An inequality is a mathematical statement containing one of the symbols <, >, or.

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1 Further Concepts for Advanced Mathematics - FP1 Unit 3 Graphs & Inequalities Section3c Inequalities Types of Inequality An inequality is a mathematical statement containing one of the symbols <, >, or. There are two types of inequality: Inequalities whose truth depends on the value of the variable concerned e.g. + 3 > is only true if is greater than. Inequalities that are always true e.g. 0 is always true for real values of. The work in this section covers the first type of inequality. Rules for manipulating inequalities 1. You may add or subtract the same value from both sides of the inequality e.g. > > The symbol means implies that. You may multiply or divide both sides of an inequality by the same positive number e.g. > 8 3 > If both sides of an inequality are multiplied or divided by a negative number, the inequality is reversed e.g. > 8 3 < You may add (but not subtract) corresponding sides of inequalities of the same type e.g. if > 8 and y > then + y > 8 +. Inequalities of the same type are transitive. e.g. if > y and y > z then > z You cannot subtract corresponding sides of inequalities of the same type. This eample shows why it should be avoided. If > 8 and y > then y > 8 giving y > 3 But what if = 10 and y = 1? These values are true for the initial inequalities but clearly show that the subtraction does not work. 1

2 Using sketch graphs to solve inequalities Eamples Solve the inequality 0 ( + 1)( ) First sketch the graph of y = ( + 1)( ) When = 0, y = = 1 1 When y = 0, = Vertical asymptotes are = 1 and = As 1 from the left, y As 1 from the right, y + As from the left, y As from the right, y + As, y = = = 0 from the negative side As +, y = = = 0 from the positive side The sketch should look like this: y 1-1 We want the values of for which the y values are 0 These are shown on the sketch on the net page.

3 1 y -1 The values for which y 0 are marked by the horizontal lines in the graph. The filled circle denotes where a value eists and may be used in the inequality. The empty circles denote values that are either undefined (as both are in this case) or are not true in the inequality. The solution to the inequality is therefore 1 < or > An algebraic method To solve the inequality 0 all we need to do is think in terms of positive and ( + 1)( ) negative numbers since 0 means either positive or 0. Think about the critical points for each part of the function For the critical point is since values of lower than will give a negative result and higher values will give a positive result. For + 1 the critical point is -1 For the critical point is We can put these on a number line and then consider where negative. ( + 1)( ) is positive or 3

4 -1 -ve -ve +ve +ve +1 -ve +ve +ve +ve -ve -ve -ve +ve ( + 1)( ) -ve +ve -ve +ve The required regions are clear from the table. The main thing to get right here is that the function is undefined for the values -1 and. The table gives the same solution 1 < or > Inequalities of the form f ( ) g( ) There are two ways to solve these: 1. By drawing the graphs of both y = f () and y = g() and looking to see where one is lower than the other.. By rearranging the inequality to f ( ) g( ) 0 and then drawing the graph of y = f ( ) g( ). Eample Solve the inequality 8 + Method 1 The sketch graph of y = 8 is simple to draw. For + y = a little more effort is needed. Intercepts: = 0, y = = 1 y = 0, = Vertical asymptote: = As from the left, y as from the right, y + 4

5 Behaviour as ± : As, y 1 from below as +, y 1 from above The two sketch graphs look like this: y + y = y = 8 The inequality + y =. + 8 is satisfied where the graph of y = 8 is above the graph of The set of values for that satisfy the inequality are marked on the diagram. One inequality includes both points of intersection so there are filled circles at each end of the line. The other does not include = as the function is undefined for that value and so an empty circle has been used. We have to find the points of intersection to solve the inequality fully: + 8 = ( 8 )( ) = = + 0 = ( 3)( 6) = 0

6 So = 3 or = 6 The solution to the inequality + 8 is 3 6 or < Method Rearrange the inequality to Using a common denominator (8 )( ) ( + ) 0 multiplying out and simplifying gives This looks a little unpleasant so to make the -1 which reverses the inequality term positive, we multiply both sides by ( 3)( 6) Factorising the top gives 0 To solve the inequality (and hence our original one), we need to draw the graph of ( 3)( 6) y = and find where the y values are negative (i.e. <0) The sketch graph of ( 3)( 6) y = 18 Intercepts: = 0, y = = 9 y = 0, = 3 or = 6 Vertical asymptote = As from the left, y as from the right, y + Behaviour as ± : As, y = = as +, y = = + There are no horizontal asymptotes (there is an oblique asymptote can you work out what is is?). 6

7 The sketch graph looks like this: y 3 6 ( 3)( 6) From the graph you can see that the solution the the inequality or < is Hence the solution to the inequality + 8 is 3 6 or < Solving the inequality algebraically Rearrange the inequality method. + ( 3)( 6) 8 to 0 as we did for the second The critical points are = for, = 3 for 3 and = 6 for 6. 7

8 Putting these on a number line gives: ( 3)( 6) So the inequality is true for values of that are less than and values of between 3 and 6. We need to check if, for any of these values, the function is undefined. For =, the function ( 3)( 6) is undefined as the denominator would be 0. This means that we don t include = in our final solution. The table shows that the solution to the inequality (yet again!) + 8 is 3 6 or < 8

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