Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -1
Functions, Equations, and Inequalities Chapter 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2. 1 Linear Equations, Functions, and Models · · Solve linear equations. Solve applied problems using linear models. Find zeros of linear functions. Solve a formula for a given variable. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Linear Equations An equation is a statement that two expressions are equal. One Variable mx + b = 0, where m and b are real numbers and m 0. Examples: 4 x + 6 = 18 5(x 2) = 7 x + 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -4
Equivalent Equations that have the same solution set. Examples: 4 x + 8 = 16 and x = 2 are equivalent because 2 is the solution of each equation. 5 y + 6 = 21 and y = 4 are not equivalent because y is equal to 3 in the first equation. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -5
Equation-Solving Principles p Addition If a = b is true, then a + c = b + c is true. You can also use the intersect feature on a graphing calculator to solve equations. y 1 = 9 x 7, y 2 = 2 Example: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -6
Equation-Solving Principles p p Multiplication If a = b is true, then ac = bc is true. Example: p We graph y 1 = 3(5 + 2 x) and y 2 = 4 2(x + 3) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -7
Applications Using Linear Models p Mathematical techniques to answer questions in realworld situations. p Often modeled by linear equations and functions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -8
Five Steps for Problem Solving 1. Familiarize yourself with the problem situation. Make a drawing Write a list Assign variables Organize into a chart or table Find further information Guess or estimate the answer 2. Translate to mathematical language or symbolism. 3. Carry out some type of mathematical manipulation. 4. Check to see whether your possible solution actually fits the problem situation. 5. State the answer clearly using a complete sentence. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -9
The Motion Formula p The distance d traveled by an object moving at rate r in time t is given by d = rt. p Example: On a bicycle tour, Dave rides his bicycle 10 mph faster than his father, Bill. In the same time that Dave travels 90 miles, his father travels 60 miles. Find their speeds. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -10
Example 1. Familiarize. We can organize the information in a table as follows. Distance Rate Time Dave 90 r + 10 Bill 60 r 2. Translate. Using the formula t t , we get two expressions for t. and Since they occur in the same time, we have the following equation: 3. Carry out. We solve the equation. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley . Slide 2 -11
Example 4. Check. If Bill travels at rate of 20 mph for a distance of 60 miles it will take him 3 hours. If Dave travels 90 miles at a rate of r + 10, or 30 mph, it will also take him 3 hours. Therefore, the answer checks. 5. State. Dave travels at a rate of 30 mph while his father, Bill, travels at a rate of 20 mph on their bicycles. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -12
Simple-Interest Formula p I = Prt n n I = the simple interest ($) P = the principal ($) r = the interest rate (%) t = time (years) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -13
Example The Robinson’s have two loans that total $15, 000. One loan is at 4% simple interest and the other is at 6. 5%. After 1 year, they owe $700 in interest. What is the amount of each loan? p Solution: 1. Familiarize. We will let x = the amount borrowed at 4% interest. Then the remainder is $15, 000 – x, borrowed at 6. 5% interest. 4% Loan 6. 5% Loan Total Amount Borrowed Interest Rate Time x 4% or 0. 04 1 yr 15, 000 – x 6. 5% or 0. 065 1 yr 15, 000 Amount of Interest 0. 04 x 0. 065(15, 000 – x) 700 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -14
Example 2. Translate. The total amount of interest on the two loans is $700. Thus we write the following equation. 0. 04 x + 0. 065(15, 000 x) = 700 3. Carry out. We solve the equation. 0. 04 x + 0. 065(15, 000 x) = 700 0. 04 x + 975 0. 065 x = 700 0. 025 x + 975 = 700 0. 025 x = 275 x = 11, 000 If x = 11, 000, then 15, 000 11, 000 = 4000. 4. Check. The interest on $11, 000 at 4% for 1 yr is $11, 000(0. 04)(1), or $440. The interest on $4000 at 6. 5% for 1 yr is $4000(0. 065)(1) or $260. Since $440 + $260 = $700, the answer checks. 5. State. The Robinson’s borrowed $11, 000 at 4% interest and $4000 at 6. 5% interest. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -15
Zeros of Linear Functions p An input c of a function f is called a zero of the function, if the output for c is 0. p f(c) = 0 p A linear function f(x) = mx + b, with m 0, has exactly one zero. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -16
Example p Find the zero of f(x) = 3 x 18. Algebraic Solution: 3 x 18 = 0 3 x = 18 x=6 p Find the zero of f(x) = 3 x – 18. Graphic Solution: The x-intercept of the graph is (6, 0). Thus, 6 is the zero of the function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -17
Formulas p p A formula is an equation that can be used to model a situation. Example: Solve y = ax + bx 2 for x. y = ax + bx 2 y + 2 = ax + bx y + 2 = x(a + b) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -18
2. 2 The Complex Numbers · Perform computations involving complex numbers. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Complex-Number System p The complex-number system is used to find zeros of functions that are not real numbers. p When looking at a graph of a function, if the graph does not cross the x-axis, it has no real-number zeros. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -20
The Number i p The number i is defined such that. Examples: Express each number in terms of i. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley i is not under the radical Slide 2 -21
Complex Numbers A complex number is a number of the form a + bi, where a and b are real numbers. The number a is said to be the real part of a + bi and the number b is said to be the imaginary part of a + bi. Imaginary Number a + bi, a 0, b 0 Pure Imaginary Number a + bi, a = 0, b 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -22
Addition and Subtraction Complex numbers obey the commutative, associative, and distributive laws. p Add: p (9 + 5 i) + (2 + 4 i) (9 + 2) + (5 i + 4 i) 11 + (5 + 4)i 11 + 9 i Subtract: (6 + 7 i) (4 3 i) (6 4) + [7 i ( 3 i)] 2 + 10 i Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -23
Multiplication When and are real numbers, . Examples: Multiply and simplify. (7 6 i)2 = 72 2 7 6 i + (6 i)2 = 49 84 i + 36 i 2 = 49 84 i 36 = 13 84 i (1 + 5 i)(1 + 7 i) = 1 + 7 i + 5 i + 35 i 2 = 1 + 7 i + 5 i 35 = 34 + 12 i Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -24
Simplifying Powers of i p Recall that 1 raised to an even power is 1, and 1 raised to an odd power is 1. p Examples: n i 47 = i 46 i = (i 2)23 i = ( 1)23 i = 1 i = i n i 72 = i 71 i = (i 2)36 i = ( 1)36 i = 1 i = i Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -25
Conjugates The conjugate of a complex number a + bi is a bi. The numbers a + bi and a bi are complex conjugates. Examples: 6 + 7 i and 6 7 i 8 3 i and 8 + 3 i 14 i and 14 i p The product of a complex number and its conjugate is a real number. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -26
Multiplying Conjugates p Example: (4 + 9 i)(4 9 i) = 42 (9 i)2 = 16 81 i 2 = 16 81( 1) = 97 p Example: (7 i)( 7 i) = 49 i 2 = 49( 1) = 49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -27
Dividing Conjugates p Example: Divide 4 3 i by 1 8 i. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -28
2. 3 Quadratic Equations, Functions, and Models Find zeros of quadratic functions and solve quadratic equations by using the principle of zero products, by using the principle of square roots, by completing the square, and by using the quadratic formula. · Solve equations that are reducible to quadratic. · Solve applied problems using quadratic equations. · Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Quadratic Equations A quadratic equation is an equation equivalent to ax 2 + bx + c = 0, a 0, where a, b, and c are real numbers. A quadratic equation written in this form is said to be in standard form. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -30
Quadratic Functions A quadratic function f is a function that can be written in the form f(x) = ax 2 + bx + c, a 0, where a, b, and c are real numbers. The zeros of a quadratic function f(x) = ax 2 + bx + c are the solutions of the associated quadratic equation ax 2 + bx + c = 0. Quadratic functions can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -31
Equation-Solving Principles p The Principle of Zero Products: If ab = 0 is true, then a = 0 or b = 0, and if a = 0 or b = 0, then ab = 0. n Example: Solve 3 x 2 5 x = 2. n Solution: 3 x 2 5 x = 2 3 x 2 5 x 2 = 0 (3 x + 1)(x 2) = 0 3 x + 1 = 0 or x 2 = 0 3 x = 1 or x=2 x= Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -32
Checking the Solutions p Check: 3(2)2 5(2) = 2 3(4) 10 = 2 12 10 = 2 2=2 The solutions are Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley and 2. Slide 2 -33
Graphical Solution p The solutions of the equation y = 3 x 2 5 x = 2, or the equivalent equation 3 x 2 5 x 2 = 0, are the zeros of the function f(x) = 3 x 2 5 x 2. They are also the first coordinates of the x-intercepts of the graph of f(x) = 3 x 2 5 x 2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -34
Equation-Solving Principles The Principle of Square Roots: If x 2 = k, then x = or x = . Example: Solve 2 x 2 6 = 0. p p Solution: 2 x 2 6 = 0 2 x 2 = 6 x 2 = 3 x= or x = Check: 2 x 2 6 = 0 2( )2 6 0 2 3 6 0 6 6 0 0=0 Solutions are Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley and . Slide 2 -35
Completing the Square To solve a quadratic equation by completing the square: p p p Isolate the terms with variables on one side of the equation and arrange them in descending order. Divide by the coefficient of the squared term if that coefficient is not 1. Complete the square by taking half the coefficient of the firstdegree term and adding its square on both sides of the equation. Express one side of the equation as the square of a binomial. Use the principle of square roots. Solve for the variable. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -36
Example Find the zeros of f(x) = x 2 + 6 x 16 by completing the square. Solution: x 2 + 6 x 16 = 0 x 2 + 6 x = 16 x 2 + 6 x + = 16 + x 2 + 6 x + 9 = 25 (x + 3)2 = 25 x+3= 5 x = 5 3 or x = 5 3 x=2 or x = 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -37
Graphical Solution: x 2 + 6 x 16 = 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -38
Quadratic Formula p The solutions of ax 2 + bx + c = 0, are given by . p This formula can be used to solve any quadratic equation. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -39
Example Solve: 4 x 2 + 3 x = 8 Solution: 4 x 2 + 3 x 8 = 0 a = 4, b = 3, c = 8 The exact solutions are: The approximate solutions are 1. 088 and 1. 838. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -40
Graphical Solution: 4 x 2 + 3 x = 8 p Graph y 1 = 4 x 2 + 3 x and y 2 = 8 The approximate solutions are 1. 088 and 1. 838. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -41
Discriminant p When you apply the quadratic formula to any quadratic equation, you find the value of b 2 4 ac, which can be positive, negative, or zero. This expression is called the discriminant. For ax 2 + bx + c = 0: b 2 4 ac = 0 One real-number solution; b 2 4 ac > 0 Two different real-number solutions; b 2 4 ac < 0 Two different imaginary-number solutions, complex conjugates. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -42
Equations Reducible to Quadratic p Some equations can be treated as quadratic, provided that we make a suitable substitution. p Example: x 4 10 x 2 + 9 = 0 Knowing that x 4 = (x 2)2, we can substitute u for x 2 and the resulting equation is then u 2 10 u + 9 = 0. This equation can then be solved for u by factoring or using the quadratic formula. Then the substitution can be reversed by replacing u with x 2, and solving for x. Equations like this are said to be reducible to quadratic, or quadratic in form. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -43
Solving an Equation Reducible to Quadratic Solve: x 4 10 x 2 + 9 = 0 u 2 10 u + 9 = 0 (substituting u for x 2) (u 9)(u 1) = 0 u 9 = 0 or u 1 = 0 u = 9 or u=1 x 2 = 9 or x 2 = 1 (substitute x 2 for u and solve for x) x = ± 3 or x = ± 1 The solutions are 3, 3, 1, and 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -44
Application Some applied problems can be translated to quadratic equations. Example: Free Fall. An acorn falls from the top of a tree that is 32 feet tall. How long will it take to reach the ground? Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -45
Solving an Application 1. Familiarize. The formula s = 16 t 2 is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds. 2. Translate. Substitute 32 for s in the formula: 32 = 16 t 2. 3. Carry out. Use the principle of square roots. 32 = 16 t 2 4. Check. In 1. 414 seconds, an acorn would travel a distance of 16(1. 414)2, or about 32 ft. The answer checks. 5. State. It would take about 1. 414 sec for an acorn to reach the ground from the top of a 32 ft tall tree. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -46
Graphical Solution: 32 = 16 t 2 p Use the Intersect method, we replace t with x, graph y 1 = 32 and y 2 = 16 x 2, and find the first coordinates of the points of intersection. Time cannot be negative in this application, so we need to find only the point of intersection with a positive first coordinate. The window [-5, 5, -10, 40], with Xscl = 1 and Yscl = 5. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -47
2. 4 Analyzing Graphs of Quadratic Functions Find the vertex, the axis of symmetry, and the maximum or minimum value of a quadratic function using the method of completing the square. · Graph quadratic functions. · Solve applied problems involving maximum and minimum function values. · Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Graphing Quadratic Functions of the Type f(x) = a(x h)2 + k The graph of a quadratic function is called a parabola. The point (h, k) at which the graph turns is called the vertex. The maximum or minimum value of f(x) occurs at the vertex. Each graph has a line x = h that is called the axis of symmetry. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -49
Example Find the vertex, the axis of symmetry, and the maximum or minimum value of f(x) = x 2 8 x + 12. Solution: Complete the square. Vertex: (4, 4) Axis of symmetry: x = 4 Minimum value of the function: 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -50
Graph f(x) = x 2 8 x + 12 x y 2 0 3 3 4 4 5 3 6 0 vertex Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -51
Example Find the vertex, the axis of symmetry, and the maximum or minimum value of f(x) = 3 x 2 + 6 x 1. Solution: Complete the square. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -52
Example continued Graph: f(x) = 3 x 2 + 6 x 1 Vertex: (1, 2) Axis of symmetry: x = 1 Maximum value of the function: 2 x y 0 1 1 2 2 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -53
Vertex of a Parabola The vertex of the graph of f(x) = ax 2 + bx + c is We calculate the x-coordinate. We substitute to find y-coordinate. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley the Slide 2 -54
Example For the function f(x) = 3 x 2 + 18 x 21: a) Find the vertex. b) Determine whethere is a maximum or minimum value and find that value. c) Find the range. d) On what intervals is the function increasing? decreasing? Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -55
Solution a) f(x) = 3 x 2 + 18 x 21 The x-coordinate of the vertex is: Since f(3) = 3(3)2 + 18(3) 21 = 6, the vertex is (3, 6). b) Since a is negative, the graph opens down, so the second coordinate of the vertex, 6, is the maximum value of the function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -56
Solution continued c) The range is ( , 6]. d) Since the graph opens down, function values increase to the left of the vertex and decrease to the right of the vertex. Thus the function is increasing on the interval ( , 3) and decreasing on (3, ) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -57
Application A farmer has enough fence to enclose a rectangular pasture with 240 ft of fencing. If the barn forms one side of the rectangle, what is the maximum area that the farmer can enclose? What should the dimensions of the fence be in order to yield this area? 1. Familiarize. Make a drawing of the situation, using w to represent the width of the fencing. 2. Translate. Since the area of a rectangle is given by length times width, we have A(w) = (240 2 w)w = 240 w 2 w 2. w w Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 240 2 w Slide 2 -58
Application continued 3. Carry out. We need to find the maximum value of A(w) and find the dimensions for which that maximum occurs. The maximum will occur at the vertex of the parabola. Thus, if w = 60, then the length l = 240 2(60) = 120 ft and the area is (60)(120) = 7200 ft 2. 4. Check. (60 + 120) = 240 feet of fencing. 5. State. The maximum possible area is 7200 ft 2 when the pasture is 60 feet wide and 120 feet long. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -59
2. 5 More Equation Solving · Solve rational and radical equations and equations with absolute value. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Rational Equations containing rational expressions are called rational equations. Solving such equations requires multiplying both sides by the least common denominator (LCD) to clear the equation of fractions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -61
Example Solve: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -62
Example continued p The possible solution is 9. Check: p The solution is 9. p Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -63
Example Solve: The LCD is x 4. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -64
Example continued p p The possible solutions are 4 and 4. Check x = 4: Division by 0 is not defined, so 4 is not a solution. p Check x = 4: The number 4 checks, so it is a solution. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -65
Radical Equations p A radical equation is an equation in which variables appear in one or more radicands. For example: p The Principle of Powers For any positive integer n: If a = b is true, then an = bn is true. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -66
Example p Solve First we isolate the power. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -67
Example continued p Check for 4 p Check for 1 p Since 4 checks but 1 does not, the only solution is 4. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -68
Example p Solve: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -69
Equations with Absolute Value For a > 0 and an algebraic expression X: |X| = a is equivalent to X = a or X = a. Solve: The solutions are 9 and 3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -70
2. 6 Solving Linear Inequalities · · Solve linear inequalities. Solve compound inequalities. Solve inequalities with absolute value. Solve applied problems using inequalities. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Inequalities p An inequality is a sentence with <, >, , or as its verb. Examples: 5 x 7 < 3 + 4 x 3(x + 6) 4(x 3) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -72
Principles for Solving Inequalities For any real numbers a, b, and c: The Addition Principle for Inequalities: If a < b is true, then a + c < b + c is true. The Multiplication Principle for Inequalities: If a < b and c > 0 are true, then ac < bc is true. If a < b and c < 0, then ac > bc is true. Similar statements hold for a b. When both sides of an inequality are multiplied or divided by a negative number, we must reverse the inequality sign. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -73
Examples Solve: {x|x < 2} or ( , 2) {x|x 4} or ( 4, ) [ ) – 5 0 5 – 5 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5 Slide 2 -74
Compound Inequalities When two inequalities are joined by the word and or the word or, a compound inequality is formed. Conjunction contains the word and. Example: 7 < 3 x + 5 and 3 x + 9 6 Disjunction contains the word or. Example: 3 x + 5 6 or 3 x + 6 > 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -75
Examples Solve: 4 x 5 3 or 4 x 5 > 3 ] ( – 5 0 ] ( 5 – 5 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5 Slide 2 -76
Inequalities with Absolute Value Inequalities sometimes contain absolute-value notation. The following properties are used to solve them. For a > 0 and an algebraic expression X: |X| < a is equivalent to a < X < a. |X| > a is equivalent to X < a or X > a. Similar statements hold for |X| a and |X| a. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -77
Example p Solve: ( ) – 5 0 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -78
Application Johnson Catering charges $100 plus $30 per hour to cater an event. Catherine’s Catering charges $50 per hour. For what lengths of time does it cost less to hire Catherine’s Catering? 1. Familiarize. Read the problem. 2. Translate. Catherine’s 50 x is less than Johnson < 100 + 30 x Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -79
Application continued 3. Carry out. 4. Check. 5. State. For values of x < 5 hr, Catherine’s Catering will cost less. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 2 -80
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