# The Modulus Function

A LevelAQAEdexcelOCREdexcel 2022OCR 2022

## The Modulus Function

The modulus of a number is the size of the number, whether it is negative or positive, for example the modulus of $6$ is $6$ and the modulus of $-6$ is $6$.

A Level   ## 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. $f(|-4|) = f(4)$
A Level   ## 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}   A Level   A Level ## 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}   A Level ## 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 $g(x)$.

A Level   ## 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  $\textcolor{red}{x = -4}$

A Level   ## Note:

You can also solve modulus inequalities using these methods. The graphical method of solving inequalities will be helpful, since there will often be a quadratic involved. Another rule that will be helpful is:

$|x-a| < b \, \iff \, a - b < x < a+b$

A Level   ## Example 1: Solving Modulus Equations Graphically – Straight Lines

Solve $|2x+3| = x+2$

[3 marks]

Step 1: Sketch the graphs of $y = |2x+3|$ and $y = x+2$ 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:

$2x+3 \geq 0$  for $x \geq - \dfrac{3}{2}$   and   $2x+3 < 0$  for $x < - \dfrac{3}{2}$

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

(1)   $2x+3 = x+2$   for $x \geq - \dfrac{3}{2}$

(2)   $- (2x+3) = x+2$   for $x < - \dfrac{3}{2}$ Step 4: Solve each equation in turn and check that the solution are valid, and remove any that are outside the range of $x$ for that equation.

Solving (1):  $x=-1$  (this is valid since $-1 \geq - \dfrac{3}{2}$)

Solving (2):  $3x = -5 \Rightarrow x = - \dfrac{5}{3}$  (this is valid since $- \dfrac{5}{3} < - \dfrac{3}{2}$)

Step 5: Check that the solutions look correct, by looking at the graph. The two solutions appear to be correct.

A Level   A Level ## Example 2: Solving Modulus Equations Graphically – Quadratics and Cubics etc.

Solve $|x^2 - 4| = 3$

[4 marks] Step 1: Sketch the graphs of $y = |x^2-4|$ and $y = 3$ 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:

$x^2 - 4 \geq 0$  for $x \leq -2$  or  $x \geq 2$

and   $x^2 - 4 < 0$  for $-2 < x < 2$

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

(1)   $x^2 - 4 = 3$   for $x \leq -2$  or  $x \geq 2$

(2)   $- (x^2 - 4) = 3$   for $-2 < x < 2$

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

Solving (1):  $x^2 = 7 \Rightarrow x = \sqrt{7}$  and  $x = - \sqrt{7}$  (this is valid since $- \sqrt{7} \leq - 2$  and  $\sqrt{7} \geq 2$)

Solving (2):  $x^2 - 1 = 0 \Rightarrow x = 1$  and  $x = -1$  (this is valid since $1$ and $-1$ both lie within $-2 < x < 2$)

Step 5: Check that the solutions look correct, by looking at the graph. The four solutions appear to be correct.

A Level ## Example Questions

a) $f(-2) = 3(-2)^2 + 4(-2) - 9 = -5$

b) $|f(-2)| = |-5| = 5$

c) $f(|-2|) = f(2) = 3(2)^2 + 4(2) - 9 = 11$

d) $- |f(2)| = - |11| = - 11$

a)i) The negative section needs to be reflected in the $x$-axis: ii) For the negative $x$-values, reflect the line in the $y$-axis: iii) Reflect $f(x)$ in the $y$-axis, and then reflect the negative section in the $x$-axis: b)

Firstly, sketch the graphs of $y = |3x - 3|$ (using part a)i)) and $y = \dfrac{3}{2}$ on the same pair of axes: $3x - 3 \geq 0$ when $x \geq 1$  and  $3x - 3 < 0$ when $x < 1$

So, we can form two equations:

(1)  $3x - 3 = \dfrac{3}{2}$  for $x \geq 1$

(2)  $-(3x-3) = \dfrac{3}{2}$  for $x < 1$

Then, we can solve these:

Solving (1):  $3x = \dfrac{9}{2} \Rightarrow x = \dfrac{3}{2}$  for $x \geq 1$  (this is valid since $\dfrac{3}{2} \geq 1$)

Solving (2):  $3x = \dfrac{3}{2} \Rightarrow x = \dfrac{1}{2}$  for $x < 1$  (this is valid since $\dfrac{1}{2} < 1$)

From looking at the graph, both solutions seem to be correct.

$2|-x-3| = x-4$

Therefore $2(-x-3) = x+4$  or  $-2(-x-3) = x+4$

Then, solve these:

\begin{aligned} 2(-x-3) &= x + 4 \\ -2x - 6 &= x + 4 \\ 3x &= - 10 \\ x &= - \dfrac{10}{3} \end{aligned}

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

So, $x = - \dfrac{10}{3}$  or  $x = - 2$

Actually, we need not do any calculation at all. On the left side of the equation, we have $|2x + 4|$, which is defined to be always positive.

However, the right side of the equation is always negative (as we have $-2$ multiplied by an always-positive term).

We can conclude, then, that there could only be a solution if both graphs meet at the $x$-axis.

However, this is not the case as the first graph touches the $x$-axis at $x=-2$ while the second graph touches the $x$-axis at $x=-1$.

Hence, there are no solutions.

\begin{aligned} |2x+2| &= |x-2| \\ (2x+2)^2 &= (x-2)^2 \\ 4x^2 + 8x + 4 &= x^2 - 4x + 4 \\ 3x^2 + 12x &= 0 \\ x^2 + 4x &= 0 \\ x(x+4) &= 0 \end{aligned}

Hence, $x = 0$  or  $x = -4$

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