2-6 Nonlinear Inequalities

Size: px

Start display at page:

Download "2-6 Nonlinear Inequalities"

Alan Beasley
5 years ago
Views:

31. Find the domain of each expression. For the expression to be defined, x 2 3x 40 0. Let f (x) = x 2 3x 40. A factored form of f (x) is f (x) = (x 8)(x + 5). f (x) has real zeros at x = 8 and x = 5.

1 31. Find the domain of each expression. For the expression to be defined, x 2 3x Let f (x) = x 2 3x 40. A factored form of f (x) is f (x) = (x 8)(x + 5). f (x) has real zeros at x = 8 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point. The solutions of x 2 3x 40 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is (, 5] [8, ). esolutions Manual - Powered by Cognero Page 1

2 33. For the expression to be defined, x Let f (x) = x 2 9. A factored form of f (x) is f (x) = (x 3)(x + 3). f (x) has real zeros at x = 3 and x = 3. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point. The solutions of x are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is (, 3] [3, ). 35. For the expression to be defined, the denominator of the radicand, 36 x 2, cannot equal 0. Let f (x) = 36 x 2. Set this function equal to 0 and solve for x. 36 x 2 = 0 at x = 6 and x = 6. Thus, the domain of the expression is (, 6) ( 6, 6) (6, ). esolutions Manual - Powered by Cognero Page 2

3 Find the solution set of f (x) g(x) f(x) g(x) 0 for values of x that result in values for f (x) that are greater than or equal to the values of g(x). This occurs when x is [ 4, 3]. Thus, the solution set is [ 4, 3]. 39. PARKS AND RECREATION A rectangular playing field for a community park is to have a perimeter of 112 feet and an area of at least 588 square feet. a. Write an inequality that could be used to find the possible lengths to which the field can be constructed. b. Solve the inequality you wrote in part a and interpret the solution. c. How does the inequality and solution change if the area of the field is to be no more than 588 square feet? Interpret the solution in the context of the situation. a. The formula for the perimeter of a rectangle is P = 2 + 2w. Substitute P = 112 and solve for w. The formula for the area of a rectangle is w = A. Substitute w = 56 and A = 588 to write an inequality that could be used to find the possible lengths to which the field can be constructed. b. Solve Let f ( ) = The factored form of f ( ) is f ( ) = ( + 14)( 42). f( ) has real zeros at = 14 and = 42. Set up a sign chart. Substitute an -value in each test interval into the factored polynomial to determine if f ( ) is positive or negative at that point. esolutions Manual - Powered by Cognero Page 3

4 The solutions of are -values such that f ( ) is positive or equal to 0. From the sign chart, you can see that the solution set is [14, 42]. The length of the playing field is at least 14 and at most 42 feet. c. If the area of the field is to be no more than 588 square feet, the inequality becomes 0 < (56 ) 588. Notice that the area of the field must be greater than 0. Start by solving 0 < (56 ). Since the length, width, and area of the rectangle must be positive, 0 < < 56. The solutions of (56 ) 588 or are -values such that f ( ) is negative or equal to 0. From the sign chart, you can see that the solution set is (, 14] [42, ). Since 0 < < 56, the solution set is (0, 14] [42, 56). The area of the playing field must be greater than 0 square feet but at most 588 square feet, so 0 < 14 or 42 < 56. Solve each inequality. (Hint: Test every possible solution interval that lies within the domain using the original inequality.) For the inequality to be defined, 4x and x 4 0 must be true. Solve both inequalities for x. The domain of x must be restricted to x 4. Solve the original inequality. Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign. esolutions Manual - Powered by Cognero Page 4

5 Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign. Let f (x) = (9x 40)(x 8). f (x) has real zeros at x = and x = 8. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point. The solutions of (9x 40)(x 8) 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is. Recall that x 4. Create a chart that includes this restriction. When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using the original inequality. Substitute an x- value in each test interval into the original inequality to determine if f (x) is a solution. esolutions Manual - Powered by Cognero Page 5

6 Thus, the solution set is. 43. < 5 For the inequality to be defined, 25 12x 0 and 16 4x 0 must be true. Solve both inequalities for x. The domain of x must be restricted to x. Solve the original inequality. esolutions Manual - Powered by Cognero Page 6

7 Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign. Let f (x) = 4(4x 7)(x + 12). f (x) has real zeros at x = and x = 12. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point. The solutions of 4(4x 7)(x + 12) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution set is. Recall that x. Create a chart that includes this restriction. When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using the original inequality. Substitute an x-value in each test interval into the original inequality to determine if f (x) is a solution. esolutions Manual - Powered by Cognero Page 7

8 Notice that x-values that lie outside of the solution set found by solving the inequality for x are also solutions. Thus, the solution set is. Solve each inequality a 4 + 7a 3 56a 2 80a < 0 Let f (a) = 3a 4 + 7a 3 56a 2 80a. f (a) can be written as f (a) = a(3a 3 + 7a 2 56a 80). By using synthetic division, it can be determined that 4 is a rational zero of 3a 3 + 7a 2 56a 80. The remaining quadratic factor (3a a + 20) can be written as (3a + 4)(a + 5). A factored form of f (a) is f (a) = a(a 4)(3a + 4)(a + 5). f (a) has real zeros at 0, 4,, and 5. Set up a sign chart. Substitute an a-value in each test interval into the factored polynomial to determine if f (a) is positive or negative at that point. esolutions Manual - Powered by Cognero Page 8

9 The solutions of 3a 4 + 7a 3 56a 2 80a < 0 are a-values such that f (a) is negative. From the sign chart, you can see that the solution set is. esolutions Manual - Powered by Cognero Page 9

10 Solve each inequality. 51. (x + 3) 2 (x 4) 3 (2x + 1) 2 < 0 Let f (x) = (x + 3) 2 (x 4) 3 (2x + 1) 2. f (x) has real zeros at x = 3 (multiplicity: 2), x = 4 (multiplicity: 3), and (multiplicity: 2). Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point. The solutions of (x + 3) 2 (x 4) 3 (2x + 1) 2 < 0 are x-values such that f (x) is negative. Since the inequality cannot equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is. esolutions Manual - Powered by Cognero Page 10

11 53. (a 3) 3 (a + 2) 3 (a 6) 2 > 0 Let f (a) = (a 3) 3 (a + 2) 3 (a 6) 2. f (a) has real zeros at a = 3 (multiplicity: 3), y = 2 (multiplicity: 3), and a = 6 (multiplicity: 2). Set up a sign chart. Substitute an a-value in each test interval into the polynomial to determine if f (a) is positive or negative at that point. The solutions of (a 3) 3 (a + 2) 3 (a 6) 2 > 0 are a-values such that f (a) is positive. Since the inequality cannot equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is (, 2) (3, 6) (6, ). esolutions Manual - Powered by Cognero Page 11

2-4 Zeros of Polynomial Functions

2-4 Zeros of Polynomial Functions Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. 33. 2, 4, 3, 5 Using the Linear Factorization Theorem and the zeros 2, 4, 3, and 5, write f