“Teach A Level Maths” Vol. 2: A2 Core Modules 16: The Modulus Function © Christine Crisp

Module C3 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

The modulus function is written as . We read it as “ mod x ” or “ abs x ”. Abs x stands for the absolute value of x; the value of x ignoring any negative signs. e.g. and can be defined formally as

The Graph of Since the graph of is the same as y = x for

The Graph of Since BUT for x < 0, the y-values have the opposite sign to those of y = x, so we need to reflect in the x-axis.

The Graph of Since BUT for x < 0, the y-values have the opposite sign to those of y = x, so we need to reflect in the x-axis. Because of the reflection, the equation of this part . . . is The domain of is  The range is

The Graph of If you use a graphical calculator for modulus functions, you will probably need to type in . On the following slides, I’ve assumed you’re not using a calculator as you will then revise transformations and practise solving equations! If you do use a calculator you will need to copy the graphs and show any working.

e.g. 1 Use transformations of to sketch graphs of the following: (b) Solution: (a) Translate by (b) Translate by y = 1

Exercise Use transformations of to sketch the following: 1. 2. 3. Solutions: 1. 2.

Exercise Use transformations of to sketch the following: 1. 2. 3. Solutions: 3.

The next section uses modulus functions in inequalities. There are 2 methods you can use, so it would be a good idea to try both on a few examples to see which you prefer.

e.g. 1 Solve the inequality Solution: Sketch the graphs of and x y We want the values of x where the red graph, is above the blue one, .

x = - 2. e.g. 1 Solve the inequality Solution: Sketch the graphs of and x y We want the values of x where the red graph, is above the blue one, . x A x B -2 2 The right hand branch of has equation y = x. So, B is where y = x meets y = 2. So, at B, x = 2. The graph is symmetrical about the y-axis, so A is x = - 2. So, or 2 regions on the graph always means 2 inequalities.

The previous example showed that is equivalent to or The same values of x are also given by So, is equivalent to Notice that we square each side of the modulus inequality. This gives us another way of solving the inequalities.

e.g. 2 Solve the inequality . Method 1: Sketch the graphs and . x y is above 1 in two regions.

e.g. 2 Solve the inequality . Method 1: Sketch the graphs and . x y is above 1 in two regions. The sketch doesn’t show the x-coordinates of A and B, so we must find them. A B x x 3 The right hand branch of the modulus graph is the same as y = x – 2 . . . so B is given by

e.g. 2 Solve the inequality . Method 1: Sketch the graphs and . x y is above 1 in two regions. The sketch doesn’t show the x-coordinates of A and B, so we must find them. A B x x 1 3 The right hand branch of the modulus graph is the same as y = x – 2 . . . so B is given by The gradient of the right hand branch is +1 and the left hand is -1, so the graph is symmetrical. So,

Method 2: Squaring both sides: To solve the quadratic inequality we need a sketch, so we find the zeros. The zeros are x = 1 and x = 3. y x y 1 3 So,

e.g. 2 Solve the inequality Solution: We can either use graphs or squaring. Method 1: Use translations from to sketch both graphs: x y To solve we want the values of x where the graph of is under the graph of . x A These are left of the point of intersection.

e.g. 2 Solve the inequality Solution: We can either use graphs or algebra. Method 1: Use translations from to sketch both graphs: x y To solve we want the values of x where the graph of is under the graph of . x A These are left of the point of intersection. Symmetry shows the point of intersection is at x = -3. So, for . N.B.

Method 2: Squaring both sides: This method was quicker for this question because we didn’t need to draw a graph. However, if the inequality is quadratic, a graph will be needed.

SUMMARY To solve inequalities involving one or more modulus functions: Method 1: Sketch the functions. ( Type “abs” if a graphical calculator is used. ) Mark the section(s) of the x-axis that give the solution. Find the points of intersection using the related functions ( without the mod. signs ). ( A negative sign is needed for reflected parts. ) Refer to the graph to write the inequality for x.

SUMMARY Method 2: Square both sides of the inequality to remove the mod. signs and solve the resulting linear or quadratic inequality. ( A graph is needed for quadratic inequalities. )

Solve the inequalities. 1. 2. Exercise Solve the inequalities. 1. 2. Solutions: 1. x y

Solve the inequalities. 1. 2. Exercise Solve the inequalities. 1. 2. Solutions: 1. x y By symmetry, Or:

(N.B. Strict inequality ) Solutions: 2. (N.B. Strict inequality ) A x B y For B: By symmetry, at A: So, ( One region so one inequality )

Method 2: Zeros: So:

e.g. 4 Solve the inequality . Solution: Sketch the graphs and A B x y I sketched by using the translation and noticing that the gradient is . This is not symmetrical about x = 1. is below in one region.

e.g. 4 Solve the inequality . Solution: Sketch the graphs and A B x y B is given by 2 A is on the reflected part of . . .

e.g. 4 Solve the inequality . Solution: Sketch the graphs and A B x y y B is given by x B A x x 2 A is on the reflected part of . . . and the original part of . . . so, for A: so,

Solve the inequalities. 1. 2. Exercise Solve the inequalities. 1. 2. Solution: 1. A x B y For B, -1 3 For A, or So,

Method 2: Zeros: So,

( Two regions so two inequalities ) Exercise 2. Solution: For B, A B x For A, or: So, ( Two regions so two inequalities )

Method 2: Zeros: So,

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

The modulus function is written as . We read it as “ mod x ” or “ abs x ”. Abs x stands for the absolute value of x; the value of x ignoring any negative signs. e.g. and can be defined formally as

The Graph of If you use a graphics calculator for modulus functions, you will probably need to type in . On the following slides, I’ve assumed you’re not using a calculator as you will then revise transformations and practice solving equations! If you do use a calculator you will still need to draw graphs and show any working.

Solution: (a) Translate by (b) Translate by e.g. 1 Use transformations of to sketch graphs of the following: (a) (b)

SUMMARY To solve inequalities involving one or more modulus functions: Sketch the functions. ( Type “abs” if a graphics calculator is used. ) Find the points of intersection using the related functions ( without the mod. signs ). ( A negative sign is needed for reflected parts. ) Refer to the graph to write the inequality for x. Mark the section(s) of the x-axis that give the solution. Method 1: Solving Inequalities

Method 2: Square both sides of the inequality to remove the mod. signs and solve the resulting linear or quadratic inequality. ( A graph is needed for quadratic inequalities. )

( -2, 2 ). e.g. 1 Solve the inequality x y e.g. 1 Solve the inequality Solution: Sketch the graphs of and We want the values of x where is above . or The graph is symmetrical about the y-axis, so A is ( -2, 2 ). So, B is the where y = x meets y = 2. The right hand branch of has equation y = x. So, at B, x = 2. So, A B is equivalent to

Solution: We can either use graphs or algebra. Symmetry shows the point of intersection is at x = -3. e.g. 2 Solve the inequality Use translations from to sketch both graphs: To solve we want the values of x where the graph of is under the graph of . So, for . These are left of the point of intersection. Method 1:

Method 2: Squaring both sides: e.g. 2 Solve the inequality It looks as though this method is much quicker because we didn’t need to draw a graph. However, if the inequality is quadratic, a graph will be needed. On the whole method 1 is more reliable.

e.g. 3 Solve the inequality . Method 1: Sketch the graphs and . is above 1 in two regions. The sketch doesn’t show the x-coordinates of A and B, so we must find them. A B x x The right hand branch of the modulus graph is the same as y = x – 2 . . . so B is given by The gradient of the right hand branch is +1 and the left hand is -1, so the graph is symmetrical. So,

e.g. 4 Solve the inequality . B x B is given by e.g. 4 Solve the inequality . so, and the original part of . . . A is on the reflected part of . . . Solution: Sketch the graphs and so, for A: