absolute Functions

Absolute functions are those functions that can be represented by formulas of the form

\[ AV(x) = A \cdot | B \, x + C| + D \]

where $A$, $B$, $C$, and $D$ are real numbers, with $A \ne 0$ and $B \ne 0$ is a nonzero real number.

Note: In the template for absolute value functions, There is a leading coefficient for the function and there is a leading coefficient for the linear function inside the absolute value bars.

Absolute Value Functions

Let $Abs(k) = |k| $ be the Core absolute value function.

Let $L_{in}(t) = B \, t + C$ be a linear function.

Let $L_{out}(y) = A \, y + D$ be a linear function.

Absolute Value functions are constructed by composing these component funcitons.

\[ (L_{out} \circ Abs \circ L_{in})(x) = A |B \,x + C| + D \]

Behavior

Rewrite $A(y) = | y - 3|$ without absolute value bars.

explanation

First, get rid of the absolute value bars.

$y - 3 < 0$ when $y < \answer {3}$
$y - 3 > 0$ when $y > \answer {3}$

\[ A(y) = \begin{cases} -(y - 3) & \text { on } (-\infty , 3) \\ y - 3 & \text { on } [3, \infty ). \end{cases} \]

Each piece is a (restricted) linear function. The leading coefficients tell us that

$|y-3|$ is decreasing on $(-\infty , 0)$
$|y-3|$ is increasing on $(0, \infty )$

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more examples can be found by following this link
More Examples of Elementary Functions

2026-05-29 21:22:09