# The Modulus Function

# The Modulus Function

**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.

**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:

**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.**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.****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**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).

**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

**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.

**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.

## Example Questions

**Question 1:** For the function f(x) = 3x^2 + 4x - 9, find the following:

**a)** f(-2)

**b)** |f(-2)|

**c)** f(|-2|)

**d)** -|f(2)|

**[4 marks]**

**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

**Question 2: **

**a)** For the function f(x) = 3x-3, x \in \mathbb{R}, sketch the graphs of:

**i)** y = |f(x)|

**ii)** y = f(|x|)

**iii)** y = |f(-x)|

**b)** Hence, or otherwise, solve the equation |3x - 3| = \dfrac{3}{2}

**[7 marks]**

**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.

**Question 3:** Solve the equation 2|-x-3| = x+4

**[3 marks]**</p

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

**Question 4:** Solve the equation |2x+4| = -2|x+1|

**[3 marks]**

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.

**Question 5:** Solve the equation |2x+2| = |x-2|

**[4 marks]**

\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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