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Rational Expressions and Functions

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1

2 Rational Expressions and Functions
Chapter 8 Rational Expressions and Functions

3 Equations with Rational Expressions and Graphs
8.4 Equations with Rational Expressions and Graphs

4 8.4 Equations with Rational Expressions and Graphs
Objectives Determine the domain of the variable in a rational expression. Solve rational equations. Recognize the graph of a rational function. Copyright © 2010 Pearson Education, Inc. All rights reserved.

5 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 1 Determining the Domains of Rational Equations Find the domain of the equation. (a) 5 x = 1 6 11 2x The domain is { x | x ≠ 0 }. The domains of the three rational terms of the equation are, in order, { x | x ≠ 0 }, (-∞, ∞), { x | x ≠ 0 }. The intersection of these three domains is all real numbers except 0, which may be written { x | x ≠ 0 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

6 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 1 Determining the Domains of Rational Equations Find the domain of the equation. (b) 3 x – 1 = 2 x + 1 6 x2 – 1 The domain is { x | x ≠ +1 }. The domains of the three rational terms are, respectively, { x | x ≠ 1 }, { x | x ≠ –1 }, { x | x ≠ +1 }. The domain of the equation is the intersection of the three domains, all real numbers except 1 and –1, written { x | x ≠ +1 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

7 8.4 Equations with Rational Expressions and Graphs
Caution on “Solutions” CAUTION When each side of an equation is multiplied by a variable expression, the resulting “solutions” may not satisfy the original equation. You must either determine and observe the domain or check all potential solutions in the original equation. It is wise to do both. Copyright © 2010 Pearson Education, Inc. All rights reserved.

8 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 2 Solving an Equation with Rational Expressions Solve 5 x = 1 6 11 2x The domain, which excludes 0, was found in Example 1(a). = 5 x 1 6 11 2x 6x Multiply by the LCD, 6x. = 5 x 1 6 11 2x 6x Distributive property = 30 x 33 Multiply. = x 3 Subtract 30. = x – 3 Divide by –1. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 2 Solving an Equation with Rational Expressions Solve 5 x = 1 6 11 2x Check: Replace x with –3 in the original equation. = 5 x 1 6 11 2x Original equation = 5 –3 1 6 11 2(–3) ? Let x = –3. = 10 6 1 11 –6 ? = 11 6 True The solution is { –3 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

10 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 3 Solving an Equation with No Solution Solve 3 x – 1 = 2 x + 1 6 x2 – 1 Using the result from Example 1(b), we know that the domain excludes 1 and –1, since these values make one or more of the denominators in the equation equal 0. Copyright © 2010 Pearson Education, Inc. All rights reserved.

11 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 3 Solving an Equation with No Solution Solve 3 x – 1 = 2 x + 1 6 x2 – 1 Multiply each side by the LCD, (x –1)(x + 1). = 3 x – 1 2 x + 1 6 x2 – 1 (x – 1)(x + 1) = 3 x – 1 2 x + 1 6 x2 – 1 (x – 1)(x + 1) Distributive property = 6 3(x + 1) 2(x – 1) Multiply. = 6 3x + 3 2x + 2 Distributive property = 6 x + 5 Combine terms. = 1 x Subtract 5. Copyright © 2010 Pearson Education, Inc. All rights reserved.

12 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 3 Solving an Equation with No Solution Solve 3 x – 1 = 2 x + 1 6 x2 – 1 Since 1 is not in the domain, it cannot be a solution of the equation. Substituting 1 in the original equation shows why. = 3 x – 1 2 x + 1 6 x2 – 1 Check: = 3 1 – 1 2 1 + 1 6 12 – 1 = 3 2 6 Since division by 0 is undefined, the given equation has no solution, and the solution set is ∅. Copyright © 2010 Pearson Education, Inc. All rights reserved.

13 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 4 Solving an Equation with Rational Expressions Solve 4 a2 – 9 = 3 2(a2 – 2a – 3) 6 a2 + 4a + 3 Factor each denominator to find the LCD, 2(a + 3)(a – 3)(a + 1). The domain excludes –3, 3, and –1. Copyright © 2010 Pearson Education, Inc. All rights reserved.

14 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 4 Solving an Equation with Rational Expressions Solve 4 a2 – 9 = 3 2(a2 – 2a – 3) 6 a2 + 4a + 3 Multiply each side by the LCD, 2(a + 3)(a – 3)(a + 1). = 4 (a + 3)(a – 3) 3 2(a – 3)(a + 1) 6 (a + 3)(a + 1) 2(a + 3)(a – 3)(a + 1) 4 · 2(a + 1) 3(a + 3) 6 · 2 (a – 3) = Distributive property 8a + 8 3a – 9 12a – 36 = Distributive property 5a – 1 12a – 36 = Combine terms 35 7a = Subtract 5a; Add 36. 5 a = Divide by 7. Copyright © 2010 Pearson Education, Inc. All rights reserved.

15 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 4 Solving an Equation with Rational Expressions Solve 4 a2 – 9 = 3 2(a2 – 2a – 3) 6 a2 + 4a + 3 Note that 5 is in the domain; substitute 5 in for a in the original equation to check that the solution set is { 5 }. 5 a = Copyright © 2010 Pearson Education, Inc. All rights reserved.

16 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 5 Solving an Equation That Leads to a Quadratic Equation Solve 4 2x + 1 2 x = 8x Since the denominator cannot equal 0, – is excluded from the domain, as is 0. 1 2 Copyright © 2010 Pearson Education, Inc. All rights reserved.

17 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 5 Solving an Equation That Leads to a Quadratic Equation Solve 4 2x + 1 2 x = 8x Multiply each side by the LCD, x(2x + 1). 4 2x + 1 = 2 x 8x x(2x + 1) 4x = 2(2x + 1) 8x2 Distributive property 4x = 4x + 2 8x2 Distributive property = –8x2 + 2 Subtract 4x; standard form. = –2(4x2 – 1) Factor. = –2(2x + 1)(2x – 1) Factor. Copyright © 2010 Pearson Education, Inc. All rights reserved.

18 8.4 Equations with Rational Expressions and Graphs
EXAMPLE 5 Solving an Equation That Leads to a Quadratic Equation Solve 4 2x + 1 2 x = 8x 2x + 1 = or x – 1 = 0 Zero-factor property x = – or x = 1 2 Because – is not in the domain of the equation, it is not a solution. Check that the solution set is 1 2 = –2(2x + 1)(2x – 1) Copyright © 2010 Pearson Education, Inc. All rights reserved.

19 8.4 Equations with Rational Expressions and Graphs
Graph of f (x) = 1 x The domain of this function includes all real numbers except x = 0. Thus, there will be no point on the graph with x = 0. The vertical line with equation x = 0 is called a vertical asymptote of the graph. Copyright © 2010 Pearson Education, Inc. All rights reserved.

20 8.4 Equations with Rational Expressions and Graphs
Graph of f (x) = 1 x The horizontal line with equation y = 0 is called a horizontal asymptote. Notice the closer positive values of x are to 0, the larger y is. Similarly, the closer negative values of x are to 0, the smaller (more negative) y is. Plot several points to verify this graph. Copyright © 2010 Pearson Education, Inc. All rights reserved.

21 8.4 Equations with Rational Expressions and Graphs
Graph of g(x) = –2 x – 3 There is no point on the graph for x = 3 because 3 is excluded from the domain. The dashed line x = 3 represents the vertical asymptote and is not part of the graph. Notice the graph gets closer to the vertical asymptote as the x-values get closer to 3. Copyright © 2010 Pearson Education, Inc. All rights reserved.

22 8.4 Equations with Rational Expressions and Graphs
Graph of g(x) = –2 x – 3 Observe the y-values as the x-values get closer to the vertical asymptote (from both sides). As the x-values get closer to the vertical asymptote from the left, the y-values get larger and as the asymptote from the right, the y-values get smaller (more negative). Copyright © 2010 Pearson Education, Inc. All rights reserved.

23 8.4 Equations with Rational Expressions and Graphs
Graph of g(x) = –2 x – 3 Again, y = 0 is a horizontal asymptote. Plot several points to verify this graph. Copyright © 2010 Pearson Education, Inc. All rights reserved.


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