Further Pure 1 Lesson 8 – Inequalities

Wiltshire Inequalities Inequalities involve the relationships >,<,≥,≤ There are two types of inequality Type 1 – Those that are only true for certain values. Example: x 2 > 9, when x > 3 or x < -3 Type 2 – Those that are always true. Example: x 2 + y 2 ≥ 0, for all x,y in the reals. In this lesson we will be looking at the first type of inequalities, and how to solve the inequality.

Wiltshire Drawing Inequalities You can either write an inequality algebraically or you can show it with a diagram. You need to be familiar with either method. x < a x > a x ≥ a x ≤ a a < x ≤ b Note that a solid end is used for ≤ & ≥ and an empty end is used for. ab

Wiltshire Manipulating Inequalities

Wiltshire Solving Inequalities We are going to look at solving inequalities of the form f(x) > 0. This can be solved 2 different ways. Method 1 - By plotting the graph of f(x) and then examining where the graph is above the x-axis. Method 2 - By solving the equation using an algebraic method.

Wiltshire Method 1 - Graphically Solve the inequality We have already learnt how to plot this graph in lesson 7. Just by studying the graph you can see where the line is above the x-axis. -3 < x < 1 & 6 < x

Wiltshire Method 2 - Algebraically We can use a table to investigate when the graph is positive and when it is negative. There are three key points on the x-axis, the two asymptotes and the value of x when y = 0. From the table you can see that -3 < x < 1 & 6 < x –––+ –+++ ––++

Wiltshire Inequalities of the form f(x) ≤ g(x) – Method 1 Solve the inequality x + 1 < 2/x We can sketch y = x + 1 and y = 2/x Now we need to find the x co-ordinates of where the two graphs meet. We need to solve x + 1 = 2/x x 2 + x = 2 x 2 + x - 2 = 0 (x + 2)(x - 1) = 0 x = -2 & 1 From the graph we can see that the solution is. x < -2 & 0 < x < 1

Wiltshire Inequalities of the form f(x) ≤ g(x) – Method 2 Solve the inequality x + 1 < 2/x x + 1 – 2/x < 0 x 2 + x – 2 < 0 x (x – 1)(x + 2) < 0 x –––+ –+++ ––++

Wiltshire Inequalities of the form f(x) ≤ g(x) – Method 3 The third method is to re- arrange the equation so that it is equal to zero and then sketch the graph. (x – 1)(x + 2) < 0 x If y = 0, x = -2 & 1 If x = 0, y is undefined As x tends to infinity y tends towards x. As x tends to minus infinity y tends towards –x. From the graph you can see that x < -2 & 0 < x < 1

Wiltshire Questions Solve the following inequalities graphically and algebraically. 1) (x+2)(x-5)(2x-3) < 0 2)x 2 ≥ x + 6 3)2/x > 0.5x 4)2x-1 ≤ 9 x-1 x+1 You can now do Ex 3B pg 92