The Modulus Function

Modulus Notation

The following notation is relevant to the modulus:

• The modulus of a number, e.g. $x$ is written as $|x|$.
• Generally, $|x| = x$ for $x geq 0$, and $|x| = -x$ for $x <0$
• Functions also have a modulus: e.g. if $f(x) = -2$, then $|f(x)| = 2$
• $|f(x)| = f(x)$ when $f(x) geq 0$ and $|f(x)| = -f(x)$ when $f(x)<0$
• If the modulus is inside the function, e.g. $f(|x|)$, then you apply the modulus to the $x$-value before applying the function, i.e. $f(|-4|) = f(4)$

Graphs of Modulus Functions – Straight Lines

There are $3$ types of modulus graphs that you may be asked to draw:

1. $y = |f(x)|$ – all negative values of $f(x)$ are made positive, by reflecting the negative section of the graph of $f(x)$ in the $x$-axis. This restricts the range to $|f(x)| geq 0$ (or a subset within $|f(x)| geq 0$, e.g. $|f(x)| geq 2$.
2. $y = f(|x|)$ – the negative $x$-values give the same result as the corresponding positive $x$-values, so the graph of $f(x)$ for $x geq 0$ is reflected in the $y$-axis, for negative $x$-values.
3. $y = |f(-x)|$ – the $x$-values swap sign (i.e. from positive to negative of from negative to positive), so the graph of $f(x)$ is reflected in the $y$-axis. Then, all negative values of $f(x)$ are made positive by reflecting the negative section of the graph of $f(x)$ in the $x$-axis. The range is restricted, as with $y = |f(x)|$.

The best and easiest way to draw these graphs is to plot the graph of $y = f(x)$ first, and then reflect it in the appropriate axis or axes.

Example: For $f(x) = 2x-1$, sketch the graphs of

$begin{aligned} y &= |f(x)| y &= f(|x|) y &= f|(-x)| end{aligned}$

Graphs of Modulus Functions – Quadratics and Cubics etc.

For modulus graphs where the function is a quadratic or cubic etc. the same rules apply as for straight lines – however, sketching them will be a little bit harder.

Example: For $f(x) = x^2-2x$, sketch the graphs of

$begin{aligned} y &= |f(x)| y &= f(|x|) y &= |f(-x)| end{aligned}$

Solving Modulus Equations Graphically

To solve modulus equations of the form $|f(x)| = n$ or $|f(x)| = |g(x)|$, you can solve them graphically, using the following method:

Step 1: Sketch the graphs of $y = |f(x)|$ and $y = n$, on the same pair of axes.

Step 2: Work out the ranges of $x$ for which $f(x) geq 0$ and $f(x) < 0$ from the graph.

e.g. $f(x) geq 0$  for $x leq textcolor{red}{a}$ or $x geq textcolor{blue}{b}$   and   $f(x) < 0$  for $textcolor{red}{a} < x < textcolor{blue}{b}$

Step 3: Use step 2 to write $2$ new equations, one that holds for each range of $x$:

$f(x) = n$   for $x leq textcolor{red}{a}$ or $x geq textcolor{blue}{b}$

$– f(x) = n$   for $textcolor{red}{a} < x < textcolor{blue}{b}$

Step 4: Solve each equation in turn and check that the solutions are valid, and remove any that are outside the range of $x$ for that equation.

Step 5: Check that the solutions look correct, by looking at the graph.

Note: Use the same method for $|f(x)| = |g(x)|$, by replacing $n$ with $g(x)$.

Solving Modulus Equations Algebraically

For equations of the form $|f(x)| = n$ and $|f(x)| = g(x)$ you can solve them algebraically instead of graphically – if you feel that you understand the topic well enough.

Example: Solve $|2x+2| = x+4$

Step 1: Solve for positive values:

$begin{aligned} 2x + 2 &= x+4 textcolor{red}{x} &= textcolor{red}{2} end{aligned}$

Step 2: Solve for negative values:

$begin{aligned} -(2x + 2) &= x+4 3x &= -6 textcolor{red}{x} &= textcolor{red}{-2} end{aligned}$

Step 3: Combine the solutions:

The solutions are $textcolor{red}{x = 2}$  and  $textcolor{red}{x = -2}$

For equations of the form $|f(x)| = |g(x)|$, it is easier to do solve them algebraically, using the following rule:

“If $|a| = |b|$, then $a^2 = b^2$ ”

So if $|f(x)| = |g(x)|$, then $[f(x)]^2 = [g(x)]^2$

Example: Solve $|x-1| = |2x+3|$

Step 1: Square both sides:

$begin{aligned} |x-1| &= |2x+3| (x-1)^2 &= (2x+3)^2 end{aligned}$

Step 2: Expand and simplify:

$begin{aligned} x^2 – 2x + 1 &= 4x^2 + 12x + 9 3x^2 + 14x + 8 &= 0 (3x+2)(x+4) &= 0 end{aligned}$

Step 3: So, the solutions are:

$textcolor{red}{x = – dfrac{2}{3}}$  and  $textcolor{red}{x = -4}$