Inequalities and Modulus Mathematics Skills Guide

This is one of a series of guides designed to help you increase your confidence in handling mathematics. This guide contains both theory and exercises which cover:1. Introduction: 2. Inequalities: (Basic facts, solution, fractional functions) 3. Modulus: (Meaning, solution of equations & inequalities) There are often different ways of doing things in mathematics and the methods suggested in the guides may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull  Ask your lecturers.  You can contact a maths Skills Adviser from the Skills Team on the email shown below.  Access more maths Skills Guides and resources at the website below.  Look at one of the many textbooks in the library.

Web: www.hull.ac.uk/skills Email: [email protected]

1. Introduction The easiest way to solve most inequalities is by sketching a simple curve and the easiest function to sketch is one that has been factorised:- y  (ax  b)(cx  d ) or y  (ax  b)(cx  d )(ex  f ) etc. In many cases functions can be factorised without having to multiply out complex expressions. For example if you had (2 x2  4 x  1)2  (2 x2  3x  4)2 you can use the difference of two squares to simplify the working. The factors of x 2  y 2 are ( x  y )( x  y ) so factorising (2 x2  4 x  1)2  (2 x2  3x  4)2 gives

2x2  4x  1 2x2  3x  4 2x2  4x  1 2x2  3x  4   2 x2  4 x  1  2 x2  3x  4 2 x2  4 x  1  2 x2  3x  4   x  5 4 x2  7 x  3  x  54 x  3x  1 Sketch Graphs (see also the Curve Sketching leaflet) Example 1. The graph of y  ( x  1)(2 x  3) .

When y  0 then ( x  1)(2 x  3)  0 which means that either x  1  0 or 2 x  3  0 x  1 or x  1.5. Also when x  0 y  3 The curve cuts the x -axis at -1 and 1.5

y

-1

x

2

so using the basic x 2 curve we have the sketch as shown. Note additional possible information –



Completing the square gives y  2 x  1

4

2  258 hence minimum at 14 ,  258 

Example 2 The graph of y  ( x  1)( x  2)(2 x  5) y  0  ( x  1)( x  2)( 2 x  5)  0

which means that either x  1  0 or x  2 = 0 or 2 x  5  0 x  1 or x  2 or x  2.5 Also when x  0 , y  10 using the basic x3 curve we have the sketch as shown.

y -2

2.5 x

-1

-10

page 1

2. Inequalities A. Basic facts (b) If (c) If (d) If

(a) If a  b then b  a a  b then k  a  k  b for all values of k a  b and k  0 then ka  kb a  b and k  0 then ka  kb

Proof of (d)

Given Then If k  0 , then as ( ()  ()  () Expanding So if

ab ab0 k ( a  b)  0 ka  kb  0 or ka  kb a  b and k  0 then ka < kb

In particular multiplying both sides of the inequality a  b by -1 gives a  b . This can also be shown on a number line. B. Solving inequalities (or inequations) (a) Find the range of values of x for which add 3 to both sides subtract 2 x from both sides multiply by - 1

x  3  2x  5 x  2x  8  x8 x  8

(b) Find the range of values of x for which x 2  1  0 x 2 is positive for all real values of x so x 2  1  0 for all real values of x .

(c) Find the range of values of x for which factorising gives

2x 2  x  3  0 ( x  1)( 2 x  3)  0

There are at least two ways of proceeding from here: Method 1 ( x  1)(2 x  3)  0  one factor is negative and the other positive. i.e. A: ( x  1)  0 and (2 x  3)  0  x  1 and x  1.5 which is impossible or B: ( x  1)  0 and (2 x  3)  0  x  1 and x  1.5 which gives 1  x  1.5 This can also be illustrated on a number line: -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 | | | | | | | | | | | | case A: case B: case A it is impossible to have x  1 and x  1.5 at the same time. case B both agree for the range  1 < x  1.5 so this is the solution

page 2

Method 2 The easiest way is to sketch the function y  2 x2  x  3  ( x  1)(2 x  3) The diagram on the right shows y the function y  ( x  1)(2 x  3) from it we can see y  0 -1 1.5 x for values of x between  1 < x  1.5 ; which agrees with the method above!

(d) Find the range of values of x for which ( x  1)( x  2)(2 x  5)  0 If you multiply 3 numbers together and get a negative number then 2 of them are positive and one negative or all 3 are negative Method 1 ( x  1)  0 and ( x  2)  0 and (2 x  5)  0  x  1 and x  2 and x  2.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

True for 1  x  2.5

Or ( x  1)  0 and ( x  2)  0 and (2 x  5)  0  x  1 and x  2 and x  2.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

Not true for any value of x Or ( x  1)  0 and ( x  2)  0 and (2 x  5)  0  x  1 and x  2 and x  2.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

Not true for any value of x Or ( x  1)  0 and ( x  2)  0 and (2 x  5)  0  x  1 and x  2 and x  2.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

True for x  2 All three inequalities agree in 1) for  1  x  2.5 and in 4) for x  2 but in no other which gives the solution x  2,  1  x  2.5

page 3

Method 2

y

Sketching the graph of y  ( x  1)( x  1)(2 x  3) (shown on the right) gives the solution x  1 , 1  x  1.5

-1

1

1.5

x

which is as before but obtained more easily than dealing with all the inequalities! The same method will also work for 4th, 5th, .... degree functions as long as they factorise. When you have a function which does not factorise you can get some approximate answers by knowing something about the general shape of the curve and/or any maximum/minimum values it has (possibly using calculus). You could draw the graph using a graph plotter but even here a lot of care needs to be taken as some packages (especially those on calculators) can be misleading. Exercise 1 Solve the following inequalities 1. x  2  4  x 2. 2 x  1  x  4 4. 2( x  1)  3( x  1) 5. 2( 2 x  1)  3 x  1 7. ( x  1)( x  2)  0 8. ( x  1)( 2 x  3)  0

3. x  5 x  2 6. 3( x  4)  x  1 9. (3  x )(3 x  5)  0

10. x 2  4 x  5  0

11. 4 x 2  x  5

13. x 2  x  2

14. ( x  1) 2  4 x 2

12. 5 x 2  3 x  2 15. x( x  1)( x  3)  0

16. x 3  x  0

17. 4 x 3  3 x 2  x  0

18. x 2  x  3  0

19. x 3  3 x 2  x  0

20. x 4  3 x 3  4 x 2  0

3 Fractional Functions Examples (a) Find the range of values for which

x2 2 x

Again there are two methods (i) If x  0 then we can multiply through If x  0 we need to change the inequality by x giving x  2  2 x  x  2 sign as we multiply by a negative number ie x  0 and x  2  x  0 giving x  2  2 x  x  2 ie x  0 and x  2  x  2 From these the solution is x  2, x  0 . This could also be done on a number line.

page 4

(ii) To save the complication of the sign of x , if we multiply through by x 2 it clears the fraction and does not affect the inequality sign as x 2 is always positive.

x  2x 2 x

 2 x2

yy==x(2+x) x(2 + x)

y

( x  2) x  2 x 2 2

x  2x  2x

xx

-2

2

0  2 x  x 2  x(2  x) x(2  x)  0 From the sketch this gives the same solution x  2 or x  0 . If you had to solve x2  3x  1  0 which does not factorise, then solving the equation x2  3x  1  0 by using the formula gives x  3 13 and a solution 2

x  3 13 or x  3 13 Check it on a graph! 2

2

( x  1) 2  1. x5 As in the example above x  5 may be positive or negative so we multiply both sides by x  52 which is positive and so does not affect the inequality. (b) Find the range of values of x for which

giving

( x  1) 2 ( x  5) 2  ( x  5) 2 x5

cancel the ( x  5)

( x  1) 2 ( x  5)  ( x  5) 2 ( x  1) 2 ( x  5)  ( x  5) 2  0

take out factor of ( x  5)





( x  5) ( x  1) 2  ( x  5)  0 ( x  5)( x 2  3 x  4)  0 ( x  5)( x  4)( x  1)  0

simplifythe secondbracket factorise

The graph of the function y  x  5x  4x  1 is

x -5

-1

4

Hence the solution of ( x  5)( x  4)( x  1)  0 is  5  x  1, x  4 Notice the ‘neat’ factorisation of ( x  1)2 ( x  5)  ( x  5)2  0 . You could multiply the brackets out and get x3  2 x2  19 x  20  0 but factorising the left hand side is not easy. Again the final part could have been done using the number line method but it is very messy compared with the above. The advantages of the sketch method is even more ‘obvious’ if you had x( x  1)( x  2)( x  3)( x  4)  0 when by the first method

page 5

you’d have to say for the product to be negative all 5 of the factors, or 3 of the five, or 1 of the five must be negative - 16 possible situations!! A sketch is much easier. If the denominator is something like x 2  1 , there would be no need to multiply by

( x 2  1) 2 as x 2  1 is always positive. (c) Find the range of values of x for which

x 1 x2  1

x

x 2  1 is positive so multiply through by 2

x  1 , giving



y = x3 - 1



y

x  1  x x2  1

x

3

x 1  x  x

1

1  x3 x3  1  0

x  1x 2  x  1  0



2 2  34  0 the solution is x  1 (see also the sketch)

Hence, as ( x 2  x  1)  x  1

x 1 1  x( x  2) x  2 common denominator is x( x  2) but it could be positive or negative so

(d) Find the range of values of x for which

2

multiply by x ( x  2) cancel down

2

x 2 ( x  2) 2 ( x  1) x 2 ( x  2) 2  x( x  2) x2 x( x + 2)( x  1)  x 2 ( x  2)

x( x + 2)( x  1)  x 2 ( x  2)  0 factorise x( x + 2)x  1  x   0 which gives x( x  2)(1)  0 dividing by - 1 gives x( x  2)  0 From the sketch y = x(x + 2) we can see the solution is 2  x  0 -2

y

0

x

page 6

Exercise 2 Find the set of values of x for which, x5 2 x 2. 2 1. 3 4x  1 x7 x 1 5  2x 5. 3 4. 1 1  7x 2x  3

x 3 2x  8 x x 6.  x  2 x 1 x2 9. 0 ( x  1)( x  3) 1 2x 12.  2 x x 2 3.

8. (1  x)(1  x) 2  0

7. ( x  1)( x  3)( x  5)  0 x 2  2 6 x 2  8x  1 10.  2 x5

11.

5x 2

x 6

1

4. The Modulus Function

if x  0 x The formal definition of the modulus of x is x    x if x  0 Or, numerically, it is the distance from 0 on a number line | -4

| -3

| -2

| -1

| 0

| 1

| 2

| 3

| 4

| 5

| 6

| 7

| 8

3 and | -3 | are the distances of the two points given by 3 and –3 on the number line from the origin. In both cases the distance is 3 ie 3  | -3 |  3 Similarly - 4  4;

6  6;

- 2  2;

6.43  6.43;

- 3.75  3.75;

0  0.

For those who have met Complex Numbers this definition fits in with the distance of, say, a  0 j or 0  bj from the origin. For those who have worked with Vectors it also fits in with the distance of, say, ci  0 j or d i  0 j from the origin. The alternative definition is to say that x is the unsigned part of the number x . That again gives | -4 | = 4, | 3 | = |+3| = 3 (as 3 is shorthand for +3). (a) Sketch the graph of y  x values of x values of y  x Values of y  x

-4 -3 -2 -1 0 -4 -3 -2 -1 0 4 3 2 1 0

1 1 1

2 2 2

3 3 3

4 4 4

5 5 5

6 6 6

6

4

2

0 - 4

- 2

0

2

4

6

Note that the part of y =- x2 below the x-axis is reflected in the x-axis. - 4

Graph of y = x shown Graph of y = | x | shown page 7

Examples (a) Find the range of values of x for which x  3  5 There are, at least two methods of solution! (i) If x  3 is positive ie x > 3 If x  3 is negative ie x < 3

then x  3  5  x  3  5  x  8

then x  3  5  ( x  3)  5  x  2

x  3 and x > 8 gives x  8

x  3 and x < 2 gives x  2

(ii) Squaring both sides gives  x  3   52 and we can remove the modulus sign to 2

get x  32 as both represent the square of the distance on the number line between the point given by -3 and the point given by x .

This gives x  32  52 y

x  32  52  0 use difference of 2 squares x  3  5x  3  5  0 x  8x  2  0

x

-2

y = (x - 8)(x + 2)

8

hence, from sketch, solution x  8 or x  2 Method (ii) is, safer and simpler to use once you’ve had some practice! (b) Find the range of values of x for which

x 2 x2

2

 x  squareboth sides    4 - two possible methods here  x  2 Multiply out, collect terms, Or use difference of two factorise squares 2 2 2 x 2  4x  22  2( x  2)2 x  4x  2  4 x  16 x  16 0  2x  22  x 2 0  2x  2  x 2x  2  x 0  3x  4x  4

0  4 x 2  16 x  16  x 2 0  3 x 2  16 x  16 0  3x  4x  4 

In both cases 3x  4x  4  0

Sketching the graph of y  ( x  4)(3x  4) helps give the solution 4x4 3

y

4

4 3

x

y

page 8

(c) Sketch the graph of y  x3  x





Factorising gives y  x3  x  x x 2  1  xx  1x  1 The graph of y  x3  x

Hence the graph of y  x3  x y

y x -1

-1

1

1

x

Those parts of y  x3  x below the x -axis are reflected in the x -axis to give y  x3  x

(d) Find the range of values of x for which 2

 

x2 1 x2

 

 x2  2 2   1  x 2  x  22  x 2  x  22  0 Square   x  2  







Using the difference of 2 squares this gives x2  x  2 x2  x  2  0 To deal with this we need to solve x 2  x  2  0 and x 2  x  2  0 so that we can find







where the function y  x 2  x  2 x 2  x  2 cuts the x -axis. x 2  x  2  0 cannot be factorised and the formula gives no values.





2 But x 2  x  2  x  1  7 2

4 y x

which is positive for all x. Also we have

-2

2

y x

 x2

1

x 2  x  2  x  2x  1 so

x2  x  2x2  x  2  0  x2  x  2  0  x  2x  1  0  2  x  1 (see sketch)







Multiplying out x2  x  2 x2  x  2 and collecting terms gives the quartic inequality x4  x2  4 x  4  0 , which is not too difficult to factorise (using the factor theorem) but, in general, the factorisation would not be easy. Exercise 3 Solve the following 1. 3 x  11  0 2. 8  7 w  6 5.

x 1 4 x2

6.

2 p2 1 3p  2

1 z  1 4. 2 x3 z 3 2 1 7.  1 8. y  2 y  2  1 x 1 y4 3.

page 9

Exercise 4 Sketch the following graphs 1. y  x  1 2. y  x 2  1

4. y  x  2x  22 x  1

3. y  x 2  2 x  3

5. y  x 3

6. y  x 4  5 x 2  4

ANSWERS Exercise 1 1. x  1 4. x  5 7. x  1,x  2 10. x  1, x  5 13. x  1, x  2 16. x  1, 0  x  1

2 . x  3 5. x  1

6.

8.  1  x  3 2

9.

11.  5  x  1 4

14. x  1, x 

19. x  3 5 , 0  x  3 5 2 2

Exercise 2 1. 7  x  23 2

4.  4  x  1 5

3. x   1

7

7. x  5,  3  x  1 10. 2  x  4,  5  x  1

17. x  0 20.  1  x  4

1 3

2 x51 2 5  x3 3

12. x  1, x   2 5

15. x  0, 1  x  3 18. 1 13  x  1 13 2

2. x  1 , x  1

3. 4  x  24

5. x 

6. x  0, 1  x  2 9. x  3, 1  x  2

4 3, 2

2

5

x8

5

8. x  1 x  1 11. x  2, x  3

12. x   2 , 0  x  2

Exercise 3 1. All x except for x   11

2. 2  w  2

3. 2  x  4

4. z  2, 6

5. 9  x  7

6.  p  22 p  1 2 p 2  3 p  2  0   1  p  2

7. x  0, x  2

8. y  2, 1, 2, 3

3

5

3

7





2

page 10

Exercise 4 1.

2.

3.

4.

-3

5.

1

-1

1

1

-2

-0.5

2

6.

-2

-1

1

2

We would appreciate your comments on this worksheet, especially if you’ve found any errors, so that we can improve it for future use. Please contact the Maths Skills Adviser by email at [email protected] The information in this leaflet can be made available in an alternative format on request using the email above. page 11