Absolute Value Function

From absolute value to a piecewise function

To evaluate, just apply the definition: $g(x)=|-x^2+4x|$ at $x=5$ gives $|-25+20|=|-5|=5$.

From a piecewise function to absolute value

Sometimes you have to go the other way: you are given the piecewise function and you want to write it compactly. You recognize the form $|E(x)|$ like this:

1. The two branches must be opposites: one is $E(x)$ and the other is $-E(x)$.

2. The branch that keeps its sign is the one in the region where $E(x)\ge 0$.

3. The point where the branch changes is where $E(x)=0$.

Two absolute values at once

If there are two absolute values, each has its own change point. Order those points on the number line, divide the domain into intervals, and in each interval decide the sign of each expression inside.

If the interval you are given already lies within a single piece, there is no need to split it: just decide the sign of each expression on that interval. For example, for $t\in[1,3]$ the expression $t^2-4t+3=(t-1)(t-3)$ is $\le0$, and $t-2$ changes sign at $t=2$, so that interval is split in two.

The vertex: domain, range and intercepts

The domain of $|x|+5$ is all real numbers, and since $|x| \ge 0$, its range is $[5, \infty)$.

If you are given the graph of $y=|x-a|$, the vertex is on the $x$-axis at $(a, 0)$: that is where you read $a$.

One absolute value inside another

For $f(x)=\big||x|-c\big|$ we work from the inside out. First we split $|x|$ at $x=0$, and then we split the outer absolute value where $|x|-c$ equals zero, that is, at $x=\pm c$. The graph is shaped like a "W": vertices at $(\pm c, 0)$ and a peak at $(0, c)$.

To find $c$ from a graph: the central peak is at $(0,c)$ and the $x$-intercepts are at $\pm c$.

Negative sign and unknown parameters

If there is a minus sign in front of the absolute value, the "V" is flipped: $f(x)=-|x-a|+k$ is shaped like a "Λ", with the vertex $(a,k)$ as a maximum and the branches pointing down (toward negative $y$).

When a graph or a problem statement gives you special points, you set up equations with the parameters.

For $f(x)=|x^2+bx+c|$ with the graph given: the roots of the parabola are the points where the graph touches the $x$-axis, and the vertex of the parabola (which is below the axis) is reflected and becomes the local maximum, with the same $x$-coordinate $x=-b/2$ and a height equal to the absolute value of the vertex's value.

Quadratic and rational functions with absolute value

Rational function with absolute value: pieces, domain and asymptotes

With a rational function you do the same thing: study the sign of the quotient (numerator and denominator together) and change the sign wherever it is negative. For the domain, exclude the values that make the denominator zero. Asymptotes are found like this:

• Vertical: at a value $x=a$ that makes the denominator zero (and not the numerator). As you approach $a$, the quotient becomes huge and, with the absolute value, tends to $+\infty$ from both sides.

• Horizontal: look at what happens when $x$ is very large, positive or negative. If the quotient approaches a number $L$, the absolute value approaches $|L|$.

Average rate of change

Equations and inequalities with absolute value

A special case: $|E(x)|=0$ holds only if $E(x)=0$. For example, $|2x-8|=0$ gives $x=4$, which is the vertex: the only point where the graph touches the $x$-axis.

That closes our opening question: the machine accepts the rods whose distance from $250$ mm is at most $1.5$ mm.

Two equal absolute values

Double root and square roots

When the graph touches the $x$-axis at only one point

$|g(x)|=0$ happens only when $g(x)=0$. If $g$ is quadratic, it has a single solution when the discriminant is zero ($b^2-4ac=0$): it is a double root, and the parabola only touches the axis with its vertex without crossing it.

Domain with a square root

A square root exists only if the radicand is $\ge0$. If the radicand has an absolute value, you get an inequality with absolute value, which is solved with the rule $|E|\le c \iff -c\le E\le c$.

Summary