The Modulus Function

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 AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR
A Level Edexcel

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 LevelEdexcel

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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 AQA Edexcel OCR

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 LevelAQAEdexcelOCR
A Level Edexcel

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 LevelEdexcel

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

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

Worksheet and Example Questions

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99
View Product

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99
View Product

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99
View Product