Solved Multiple Choice Questions on College Algebra - Final Exam | MATH 110, Exams of Algebra

Download Solved Multiple Choice Questions on College Algebra - Final Exam | MATH 110 and more Exams Algebra in PDF only on Docsity! 1 MATH 110 REVIEW to accompany Sullivan: College Algebra, 8th Ed. edited by JOHN A. BEACHY Northern Illinois University 2 Preface This set of review materials contains a brief summary of each section, together with sample test questions that have been taken from previous exams actually given in Math 110. You can find the solutions to odd numbered problems from the text in the Student Solutions Manual. John Beachy Spring 2006 The review material has been updated for the Eighth Edition of College Algebra. Note that there are now solutions to the problems that are not in the Student Solutions Manual. Fall 2007 5 6 PREFACE Chapter 0 Review Section summaries Section R.1: Real Numbers You should review the rules for working with numbers, especially those for fractions (see page 13): a b + c d = ad + bc bd a b · c d = ac bd a b ÷ c d = a b · d c where b, c, d are nonzero Review problems: p16 #71,73,77,81,83 Section R.2: Algebra Essentials In this section the rules for exponents are particularly important. aman = am+n a0 = 1 a−n = 1 an am an = am−n (am)n = amn (ab)n = anbn (a b )n = an bn (a b ) −n = bn an Explanation: The shorthand notation that we use in writing a3 instead of aaa dates back only to the 1600’s, so it is a fairly recent invention (in the long history of math). When we multiply powers in this form, we can expand the powers and just count how many terms we have in the result. For example, a4 · a3 = aaaa · aaa = a7, This method works for all positive exponents and leads to the general rule am · an = am+n. It’s very useful to allow the exponents to be 0 or a negative number. But then there’s no way to interpret the shorthand notation as a repeated product of a. We should at least make sure that any definition we give is consistent with the general formula am ·an = am+n. To define a0, we note that by the general rule for exponents we should have am · a0 = am+0 = am. Dividing both sides of the equation by am gives us am · a0 am = am am , and so we should have a0 = 1. This shows that to be consistent with the rule for exponents we must define a0 = 1. 7 10 CHAPTER 0. REVIEW Sample Questions R.2 #92. After simplifying, the numerator of 4x−2(yz)−1 23x4y is (a) 2x2 (d) 1 (b) 2x2z (e) 4 (c) yz R.3 #15. The lengths of the legs of a right triangle are 7 and 24. Find the length of the hypotenuse. (a) 84 (b) 31 (c) 25 (d) √ 97 (e) √ 62 R.3 A. Find the volume V and the surface area S of a rectangular box with length 4 feet, width 2 feet, and height 5 feet. (a) V = 40 cubic feet and S = 76 square feet (b) V = 40 cubic feet and S = 11 square feet (c) V = 20 cubic feet and S = 76 square feet (d) V = 20 cubic feet and S = 11 square feet (e) None of these R.3 #31. Find the volume V and surface area S of a rectangular box with length 8 feet, width 4 feet, and height 7 feet. (a) V = 214 cubic feet and S = 116 square feet (b) V = 214 cubic feet and S = 232 square feet (c) V = 224 cubic feet and S = 116 square feet (d) V = 224 cubic feet and S = 232 square feet (e) None of these R.4 A. Perform the indicated operation and simplify: (2x − 5)(3x + 4) (a) 6x2 − 20 (d) 5x2 − 2x − 20 (b) 6x2 − 7x − 20 (e) 5x − 1 (c) 5x2 − 7x − 20 R.4 B. Perform the indicated operation and simplify: (2x2)3(4x3) (a) 32x9 (d) 6x8 (b) 32x8 (e) None of these (c) 8x8 11 R.4 C. Simplify: (x2 − 3x + 1) − (2x − 5) (a) −2x3 + 11x2 − 17x + 5 (d) x2 − 5x + 6 (b) x2 − x − 4 (e) x2 − x + 6 (c) x2 − 5x − 4 R.4 Example 9b. Expand and simplify: (x − 1)3 (a) x3 − 1 (b) x3 − x2 − x + 1 (c) x3 − 3x2 − x + 1 (d) x3 − 3x2 + 3x − 1 (e) None of these R.4 #64. Multiply and simplify: (x − 3y)(−2x + y) (a) −2x2 − 5xy − 3y2 (d) −2x2 − 7xy + 3y2 (b) −2x2 − 5xy + 3y2 (e) None of these (c) −2x2 − 7xy − 3y2 R.4 #84. Expand and simplify: (2x + 3y)2 (a) 2x2 + 3y2 (d) 4x2 + 6xy + 9y2 (b) 4x2 + 9y2 (e) None of these (c) 4x2 + 12xy + 9y2 R.4 #87. Expand and simplify: (2x + 1)3 (a) 8(x3 + 1) (b) 8x3 + 1 (c) 8x3 + 4x2 + 2x + 1 (d) 8x3 + 12x2 + 6x + 1 (e) None of these R.4 D. What is the remainder when x2 + x + 1 is divided by x − 2? (a) 7 (d) 4 (b) 6 (e) None of these (c) 5 12 CHAPTER 0. REVIEW R.4 #93. When 5x4 − 3x2 + x + 1 is divided by x2 + 2 the remainder is (a) x + 27 (b) x + 25 (c) x + 15 (d) x + 13 (e) None of these R.4 #97. When 2x4 − 3x3 + x + 1 is divided by 2x2 + x + 1 the quotient is (a) x2 − 2x (b) x2 + 2x (c) x2 − 2x + 12 (d) x2 + 2x − 12 (e) None of these R.4 #103. When x3 − a3 is divided by x − a the quotient is (a) x2 + ax + a2 (b) x2 − ax + a2 (c) x2 + a2 (d) x2 − a2 (e) None of these R.5 Example 11. Factor the following expression completely: x2 − x − 12 (a) (x − 6)(x + 2) (d) (x − 4)(x + 3) (b) (x + 4)(x − 3) (e) None of these (c) (x + 6)(x − 2) R.5 A. Which one of the following is a factor of 4x2 + 5x − 6? (a) 4x + 3 (d) 2x − 3 (b) x − 2 (e) 4x + 5 (c) 4x − 3 R.5 B. Factor completely: ac2 + 5bc2 − a − 5b (a) (a + 5b)(c + 1)(c − 1) (d) (a − 5b)(c + 1)2 (b) (a − 5b)(c + 1)(c − 1) (e) None of these (c) (a + 5b)(c + 1)2 15 R.7 #72. After simplifying, the numerator of 1 h [ 1 (x + h)2 − 1 x2 ] is (a) −2x − h (b) −2x + h (c) 2x − h (d) 2x + h (e) None of these R.7 #76. After simplifying, the denominator of 1 − x x + 1 2 − x − 1 x is (a) x + 1 (b) (x + 1)2 (c) x2 − 1 (d) (x − 1)2 (e) None of these R.7 D. Simplify and factor: 4 + 1 x2 25 − 1 x2 . (a) 4 25 (d) (x + 2)(x − 2) (x + 5)(x − 5) (b) 4x2 + 1 (x + 5)(x − 5) (e) 4x2 + 1 (5x + 1)(5x − 1) (c) (2x + 1)(2x − 1) (5x + 1)(5x − 1) R.8 A. Simplify: 2 √ 3 − √ 48 (a) 2 √ 3 (d) 3 √ 5 (b) −14 √ 3 (e) −6 (c) −2 √ 3 R.8 B. Simplify: 2 √ 3 + 2 √ 12 (a) 2 √ 15 (d) 30 (b) 6 √ 3 (e) None of these (c) 10 √ 3 16 CHAPTER 0. REVIEW R.8 Example 8a. Simplify: (x2/3y)(x−2y)1/2 (a) x8/3 (d) y3/2 x1/3 (b) y x2/3 (e) None of these (c) y3/2 x4/3 R.8 Example 8c. Simplify: ( 9x2y1/3 x1/3y )1/2 (a) 3x (d) 9x9/5 y1/3 (b) 3x9/5 y1/3 (e) None of these (c) 3x5/6 y1/3 R.8 #18. Simplify (assuming that all variables are positive: 3 √ 3xy2 81x4y2 (a) 1√ 3x (d) 1 3 √ x (b) √ 3 x (e) None of these (c) x√ 3 R.8 #30. Simplify: 2 √ 12 − 3 √ 27 (a) − √ 15 (d) −6 √ 324 (b) −19 √ 3 (e) None of these (c) −5 √ 3 R.8 C. Simplify: 4 3 √ 7 − 3 3 √ 56 (a) 2 (d) 2 3 √ 7 (b) −3 3 √ 49 (e) None of these (c) −4 3 √ 7 17 R.8 D. Rationalize the denominator: 10 4 − √ 2 (a) 5 √ 2 (d) 5(4 − √ 2) 9 (b) 5(4 − √ 2) 7 (e) 5(4 + √ 2) 9 (c) 5(4 + √ 2) 7 R.8 E. Simplify ( 1 64 ) −2/3 (a) 16 (d) − 196 (b) 512 (e) None of these (c) − 116 R.8 F. Simplify ( 27 8 ) −2/3 (a) 94 (d) −16 81 (b) 49 (e) None of these (c) −49 R.8 G. Factor the expression x1/2(x2 + x) + x3/2 − 24x1/2 (where x ≥ 0). (a) x1/2(x + 1)(x − 3) (b) x3/2(x + 1)(x − 3) (c) x3/2(x + 2)(x − 6) (d) x1/2(x + 6)(x − 4) (e) (x + 6)(x − 4) R.8 H. Multiply and simplify (2 √ x − 3)(2√x + 5) (a) 4x + 4 √ x − 15 (b) 2x + 4 √ x − 15 (c) 4x − 15 (d) 4 √ x − 15 (e) None of these 20 CHAPTER 0. REVIEW R.8 A. (c) R.8 B. (b) R.8 Example 8a. (d) R.8 Example 8c. (c) R.8 #18. (e) R.8 #30. (c) R.8 C. (e) R.8 D. (c) R.8 E. (a) R.8 F. (b) R.8 G. (d) R.8 H. (a) R.8 #37. (d) R.8 #53. (b) Solutions R.2 #92. 4x−2(yz)−1 23x4y = 4 8x2x4y(yz) = 1 2x6y2z R.3 A. Find the volume V and the surface area S of a rectangular box with length 4 feet, width 2 feet, and height 5 feet. Solution: The volume is length times width times height, so we get 4 · 2 · 5 = 40 cubic feet. To find the surface area: there are 4 sides, two that are 4 × 5 and two that are 2 × 5. The total area for the sides is 20 + 20 + 10 + 10 = 60 square feet. The top and bottom of the box are 4× 2, adding 8 + 8 = 16 more square feet. The total surface area is 76 square feet. The answer is V = 40 cubic feet and S = 80 square feet. R.4 A. (2x− 5)(3x + 4) = (2x)(3x + 4)− (5)(3x + 4) = 6x2 + 8x− 15x− 20 = 6x2 − 7x− 20 R.4 B. (2x2)3(4x3) = 23(x2)3(4x3) = 8x6 · 4x3 = (8 · 4)(x6x3) = 32x9 R.4 C. (x2−3x+1)−(2x−5) = x2−3x+1−2x+5 = x2+(−3x−2x)+(1+5) = x2−5x+6 R.4 #64. (x − 3y)(−2x + y) = x(−2x + y) + (−3y)(−2x + y) = −2x2 + xy + 6xy − 3y2 = −2x2 + 7xy − 3y2 R.4 #84. (2x + 3y)2 = (2x + 3y)(2x + 3y) = 4x2 + 6xy + 6xy + 9y2 = 4x2 + 12xy + 9y2 21 R.4 D. What is the remainder when x2 + x + 1 is divided by x − 2? Solution: The remainder is +7. x +3 x −2 x2 +x +1 x2 −2x +3x +1 +3x −6 +7 R.5 A. 4x2 + 5x − 6 = (4x − 3)(x + 2) R.5 B. ac2+5bc2−a−5b = c2(a+5b)−a−5b = c2(a+5b)+(−1)(a+5b) = (c2−1)(a+5b) = (c + 1)(c − 1)(a + 5b) R.5 #122. 4(x + 5)3(x − 1)2 + 2(x + 5)4(x − 1) = (2 · 2)(x + 5)3(x − 1)(x − 1) + 2(x + 5)3(x + 5)(x − 1) = (find common terms) 2(x + 5)3(x − 1) · (2(x − 1) + (x + 5)) = (factor out common terms) 2(x + 5)3(x − 1)(2x − 2 + x + 5) = 2(x + 5)3(x − 1)(3x + 3) = 6(x + 5)3(x − 1)(x + 1) R.7 A. x2 + x − 6 x2 − 4 = (x + 3)(x − 2) (x + 2)(x − 2) = x + 3 x + 2 R.7 B. 1 x − 1− 2 x + 2 = x + 2 (x − 1)(x + 2) − 2(x − 1) (x − 1)(x + 2) = x + 2 − 2x + 2 (x − 1)(x + 2) = −x + 4 (x − 1)(x + 2) R.7 C. ( x x + 1 ) ( 2x + 2 x2 ) = ( x x + 1 )( x2 2(x + 1) ) = x3 2(x + 1)2 R.7 #66. 2 (x + 2)2(x − 1) − 6 (x + 2)(x − 1)2 = 2(x − 1) (x + 2)2(x − 1)2 − 6(x + 2) (x + 2)2(x − 1)2 = 2x − 2 − 6(x + 2) (x + 2)2(x − 1)2 = 2x − 2 − 6x − 12 (x + 2)2(x − 1)2 = −4x − 14 (x + 2)2(x − 1)2 = −2(2x + 7) (x + 2)2(x − 1)2 R.7 #72. 1 h [ 1 (x + h)2 − 1 x2 ] = 1 h [ x2 − (x + h)2 (x + h)2x2 ] = 1 h [ x2 − (x2 + 2xh + h2) (x + h)2x2 ] = 1 h [ x2 − x2 − 2xh − h2 (x + h)2x2 ] = 1 h [−2xh − h2 (x + h)2x2 ] = 1 h [ h(−2x − h) (x + h)2x2 ] = −2x − h (x + h)2x2 22 CHAPTER 0. REVIEW R.7 #76. 1 − x x + 1 2 − x − 1 x = x + 1 − x x + 1 2x − (x − 1) x = 1 x + 1 x + 1 x = 1 x + 1 · x x + 1 = x (x + 1)2 R.7 D. 4 + 1 x2 25 − 1 x2 = 4x2 x2 + 1 x2 25x2 x2 − 1 x2 = 4x2 + 1 x2 25x2 − 1 x2 = 4x2 + 1 x2 · x 2 25x2 − 1 = x2(4x2 + 1) x2(25x2 − 1) = 4x2 + 1 25x2 − 1 = 4x2 + 1 (5x + 1)(5x − 1) R.8 A. Simplify: 2 √ 3− √ 48 = 2 √ 3− √ 16 · 3 = 2 √ 3− √ 16 · √ 3 = 2 √ 3−4 · √ 3 = −2 √ 3 R.8 B. 2 √ 3 + 2 √ 12 = 2 √ 3 + 2 √ 4 · 3 = 2 √ 3 + 2 · 2 √ 3 = 2 √ 3 + 4 √ 3 = 6 √ 3 R.8 #18. 3 √ 3xy2 81x4y2 = 3 √ 1 27x3 = 3 √ 1 3 √ 27 3 √ x3 = 1 3x R.8 #30. 2 √ 12 − 3 √ 27 = 2 √ 4 · 3 − 3 √ 9 · 3 = 2 · 2 √ 3 − 3 · 3 √ 3 = 4 √ 3 − 9 √ 3 = −5 √ 3 R.8 C. 4 3 √ 7 − 3 3 √ 56 = 4 3 √ 7 − 3 3 √ 8 · 7 = 4 3 √ 7 − 3 · 2 3 √ 7 = 4 3 √ 7 − 6 3 √ 7 = −2 3 √ 7 R.8 D. 10 4 − √ 2 = 10(4 + √ 2) (4 − √ 2)(4 + √ 2) = 10(4 + √ 2) 16 − 2 = 2 · 5 · (4 + √ 2) 2 · 7 = 5(4 + √ 2) 7 R.8 E. ( 1 64 ) −2/3 = ( 64 1 )2/3 = 642/3 = ( 3 √ 64)2 = 42 = 16 R.8 F. ( 27 8 ) −2/3 = ( 8 27 )2/3 = 82/3 272/3 = ( 3 √ 8)2 ( 3 √ 27)2 = 22 32 = 4 9 R.8 G. x1/2(x2+x)+x3/2−24x1/2 = x1/2(x2+x)+x1/2 ·x−24x1/2 = x1/2(x2+x+x−24) = x1/2(x2 + 2x − 24) = x1/2(x + 6)(x − 4) R.8 H. (2 √ x − 3)(2√x + 5) = (2√x)2 − 6√x + 10√x − 15 = 4x + 4√x − 15 25 Sample Questions 1.1 A. Solve for x: 7 − 2x = 9 + 3x (a) x = 2 (d) x = − 25 (b) x = −2 (e) x = −3 (c) x = 25 1.1 B. Solve the equation: 1 − 12x = 6 + x. (a) x = 103 (d) x = −3 (b) x = −103 (e) None of these (c) x = 2 1.1 Example 6. Solve the equation: 3x x − 1 + 2 = 3 x − 1 (a) x = 1 (b) x = 5 (c) x = 15 (d) There is no solution (e) None of these 1.1 #51. Solve this equation: 2x x2 − 4 = 4 x2 − 4 − 3 x + 2 (a) x = 2 or x = −2 (b) x = −1 (c) x = 2 (d) There is no solution (e) None of these 1.1 #59. Solve this equation: 4 x − 2 = −3 x + 5 + 7 x2 + 3x − 10 (a) x = 2 (b) x = 1 (c) x = −1917 (d) There is no solution (e) None of these 26 CHAPTER 1. EQUATIONS AND INEQUALITIES 1.1 C. Going into the final exam, which will count as two tests, Brooke has test scores of 80, 83, 71, 61, and 89. What score does Brooke need on the final in order to have an average score of 80? (a) 90 (d) 82 (b) 88 (e) None of these (c) 85 1.1 #96. A wool suit, discounted by 30% for a clearance sale, has a price tag of $399. What was the suit’s original price? (a) Not enough information to determine (d) $532 (b) $306.92 (approximately) (e) $570 (c) $518.70 1.2 A.Solve for x: x2 − 3x + 2 = 0 (a) x = 1 or x = 2 (d) x = 2 or x = −3 (b) x = 1 or x = −2 (e) None of these (c) x = 2 or x = 3 1.2 B. Solve for x: x2 − 2x = 4 (a) x = 4 or x = 2 (b) x = 4 + √ 5 2 or x = 2 − √ 5 2 (c) x = 2 + √ 5 2 or x = 4 − √ 5 2 (d) x = 1 + √ 5 or x = 1 − √ 5 (e) None of these 1.2 C. Find the value of a so that x2 + ax + 19 is a perfect square. (a) a = 13 (b) a = 2 3 (c) a = 1 9 (d) a = 2 9 (e) None of these 1.2 D. Find the value of k so that x2 − 32x + k is a perfect square. (a) 34 (b) −34 (c) 916 (d) − 916 (e) None of these 27 1.2 #51. Use the quadratic formula to solve this equation: 23x 2 − 53x + 1 = 0 (a) x = −1 and x = −2 (b) x = −1 and x = − 32 (c) x = 1 and x = − 32 (d) x = 1 and x = 2 (e) None of these 1.2 #67. Find the real solutions, if any: 4 − 1 x − 2 x2 = 0 (a) x = −1 ± √ 17 8 (b) x = 1 ± √ 33 8 (c) x = 1 2 or x = −1 4 (d) There is no solution (e) None of these 1.2 #87. Use the quadratic formula to find the real solutions: x2 + √ 2x = 12 (a) x = √ 2 ± 2 2 (d) x = − √ 2 ± √ 3 2 (b) x = − √ 2 ± 2 2 (e) None of these (c) x = √ 2 ± √ 3 2 1.2 #105. An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, then the dimensions of the sheet metal should be (a) 1 foot by 1 foot (b) 2 feet by 2 feet (c) 4 feet by 4 feet (d) 8 feet by 8 feet (e) None of these 30 CHAPTER 1. EQUATIONS AND INEQUALITIES 1.6 A. The solution set of the combined inequality −1 < 3 − 2x ≤ 15 is (a) (−6, 2] (d) [2, 6) (b) [−6, 2) (e) [−13/2,∞) (c) (2, 6] 1.6 B. Solve this inequality: |x − 2| < 3 (a) −2 < x < 1 (d) 0 < x < 1 (b) −2 < x < 5 (e) None of these (c) 2 < x < 5 1.6 Example 6. Solve this inequality: |2x − 5| > 3 (a) x < 1 or x > 4 (d) x < 4 and x > −1 (b) x < −1 or x > 4 (e) None of these (c) x < 4 and x > 1 1.6 C. Find the solution set of this inequality: |5 − 2x| < 9 (a) The empty set (no solutions) (d) (−2,∞) (b) {x | x > 7 or x < −2} (e) (−∞, 7) (c) (−2, 7) 1.6 D. Find the solution set of this inequality: |3 − 2x| ≥ 7 (a) {x | x ≤ 5 or x ≥ −2} (d) all real numbers (b) { x | x ≥ 5 or x ≤ −2} (e) (−∞,−2] (c) [−2, 5] 1.6 #43. The solution set of the inequality | 2x − 3 |≥ 2 is (a) {x | 12 ≤ x ≤ 52} (d) {x | −12 ≤ x ≤ 52} (b) {x | x ≤ −12 or x ≥ 52} (e) None of these (c) {x | x ≤ 12 or x ≥ 52} 1.6 #45. Solve this inequality: |1 − 4x| − 7 < −2 (a) −32 < x < 1 (d) −1 < x < 32 (b) −32 < x < −1 (e) None of these (c) 1 < x < 32 31 1.7 #27. A motorboat maintains a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. If the total time for the trip is 1.5 hours, then the speed of the current must be (a) 2 miles per hour (d) 10 miles per hour (b) 4 miles per hour (e) None of these (c) 5 miles per hour 1.7 A. (see #28) Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 AM, each heading for Wildwood. One car’s average speed is 5 miles per hour more than the other’s. The slower car arrives at Wildwood at 11:00 AM, 15 minutes after the other car. What is the average speed of the slower car? (a) 50 mph (d) 70 mph (b) 55 mph (e) None of these (c) 60 mph 1.7 #33. Trent can deliver his newspapers in 30 minutes. It takes Lois 20 minutes to do the same route. How long would it take them to deliver the newspapers if they work together? (a) 50 minutes (d) 10 minutes (b) 25 minutes (e) None of these (c) 12 minutes 1.7 B. If x gallons of cherry juice costing $3.00 per gallon are to be combined with y gallons of apple juice costing $1.00 per gallon to make a fruit juice mix costing $2.50 per gallon, then what is x y ? (a) 3 (d) 85 (b) 10 (e) It cannot be determined (c) 13 32 CHAPTER 1. EQUATIONS AND INEQUALITIES Answer Key 1.1 A. (d) 1.1 B. (b) 1.1 Example 6. (d) 1.1 #51. (d) 1.1 #59. (e) 1.1 C. (b) 1.1 #96. (e) 1.2 A.(a) 1.2 B. (d) 1.2 C. (b) 1.2 D. (c) 1.2 #51. (e) 1.2 #67. (b) 1.2 #87. (b) 1.2 #105. (c) 1.2 #107. (d) 1.2 E. (d) 1.4 A. (c) 1.4 #25. (d) 1.4 #28. (b) 1.4 #31. (e) 1.5 A. (a) 1.5 #75. (a) 1.5 #80. (b) 1.5 #87. (a) 1.5 B. (c) 1.6 A. (b) 1.6 B. (e) 1.6 Example 6. (a) 1.6 C. (c) 35 1.5 B. The solution set of the inequality 0 < (x − 4)−1 < 12 is Solution: First note that we must have x − 4 > 0. Then we can invert the inequality 1 x − 4 < 1 2 to get x − 4 > 2. (Remember that inverting the terms reverses the inequality. For example, 14 < 1 2 , but 4 > 2.) The final answer is (6,∞). 1.6 A. The solution set of the combined inequality −1 < 3 − 2x ≤ 15 is Solution: −1 < 3 − 2x ≤ 15 −4 < −2x ≤ 12 4 > 2x ≥ −12 2 > x ≥ −6 Now −6 ≤ x < 2, so the final answer (in interval form) is [−6, 2). 1.6 B. Solve this inequality: |x − 2| < 3 Solution: |x − 2| < 3 −3 < x − 2 < 3 −1 < x < 5 1.6 C. Find the solution set of this inequality: |5 − 2x| < 9 Solution: |5 − 2x| < 9 −9 < 5 − 2x < 9 −14 < −2x < 4 −4 < 2x < 14 −2 < x < 7 Answer (in interval form): (−2, 7) 1.6 D. Find the solution set of this inequality: |3 − 2x| ≥ 7 Solution: We need to replace |3−2x| ≥ 7 with two inequalities: 3−2x ≥ 7 or 3−2x ≤ −7. −2x ≥ 4 or −2x ≤ −10 x ≤ −2 or x ≥ 5 In set notation: { x | x ≥ 5 or x ≤ −2} 1.7 A. (see #28) Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 AM, each heading for Wildwood. One car’s average speed is 5 miles per hour more than the other’s. The slower car arrives at Wildwood at 11:00 AM, 15 minutes after the other car. What is the average speed of the slower car? Solution: Let x be the average speed of the slower car. Use the fact that both cars travel the same distance and the formula rate · time = distance, with distance in miles and time in hours. Distance for the slower car: 3x. Distance for the faster car: (2 34)(x + 5). We get this equation: 3x = (2 34)(x + 5). Solve: 3x = (234)x + ( 11 4 )(5) (3 − 234)x = 554 14x = 554 x = 55 1.7 B. If x gallons of cherry juice costing $3.00 per gallon are to be combined with y gallons of apple juice costing $1.00 per gallon to make a fruit juice mix costing $2.50 per gallon, then what is x y ? Solution: The equation that describes the total cost of the mixture is 3x+y = 2.5(x+y). Divide both sides by y to find the ratio: 3x y + y y = 2.5x y + 2.5y y . Simplify: (3) x y + 1 = (2.5) x y + 2.5 (.5) x y = 1.5 x y = 3 36 CHAPTER 1. EQUATIONS AND INEQUALITIES Chapter 2 Graphs Section summaries Section 2.1 The Distance and Midpoint Formulas You need to know the distance formula d = √ (x2 − x1)2 + (y2 − y1)2 and the midpoint formula ( x1 + x2 2 , y1 + y2 2 ) . The distance formula comes from the Pythagorean theorem (review page 30); you may also need to use the Pythagorean theorem to verify that three points are the vertices of a right triangle. Review problems: p161 #19,29,35,45 Section 2.2 Graphs of Equations Review the procedure for finding x and y-intercepts on page 166. Review the tests for symmetry on page 168. A function is even precisely when its graph is symmetric with respect to the y-axis; it is odd precisely when its graph is symmetric with respect to the origin. (Compare the tests on page 168 to the tests on pages 231 and 232.) Review problems: p171 #41,43,63,65,67 37 40 CHAPTER 2. GRAPHS 2.2 #61. The graph of the equation 9x2 + 4y2 = 36 has (a) x-intercept (0, 0) and y-intercept (0, 0) (b) x-intercept (2, 0) and y-intercept (0, 3) (c) x-intercept (3, 0) and y-intercept (0, 2) (d) x-intercepts (2, 0) and (−2, 0) and y-intercepts (0, 3) and (0,−3) (e) x-intercepts (3, 0) and (−3, 0) and y-intercepts (0, 2) and (0,−2) 2.2 #69. The graph of y = −x3 x2 − 9 is symmetric with respect to (a) the x-axis and y-axis, but NOT the origin. (b) the origin, but NOT the x-axis or y-axis. (c) the x-axis and the origin, but NOT the y-axis. (d) the y-axis and origin, but NOT the x-axis. (e) the x-axis, the y-axis and the origin. 2.3 A. The equation of the vertical line passing through the point (4, 7) is (a) x = 4 (d) y = 7 (b) x = 7 (e) 4x = 7y (c) y = 4 2.3 B. Find the slope of the line through the points (−3,−1) and (1, 7). (a) 3 (d) 2 (b) −3 (e) None of these (c) 12 2.3 C. Find an equation for the line through (0, 3) and (−2, 0). (a) 2x − 3y + 6 = 0 (d) 2x + 3y − 6 = 0 (b) 3x + 2y − 6 = 0 (e) 3x + 2y + 6 = 0 (c) 3x − 2y + 6 = 0 2.3 Example 8. Find the slope m and y-intercept b of the equation 2x + 4y = 8. (a) m = 12 and b = 2 (d) m = −2 and b = 4 (b) m = −12 and b = 2 (e) None of these (c) m = 2 and b = 4 41 2.3 #49. The equation of the line containing the points (1, 3) and (−1, 2) is (a) y = 2x + 1 (d) y = −2x + 5 (b) y = −12x + 72 (e) This is a vertical line, so there is no equation. (c) y = 12x + 5 2 2.3 D. Which of the following is an equation of the line passing through the point (5,−4) and parallel to the line with equation 3x − 5y + 2 = 0? (a) y = 3x − 4 (d) y = 35x − 4 (b) y = 3x − 19 (e) y = − 53x − 9 (c) y = 35x − 7 2.3 #65. Find an equation for the line perpendicular to y = 1 2 x + 4 containing (1,−2). (a) y = 2x + 4 (d) y = −2x (b) y = −2x − 4 (e) None of these (c) y = 2x 2.3 #67. Find an equation for the line perpendicular to 2x+y = 2 and containing (−3, 0). (a) y = 2(x + 3) (d) y = − 12(x + 3) (b) y = −2(x + 3) (e) None of these (c) y = 12(x + 3) 2.3 E. The line which is perpendicular to the line given by y = 4x − 3 and which passes through the point (0, 5) also passes through which of the following points? (a) (4, 0) (d) (4, 6) (b) (4, 13) (e) (4,−11) (c) (4, 4) 2.3 #97. The graph of the line with equation 12x + 1 3y = 1 has (a) x-intercept (1/2, 0) and y-intercept (0, 1/3) (b) x-intercept (1/3, 0) and y-intercept (0, 1/2) (c) x-intercept (3, 0) and y-intercept (0, 2) (d) x-intercept (2, 0) and y-intercept (0, 3) (e) None of these 42 CHAPTER 2. GRAPHS 2.4 A. The standard form of the equation of the circle with radius 6 and center (−3,−6) is (a) (x + 3)2 + (y + 6)2 = 36 (b) (x − 3)2 + (y − 6)2 = 36 (c) (x + 6)2 + (y + 3)2 = 36 (d) (x − 6)2 + (y − 3)2 = 36 (e) None of these 2.4 #25. The circle x2 + y2 − 2x + 4y − 4 = 0 has (a) center (1,−2) and radius 9 (d) center (−1, 2) and radius 3 (b) center (1,−2) and radius 3 (e) center (−1, 2) and radius 9 (c) center (−2, 4) and radius 16 2.4 #29. The graph of the equation x2 + y2 − x + 2y + 1 = 0 is (a) a circle with center (1,−2) and radius 1. (b) a circle with center (−1, 2) and radius 1. (c) a circle with center ( 12 ,−1) and radius 1. (d) a circle with center ( 12 ,−1) and radius 14 . (e) None of these 2.4 B. The graph of the equation x2 + y2 − 6x + 2y + 7 = 0 is (a) a circle with center (3,−1) and radius 3. (b) a circle with center (3,−1) and radius √ 3. (c) a circle with center (3, 1) and radius √ 7. (d) a circle with center (1, 3) and radius √ 3. (e) None of these 45 2.4 A. The standard form of the equation of the circle with radius 6 and center (−3,−6) is Solution: Use the standard form of an equation of a circle: (x − h)2 + (y − k)2 = r2. You get (x + 3)2 + (y + 6)2 = 36. 2.4 B. The graph of the equation x2 + y2 − 6x + 2y + 7 = 0 is Solution: The answer is found by completing the square. x2 − 6x + ?? + y2 + 2y + ?? = −7 x2 − 6x + 9 + y2 + 2y + 1 = −7 + 9 + 1 (x − 3)2 + (y + 1)2 = 3 (x − 3)2 + (y − (−1))2 = ( √ 3)2 This is a circle with center (3,−1) and radius √ 3. 46 CHAPTER 2. GRAPHS Chapter 3 Functions and Their Graphs Section summaries Section 3.1 Functions A function from a set X to a set Y is a rule or correspondence that associates with each element of X exactly one element of Y . (In Math 110, these sets usually consist of real numbers.) With the function notation y = f(x), each x value has only one corresponding y value. You can think of a function as being like a program. The x-values are the inputs, and the y-values are the outputs. The possible inputs form the domain of the function, and the possible outputs form its range. For the functions that we are dealing with, the numbers that we need to exclude from the domain are numbers that lead to division by zero, or the square root of a negative number. (If the function was a program, trying to divide by zero or take the square root of a negative number would give an error message.) Given two functions, we can make a new function from their sum, difference, product, or quotient. (See pages 217–218.) Review problems: p219 #39,41,51,59,61,75,79,89 Section 3.2 The Graph of a Function The vertical line test: A set of points in the (x, y)-plane is the graph of a function precisely when every vertical line intersects the set in at most one point. Review problems: p226 #9,25,27,29 Section 3.3 Properties of Functions A function is called even if f(−x) = f(x) (the graph is symmetric about the y-axis) and odd if f(−x) = −f(x) (the graph that is symmetric about the origin). A function is increasing on an interval if its values keep going up, and decreasing on an interval if its values keep going down. A high point on the graph is called a local 47 50 CHAPTER 3. FUNCTIONS AND THEIR GRAPHS Sample Questions 3.1 A. For the function f(x) = x3 + x, find f(−2). (a) 6 (d) −10 (b) 10 (e) None of these (c) −6 3.1 Example 6. For the function f(x) = 2x2 − 3x, find f(3x). (a) 36x2 − 9x (d) 18x2 − 3x (b) 36x2 − 3x (e) None of these (c) 18x2 − 9x 3.1 B. For the function f(x) = x2 − 2, find f(y + 2). (a) x2y + 2x2 − 2y − 4 (d) y2 + 2 (b) y2 + 4y + 2 (e) None of these (c) y2 3.1 #51. What is the domain of the function f(x) = x x2 − 16? (a) All real numbers (d) All real numbers except 16 (b) All real numbers except 0 (e) None of these (c) All real numbers except 4,−4 3.1 C. What is the domain of the function f(x) = x + 1 x − 1? (a) All real numbers except −1 (b) All real numbers except 1,−1 (c) All real numbers except 1 (d) All real numbers (e) None of these. 3.1 D. What is the domain of the function g(x) = x2 + 1 x3 − 4x? (a) All real numbers except −4 (d) {0, 2,−2} (b) All real numbers except 2,−2 (e) {0, 2,−2,−4} (c) All real numbers except 0, 2,−2 51 3.1 #57. Find the domain of the function f(x) = 4√ x − 9 ? (a) (9,∞) (d) (−∞, 9) (b) [9,∞) (e) (−∞, 9] (c) (4/9,∞) 3.1 E. What is the domain of the function f(x) = x + 2√ 5 + 3x ? (a) (−∞, 0] (d) (−5/3,∞) (b) (−∞, 0) (e) (−∞,−5/3) (c) (−2,∞) 3.1 F. Let f(x) = 2x2 − 4. Find f(x − 3). (a) 2x2 − 12x + 18 (d) 2x2 + 14 (b) 2x2 − 12x + 14 (e) None of these (c) 2x2 − 22 3.1 G. If f(x) = 2x + 1 3x − 5, then what is f(3x + 1)? (a) f(3x + 1) = 3 · ( 2x + 1 3x − 5 ) + 1 (d) f(3x + 1) = (2x + 1)(3x + 1) (3x − 5) (b) f(3x + 1) = 6x + 2 9x − 4 (e) None of these (c) f(3x + 1) = 6x + 3 9x − 2 3.1 #75. For f(x) = x2 − x + 4, find and simplify the difference quotient f(x + h) − f(x) h , where h 6= 0. (a) h + 1 (d) 2x + h − 1 (b) h − 1 (e) None of these (c) 2x + h + 1 3.1 #79. For f(x) = x3 − 2, find and simplify the difference quotient f(x + h) − f(x) h , where h 6= 0 (a) h2 (d) 3x2 + 3xh2 + h3 (b) x2 + xh + h2 (e) None of these (c) 3x2 + 3xh + h2 52 CHAPTER 3. FUNCTIONS AND THEIR GRAPHS 3.2 #25c. For the function f(x) = x + 2 x − 6, if f(x) = 2, then x = (a) −1 (d) There is no answer (b) 8 (e) None of these (c) 14 3.2 #25e. Find the x-intercept(s) of the graph of the function f(x) = x + 2 x − 6. (a) 2 (d) −1/3 (b) −2 (e) None of these (c) 2,−6 3.2 #25f. Find the y-intercept(s) of the graph of the function f(x) = x + 2 x − 6. (a) 2 (d) −1/3 (b) −2 (e) None of these (c) 2,−6 3.2 #27c. For the function f(x) = 2x2 x4 + 1 , if f(x) = 1, then x = (a) 1 (d) There is no answer (b) −1 (e) None of these (c) 1,−1 3.2 #27e. Find the x-intercept(s) of the graph of the function f(x) = 2x2 x4 + 1 . (a) 0 (d) There is no x-intercept (b) 1 (e) None of these (c) 2 3.2 #27f. Find the y-intercept(s) of the graph of the function f(x) = 2x2 x4 + 1 . (a) 0 (d) There is no y-intercept (b) 1 (e) None of these (c) 2 55 3.5 B. Using the function f(x) = x3, find the equation of the corresponding function whose graph is shifted left 1 unit, reflected about the x-axis, and shifted up 2 units. (a) f(x) = −(x + 2)3 + 1 (d) f(x) = −(x − 1)3 + 2 (b) f(x) = −(x − 2)3 + 1 (e) None of these (c) f(x) = −(x + 1)3 + 2 3.5 C. Using the function f(x) = x3, find the equation of the corresponding function whose graph is reflected about the x-axis and then shifted down 4 units. (a) f(x) = −(x − 4)3 (d) f(x) = −x3 − 4 (b) f(x) = −(x + 4)3 (e) None of these (c) f(x) = −x3 + 4 3.5 D. If the graph of y = 3x2 + 4x − 5 is reflected about the y-axis, the new equation is (a) y = −3x2 − 4x + 5 (d) y = 3x2 − 4x − 5 (b) y = −3x2 + 4x + 5 (e) None of these (c) y = 3x2 − 4x + 5 3.5 E. Which function represents the graph at the right? (a) f(x) = −(x − 1)2 + 2 (b) f(x) = −(x − 2)2 + 1 (c) f(x) = −(x − 1)2 − 2 (d) f(x) = −(x − 2)2 − 1 (e) None of these (1, 2) x y −5 5 3.6 #7. A rectangle has one corner on the graph of y = 16 − x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. Express the area as a function of x. (a) A(x) = −x2 + x + 16 (d) A(x) = −x2 + 16 (b) A(x) = −x3 + 16x (e) None of these (c) A(x) = −x3 + 8x 56 CHAPTER 3. FUNCTIONS AND THEIR GRAPHS 3.6 #13. A wire of length x is bent into the shape of a circle. Express the area A(x) as a function of x. (a) A(x) = x2 (d) A(x) = x2 4π (b) A(x) = πx2 (e) None of these (c) A(x) = x2 4 3.6 A. Alex has 400 feet of fencing to enclose a rectangular garden. One side of the garden lies along the barn, so only three sides require fencing. Express the area A(x) of the rectangle as a function of x, where x is the length of the side perpendicular to the side of the barn. (a) A(x) = −x2 + 200x (b) A(x) = −x2 + 400x (c) A(x) = −2x2 + 200x (d) A(x) = −2x2 + 400x (e) None of these 3.6 B. Germaine has 40 feet of fencing to enclose a rectangular pool. One side of the pool lies along the house, so only three sides require fencing. Express the area A(x) of the rectangle as a function of x, where x is the length of the side perpendicular to the side of the house. (a) A(x) = −x2 + 40x (b) A(x) = −x2 + 20x (c) A(x) = −2x2 + 40x (d) A(x) = −2x2 + 20x (e) None of these 57 Answer Key 3.1 A. (d) 3.1 Example 6. (c) 3.1 B. (b) 3.1 #51. (c) 3.1 C. (c) 3.1 D. (c) 3.1 #57. (a) 3.1 E. (d) 3.1 F. (b) 3.1 G. (c) 3.1 #75. (d) 3.1 #79. (c) 3.2 #25c. (c) 3.2 #25e. (b) 3.2 #25f. (d) 3.2 #27c. (c) 3.2 #27e. (a) 3.2 #27f. (a) 3.3 #39. (c) 3.3 #41. (b) 3.3 #43. (a) 3.3 A. (b) 3.3 #55. (c) 3.3 B. (c) 3.3 C. (a) 3.4 A. (e) 3.4 #25. (d) 3.4 #35. (d) 3.5 A. (b) 3.5 B. (c) 60 CHAPTER 3. FUNCTIONS AND THEIR GRAPHS 3.5 D. If the graph of y = 3x2 + 4x − 5 is reflected about the y-axis, the new equation is Solution: (d) Substitute −x to get y = 3(−x)2 + 4(−x) − 5 = 3x2 − 4x − 5 3.5 E. Which function represents the graph at the right? (a) f(x) = −(x − 1)2 + 2 (b) f(x) = −(x − 2)2 + 1 (c) f(x) = −(x − 1)2 − 2 (d) f(x) = −(x − 2)2 − 1 (e) None of these (1, 2) x y −5 5 Solution: (a) The graph is that of y = x2 shifted 1 unit to the right, reflected about the x-axis, and then shifted up 2 units. 3.6 A. Alex has 400 feet of fencing to enclose a rectangular garden. One side of the garden lies along the barn, so only three sides require fencing. Express the area A(x) of the rectangle as a function of x, where x is the length of the side perpendicular to the side of the barn. Solution: (d) The three sides are x, x, and 400−2x, so the total area is length × width, giving A(x) = x(400 − 2x) = 400x − 2x2 = −2x2 + 400x. 3.6 B. Germaine has 40 feet of fencing to enclose a rectangular pool. One side of the pool lies along the house, so only three sides require fencing. Express the area A(x) of the rectangle as a function of x, where x is the length of the side perpendicular to the side of the house. Solution: (c) The three sides are x, x, and 40 − 2x, giving A(x) = x(40 − 2x) = 40x − 2x2 = −2x2 + 40x. Chapter 4 Linear and Quadratic Functions Section summaries Section 4.1 Linear Functions and Their Properties A linear function is one of the form f(x) = mx + b , where m gives the slope of its graph, and b gives the y-intercept of its graph. The slope m measures the rate of growth of the function, so a linear function is increasing if m > 0 and decreasing if m < 0. Review problems: p284 #17,21,25,37,43,49 Section 4.2 Building Linear Functions from Data In this section linear functions are constructed from data presented in various ways. Review problems: p290 #3,5,7,15,19,21 Section 4.3 Quadratic Functions and Their Properties The general form of a quadratic function is f(x) = a(x − h)2 + k , where (h, k) is the vertex of the graph (which is a parabola). You can see from the formula that h gives the left/right shift while k gives the up/down shift. The coefficient a represents a vertical stretch or compression. Since the basic member of this family is f(x) = x2, whose graph opens up, the graph of f(x) = a(x − h)2 + k will open up if a is positive, and down if a is negative. If the graph opens up, its height is minimum at the vertex; if the graph opens down, its height is maximum at the vertex. 61 62 CHAPTER 4. LINEAR AND QUADRATIC FUNCTIONS If a quadratic is given in the form f(x) = ax2 + bx + c, then the x-coordinate of its vertex is x = − b 2a . Since you already know the quadratic formula, you can remember it as part of the formula: x = − b 2a ± √ b2 − 4ac 2a . This way of looking at the quadratic formula shows that if the graph has x-intercepts, they occur as points on either side of the line x = − b 2a , which is the axis of symmetry of the graph. The summary on page 301 explains the steps in graphing a quadratic function. Review problems: p302 #13,15,27,31,37,43,55,61,81,83 65 Answer Key 4.1 #29. (d) 4.1 #37c. (b) 4.3 A. (b) 4.3 #42 (c) 4.3 B. (d) 4.3 C. (a) 4.3 D. (b) 4.3 #55. (c) 4.3 #61. (c) 4.3 #62. (b) 4.3 #81. (b) 4.3 E. (d) Solutions 4.3 A. If f(x) is a quadratic function whose graph has the vertex (h, k), which one is the correct form of the function? Solution: (b) f(x) = a(x − h)2 + k 4.3 B. Find the vertex of the quadratic function f(x) = 2x2 − 4x + 9. Solution: (d) The text gives this formula: the x-coordinate of the vertex of the graph of f(x) = ax2 + bx + c is x = − b2a . In this example, a = 2 and b = −4, so the vertex occurs at x = −−42·2 = 1. Then f(1) = 2 − 4 + 9 = 7 gives the y-coordinate. If you forget the formula, you can always complete the square: f(x) = 2x2 − 4x + 9 = 2(x2 − 2x) + 9 = 2(x2 − 2x + 1) + 9 − 2 = 2(x − 1)2 + 7 so h = 1 and k = 7 and the vertex is (1, 7). 4.3 C. Find the axis of symmetry of the graph of f(x) = 4x2 − 8x + 3. Solution: (a) The axis of symmetry passes through the vertex, which has x-coordinate −−82·4 = 1. The axis of symmetry is the line x = 1. Again, if you forget the formula, complete the square: f(x) = 4x2 − 8x + 3 = 4(x2 − 2x) + 3 = 4(x2 − 2x + 1) + 3 − 4 = 4(x − 1)2 − 1 This shows that the vertex is at (1,−1). 4.3 D. Let f(x) = 4x2 − 8x + 3. Find the x and y-intercepts, if any. Solution: (b) Since f(0) = 3, the y-intercept is (0, 3). To find the x-intercept, solve 4x2 − 8x + 3 = 0. This can be factored as 4x2 − 8x + 3 = (2x − 1)(2x − 3), so 2x − 1 = 0 or 2x − 3 = 0, giving the x-intercepts ( 32 , 0) and (12 , 0). 66 CHAPTER 4. LINEAR AND QUADRATIC FUNCTIONS 4.3 #62. Find the minimum value of the quadratic function f(x) = 4x2 − 8x + 3. Solution: (b) The minimum value occurs at the vertex, which has x-coordinate −−82·4 = 1. Then f(1) = −1 is the minimum height. 4.3 E. A store selling calculators has found that, when the calculators are sold at a price of p dollars per unit, the revenue R (in dollars) as a function of the price p is R(p) = −750p2 + 15000p. What is the largest possible revenue? That is, find the maximum value of the revenue function. Solution: (d) The graph is a parabola, opening down, so to find the maximum value we need to find the y-coordinate of the vertex. We get x = − b2a = − 150002·(−750) = − 15000 −1500 = 10. The maximum revenue is R(10) = −750(10)2 + 15000(10) = −75000 + 150000 = 75000. Chapter 5 Polynomial and Rational Functions Section summaries Section 5.1 Polynomial Functions The general form of a polynomial function is f(x) = anx n +an−1x n−1 + · · ·+a1x+a0. The degree of f(x) is the largest exponent in the formula. Linear functions f(x) = mx+ b and quadratic functions f(x) = ax2 + bx + c are the simplest cases. If |x| is large, then the term anx n is much larger than the others, so the “big picture” of f(x) is that its graph follows the pattern of xn, flipped over if an is negative. The number of different x-intercepts of a polynomial of degree n is at most n, because a polynomial equation of degree n has at most n roots. The same is true of any horizontal line–the graph of a polynomial of degree n can cross the line at most n times. This means that the graph has at most n − 1 “turning points” (see the discussion on the top of page 321), and this helps you in graphing. Finally, a polynomial has two types of behavior at an x-intercept. It may cross the x-axis, like y = x3, or it may just touch the x-axis, like y = x2. If you can factor the function completely, you can tell whether it crosses or touches by looking at the exponent of the factor that corresponds to the root you are interested in. If the exponent is odd, the graph will cross the axis because the y-values will change sign, but if the exponent is even, the graph will just touch the axis and stay on the same side. You should review the summaries on pages 336 and 338 very carefully. Review problems: p340 #31,37,45,55,63,71,75 Section 5.2 Properties of Rational Functions We have been building up to more and more complicated functions. This section deals with some basic properties of functions of the form f(x) = p(x) q(x) , where p(x) and q(x) are polynomials. These are called rational functions. The functions f(x) = 1 x and f(x) = 1 x2 are two familiar examples. 67 70 CHAPTER 5. POLYNOMIAL AND RATIONAL FUNCTIONS 5.1 E. Which one of these functions might have the given graph? (a) f(x) = x(x − 1)(x − 2)2 (d) f(x) = x2(x − 1)(x − 2) (b) f(x) = −x(x − 1)(x − 2) (e) f(x) = −x2(x − 1)(x − 2) (c) f(x) = x(x − 1)(x − 2) x y 1 2 5.2 A. Find the domain of the function f(x) = x − 2 x + 1 . (a) All real numbers except −1 (d) All real numbers except 2 (b) All real numbers except 1 (e) None of these (c) All real numbers except −2 5.2 B. What is the domain of the function G defined by G(x) = x + 4 x3 − 4x? (a) all reals except −4 (d) {0, 2,−2} (b) all reals except 2,−2 (e) {0, 2,−2,−4} (c) all reals except 0, 2,−2 5.2 C. The graph of y = 1 (x − 4)2 looks like that of y = 1 x2 but is shifted (a) left 4 units (d) up 4 units (b) right 4 units (e) None of these (c) down 4 units 71 5.2 D. Find the vertical asymptotes of the graph of f(x) = x2 − 3x x2 − 2x − 8. (a) x = −4 and x = 2 (b) x = 4 and x = −2 (c) x = 0 and x = 3 (d) x = 4, x = −2, x = 0 and x = 3 (e) None of these 5.2 E. The line x = 4 is a vertical asymptote of the graph of which of the following functions? (a) f(x) = x − 4 (d) f(x) = x − 4 x + 3 (b) f(x) = 2x − 8 x − 4 (e) f(x) = 4x + 1 x + 2 (c) f(x) = 1 x2 − 16 5.2 F. Find the x-intercepts and vertical asymptotes of the graph of f(x) = x2 − 3x x2 − 2x − 8 (a) x-intercepts (4, 0), (3, 0), (−2, 0), (0, 0), vertical asymptotes x = 4, x = 3, x = −2, x = 0 (b) x-intercepts (−4, 0), (2, 0), vertical asymptotes x = 0, x = 3 (c) x-intercepts (4, 0), (−2, 0), vertical asymptotes x = 0, x = 3 (d) x-intercepts (0, 0), (3, 0), vertical asymptotes x = 4, x = −2 (e) x-intercepts (0, 0), (3, 0), vertical asymptotes x = −4, x = 2 5.2 G. Find the vertical asymptotes of f(x) = (x − 1)(x + 2)(x − 3) x(x − 4)2 . (a) x = 1,−2, 3 (d) x = 4 (b) x = 0, 1,−2, 3, 4 (e) None of these (c) x = 0, 4 5.2 H. Find the horizontal asymptote of the graph of f(x) = x − 3 5x + 2 . (a) y = 15 (d) x = −13 (b) y = −32 (e) None of these (c) x = −25 72 CHAPTER 5. POLYNOMIAL AND RATIONAL FUNCTIONS 5.2 K. Find the horizontal asymptote for the graph of f(x) = 5x − 1 2x + 3 . (a) x = −32 (d) y = 52 (b) x = 52 (e) None of these (c) y = −32 5.2 #48. Find the horizontal asymptote (if any) of f(x) = −2x2 + 1 2x3 + 4x2 . (a) y = −2 (d) y = 0 (b) y = −1 (e) None of these (c) y = 1 5.2 L. Which one of these functions does not have a horizontal asymptote? (a) f(x) = 2 3x − 5 (d) f(x) = 2x 3x2 − 5 (b) f(x) = 2x2 + 1 3x − 5 (e) f(x) = 2 + 6 3x2 − 5 (c) f(x) = 2x2 + 1 3x2 − 5 5.2 M. Find the asymptotes of the following function. f(x) = 3x2 − 3x x2 + x − 12 (a) The horizontal asymptote is y = 0; the vertical asymptotes are x = −4 and x = 3. (b) The horizontal asymptote is y = 3; the vertical asymptotes are x = 4 and x = −3. (c) The horizontal asymptote is y = 3; the vertical asymptotes are x = −4 and x = 3. (d) The horizontal asymptote is y = −4; the vertical asymptote is x = 3. (e) None of these 5.4 #5. Solve the inequality x3 − 4x2 > 0. (a) (4,∞) (d) (−∞, 0) (b) (0,∞) (e) None of these (c) (−∞, 4) 75 5.1 C. Which of the following is the graph of f(x) = (x + 1)2(x − 2)? (a) (b) (c) (d) x y −5 5 x y −5 5 x y −5 5 x y −5 5 Solution: (c) The function has roots −1 and 2, so these must be x-intercepts on the graph. All four graphs pass this test, so we have to look at it more deeply. The multiplicity of −1 is 2, so the graph should only touch the axis at x = −1. Since the multiplicity of 2 is just 1, the graph should cross the axis at x = 2. This eliminates answers (a) and (d). We could plot one more point: since f(0) = −2, the y=intercept must be −2, and this eliminates answer (b). 5.1 D. The function f(x) = x2(x − 2)(x + 3)2 has Answer: (b) one zero of multiplicity one and two zeros of multiplicity two. Solution: x = 0 has multiplicity 2; x = 2 has multiplicity 1; x = −3 has multiplicity 2. 5.1 E. Which one of these functions might have the given graph? (a) f(x) = x(x − 1)(x − 2)2 (d) f(x) = x2(x − 1)(x − 2) (b) f(x) = −x(x − 1)(x − 2) (e) f(x) = −x2(x − 1)(x − 2) (c) f(x) = x(x − 1)(x − 2) x y 1 2 Solution: (e) The roots must be x = 0, x = 1, and x = 2, and this is true for every formula. Since that graph touches the x-axis at x = 0 but does not cross, the root x = 0 must have even multiplicity. The graph crosses the x-axis at x = 1 and x = 2, so these roots must have odd multiplicity. This eliminates the formulas in (a), (b), and (c). Finally, to tell the difference between (d) and (e), check the value of the function at x = −2. In (d) we get f(−2) = (−2)2(−2− 1)(−2− 2) = 48, so this doesn’t agree with the graph, so the answer must be (e). 76 CHAPTER 5. POLYNOMIAL AND RATIONAL FUNCTIONS Alternatively, you can see that for large values of x the formula in (d) behaves like f(x) = x4, while the formula in (e) behaves like f(x) = −x4. The behavior of the graph is like f(x) = −x4, so the answer must be (e). 5.2 A. Find the domain of the function f(x) = x−2x+1 . Solution: (a) Set the denominator equal to 0. Exclude x = −1 to avoid division by 0. 5.2 B. What is the domain of the function G defined by G(x) = x+4 x3−4x ? Solution: (c) Set x3 − 4x = 0. This gives x(x2 − 4) = 0, or x(x − 2)(x + 2) = 0. The numbers 0, 2, and −2 must be excluded. Answer: all real numbers except 0, 2,−2. 5.2 C. The graph of y = 1 (x−4)2 looks like that of y = 1 x2 but is shifted Solution: (b) right 4 units. One way to remember which way it is shifted is to observe that the graph of y = 1 x2 has a vertical asymptote at x = 0, while the graph of y = 1 (x−4)2 has a vertical asymptote at x = 4. 5.2 D. Find the vertical asymptotes of the graph of f(x) = x 2 −3x x2−2x−8 . Solution: (b) Set the denominator equal to 0. x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 The vertical asymptotes occur at x = 4 and x = −2. 5.2 E. The line x = 4 is a vertical asymptote of the graph of which of the following functions? (a) f(x) = x−4 (b) f(x) = 2x−8x−4 (c) f(x) = 1x2−16 (d) f(x) = x−4 x+3 (d) f(x) = 4x+1 x+2 Solution: (c) To have a vertical asymptote at x = 4, the denominator must have a factor of x − 4. This seems to be true for both (b) and (c). Actually, f(x) = 2x−8x−4 = 2(x−4) x−4 = 2 for all values except x = 4. In the function in (b), the graph is a horizontal line at y = 2, with the point (4, 2) missing, so it has no vertical asymptote. The correct answer is that only f(x) = 1 x2−16 has a graph with a vertical asymptote at x = 4. 5.2 F. Find the x-intercepts and vertical asymptotes of the graph of f(x) = x 2 −3x x2−2x−8 Solution: (d) x-intercepts (0, 0), (3, 0), vertical asymptotes x = 4, x = −2 Set the numerator equal to 0 to find the x-intercepts. x2 − 3x = 0 x(x − 3) = 0 x = 0 and x = 3. Set the denominator equal to 0 to find the vertical asymptotes. x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 x = 4 and x = −2 5.2 G. Find the vertical asymptotes of f(x) = (x−1)(x+2)(x−3) x(x−4)2 . Solution: (c) Set the denominator equal to 0 to get x = 0 and x = 4. 5.2 H. Find the horizontal asymptote of the graph of f(x) = x−35x+2 . Solution: (a) Multiplying both the numerator and denominator by 1x gives f(x) = 1− 3 x 5+ 2 x , and in this form we see that as x increases the function gets closer and closer to 15 . The shortcut is to remember that the highest powers of x dictate the behavior, so f(x) will behave like x5x = 1 5 for large values of x. 77 5.2 K. Find the horizontal asymptote for the graph of f(x) = 5x−12x+3 . Solution: (d) For large values of x the function behaves like f(x) = 5x2x , so the horizontal asymptote is y = 52 . 5.2 #48. Find the horizontal asymptote (if any) of f(x) = −2x 2+1 2x3+4x2 . Solution: (d) For large values of x, the function behaves like f(x) = −2x 2 2x3 = − 1x , so y = 0 is a horizontal asymptote. Remember that if the denominator has higher degree than the numerator, then y = 0 will be a horizontal asymptote. 5.2 L. Which one of these functions does not have a horizontal asymptote? (a) f(x) = 23x−5 (b) f(x) = 2x2+1 3x−5 (c) f(x) = 2x2+1 3x2−5 (d) f(x) = 2x 3x2−5 (e) f(x) = 2 + 6 3x2−5 Solution: (b) If the degree of the numerator is larger than the degree of the denominator, then there is no horizontal asymptote. This happens for (b). 5.2 M. Find the asymptotes of the following function. f(x) = 3x 2 −3x x2+x−12 Solution: (c) For large values of x, the function behaves like f(x) = 3x 2 x2 , so y = 3 is a horizontal asymptote. To find the vertical asymptotes, solve x2 + x − 12 = 0. (x + 4)(x − 3) = 0 The vertical asymptotes are x = −4 and x = 3. 80 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Not all functions have an inverse function. If two x-values produce the same y-value in the formula y = f(x), then given y there is no unique way to recover x. In order to have an inverse, the graph of y = f(x) must pass the horizontal line test (see page 411). If y = f(x) passes the test, then we simply interchange x and y in the formula y = f(x), and solve for y to get the formula for the inverse. See page 416 for this procedure that actually finds the formula for the inverse of a function. A function and its inverse are closely connected. The domain of f−1 is the range of f ; the range of f−1 is the domain of f (see page 413). The graph of f−1 is the mirror image of the graph of f , in the line y = x. This happens since if the point (x, y) is on the graph of f , then the symmetric point (y, x) must be on the graph of f−1 (see page 415). Review problems: p420 #35,39,43,49,57,81 Section 6.3 Exponential Functions An exponential function is one of the form f(x) = ax where a is a positive real number and a 6= 1. (We will usually assume that a > 1.) The domain of an exponential function is the set of all real numbers. Its graph has the x-axis as a horizontal asymptote. The points (0, 1), (1, a), and (−1, 1/a) are easy ones to plot. Note that if x increases by 1, then f(x) is multiplied by a, since f(x + 1) = ax+1 = ax · a = af(x) (see the theorem on page 425). You need to be familiar with the basic shape of an exponential function. See Figure 18 on page 426 for the graph of y = 2x and Figure 27 on page 430 for the graph of y = ex. The base e is important because it makes calculations easier when doing calculus. (It is the one exponential function whose graph crosses the y-axis at a 45 degree angle, making the slope of the graph equal to 1 when x = 0.) For our class, the only thing you need to remember about e is its approximate value of 2.7 and the fact that the graph of y = ex lies between the graphs of y = 2x and y = 3x. Building on the basic shape of y = ax, we can graph other functions in the family by using transformations (as we did in Section 3.5). Review problems: p433 #21,25,43,53,65,71,89 Section 6.4 Log Functions A logarithmic function (or log function for short) is one of the form f(x) = loga(x) where a > 0 and a 6= 1. If a = 10, we usually write log(x) instead of log10(x). If a = e, we write ln(x) instead of loge(x), and call this the natural log function. 81 The log function loga(x) is defined to be the inverse of the exponential function ax. First, this tells us the basic shape of the graph (see Figure 30 on page 440). It also guarantees that the graph has the y-axis as a vertical asymptote, and that the domain of loga(x) is (0, +∞), the same as the range of ax. Now, when finding the domain of a function, you not only need to watch out for division by zero, or the square root of a negative number, but also for the log of a negative number. All of these are undefined for real numbers. To express the inverse relationship, we can say that y = loga(x) if and only if x = a y (see the top of page 438). We also have the following equations, which summarize the inverse relationship (see the theorem at the top of page 451): aloga(x) = x and loga(a x) = x . These identities are important in solving equations that involve logs. For example, to solve the equation log2(2x + 1) = 3 we need to simplify the left hand side. Since 2log2(2x+1) = 2x + 1, the first step is to make both sides of the equation into an exponent with base 2, to get 2log2(2x+1) = 23, which simplifies to 2x + 1 = 8. To solve the equation ln(e−2x) = 8, just note that the left hand side is equal to −2x, so the equation simplifies immediately to −2x = 8. To solve the equation e2x+5 = 8, we need to get rid of the base e on the left hand side. This is done by substituting both sides into the natural log function, to get ln(e2x+5) = ln(8), or simply 2x + 5 = ln 8. Review problems: p446 #21,33,37,43,63,71,81,89,99 Section 6.5 Properties of Logarithms Since logs represent exponents, they should behave like exponents. For example, if we write two numbers M and N in scientific notation as powers of 10, then to multiply M and N we only need to add the exponents. To find the square root of M , we only need to divide the exponent of M by 2. The crucial properties of logs are summarized in the following equations (see page 451 and 452). log a (MN) = log a (M) + log a (N) log a (M/N) = log a (M) − log a (N) log a (Mr) = r log a (M) There is also a formula to change the base: loga(M) = logb(M) logb(a) . (See the section summary on page 456.) Review problems: p457 #13,17,29,41,49,57,83,85 Section 6.6 Log and exponential equations In this section, the properties of logarithms are used to solve various kinds of equations. Review problems: p463 #17,31,35,77,81 82 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Sample Questions 6.1 #11a. Let f(x) = 2x and g(x) = 3x2 + 1. Find (f ◦ g)(4). (a) 337 (d) 24x3 + 8x (b) 193 (e) None of these (c) 98 6.1 #11. Let f(x) = 2x and g(x) = 3x2 + 1. Find the composite function (g ◦ f)(x). (a) 12x2 + 1 (d) 6x3 + 2x (b) 6x2 + 2 (e) 6x3 + 1 (c) 6x2 + 1 6.1 #15a. Let f(x) = √ x and let g(x) = 2x. Find (f ◦ g)(4). (a) √ 2 (d) 16 (b) 2 √ 2 (e) None of these (c) 4 6.1 A. Let f(x) = 2x2 + 1 and let g(x) = x + 3. Find the composite function (f ◦ g)(x). (a) 2x2 + 18 (d) 2x2 + 12x + 19 (b) 2x2 + 19 (e) None of these (c) 2x2 + 12x + 18 6.1 #31b. Let f(x) = 3x + 1 and g(x) = x2. Find the composite function (g ◦ f)(x). (a) x2 + 3x + 1 (d) 9x2 + 6x + 1 (b) 9x2 + 1 (e) None of these (c) 3x3 + x2 6.1 Example 4a. Find the domain of f ◦ g if f(x) = 1 x + 2 and g(x) = 4 x − 1. (a) {x | x 6= ±1} (d) {x | x 6= −2} (b) {x | x 6= 1} (e) None of these (c) {x | x 6= −1} 85 6.2 #59a. The function f(x) = 2 x + 3 is a one-to-one function. Find the inverse function. (a) f−1(x) = 2 x − 23 (d) f−1(x) = 12x + 32 (b) f−1(x) = 2 x + 23 (e) None of these (c) f−1(x) = 2 x − 3 6.2 #59b. Find the range of the function f(x) = 2 x + 3 . (See the previous problem.) (a) {y | y 6= 0} (d) All real numbers (b) {y | y 6= 2/3} (e) None of these (c) {y | y 6= 1/2} 6.2 #60. The function f(x) = 4 2 − x , for x 6= 2, is a one-to-one function. Find the inverse function f−1. (a) f−1(x) = 2 − 4 x (d) f−1(x) = 4 x − 2 (b) f−1(x) = 1 2 − 1 4 x (e) None of these (c) f−1(x) = −4 2 − x 6.2 #63a. The function f(x) = 2x 3x − 1 is a one-to-one function. Find the inverse f −1. (a) f−1(x) = 3x − 1 2x (d) f−1(x) = x 2 − 3x (b) f−1(x) = x 3 (e) None of these (c) f−1(x) = x 3x − 2 6.2 #63b. Find the range of the function f(x) = 2x 3x − 1. (See the previous problem.) (a) all real numbers except 13 (d) all real numbers except − 23 (b) all real numbers except − 13 (e) all real numbers except 0 (c) all real numbers except 23 86 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.2 C. The function f(x) = 3x − 2 is a one-to-one function. Find the inverse function f−1. (a) f−1(x) = 1 3x − 2 (d) f −1(x) = 13x + 2 3 (b) f−1(x) = 12x + 3 2 (e) None of these (c) f−1(x) = 12x − 32 6.2 #94. The period T of a simple pendulum is T = 2π √ x g , where x is its length and g is a constant (the acceleration due to gravity). Solve for x as a function of T . (a) x = 2π √ T g (d) x = gT 2 4π2 (b) x = gT 2π (e) None of these (c) x = gT 2 2π 6.3 A. Which answer describes the graph of the exponential function f(x) = ex? (a) The graph goes through (0, e) and decreases as x increases. (b) The graph goes through (0, e) and increases as x increases. (c) The graph goes through (0, 1) and decreases as x increases. (d) The graph goes through (0, 1) and increases as x increases. (e) The graph is a straight line through (1, e). 6.3 B. Which of the following is the graph of y = 2x? (a) (b) (c) (d) (1, 0) x y −5 5 (0, 1) x y −5 5 (0, 0) x y −5 5 (0, 1) x y −5 5 87 6.3 C. Which exponential function is represented by this graph? (0, 0) (1,−1) x5 y 5 (a) f(x) = −2x2 + x (d) f(x) = 1 + 2x (b) f(x) = 1 − 2−x (e) f(x) = 1 + ex (c) f(x) = 1 − 2x 6.3 #43. Find the horizontal asymptote of f(x) = 3−x − 2. (a) y = 2 (d) x = 0 (b) y = 0 (e) x = −2 (c) y = −2 6.3 #52a. The horizontal asymptote of the graph of y = ex − 1 is (a) y = −1 (d) y = e (b) y = 0 (e) There is no horizontal asymptote (c) y = 1 6.3 #52b. The vertical asymptote of the graph of y = ex − 1 is (a) x = −1 (d) x = e (b) x = 0 (e) There is no vertical asymptote (c) x = 1 6.3 #55. Find the horizontal asymptote of the graph of f(x) = 2 − e−x/2. (a) y = −2 (d) x = 0 (b) y = 0 (e) None of these (c) x = −2 90 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.4 #29. log 1 2 16 = (a) 8 (d) −14 (b) 4 (e) −4 (c) 14 6.4 #33. log√2 4 = (a) 0 (d) 3 (b) 1 (e) 4 (c) 2 6.4 #35. ln √ e = (a) −1 (d) 2.718 (b) .5 (e) None of these (c) 1.359 6.4 F. The domain of f(x) = log(1 − 5x) is (a) (15 ,∞) (d) (−∞, 15) (b) [15 ,∞) (e) None of these (c) (−∞, 15 ] 6.4 #43. The domain of f(x) = ln ( 1 x + 1 ) is (a) {x | x ≥ −1} (d) {x | x > −1} (b) {x | x 6= −1} (e) None of these (c) {x | x < −1} 6.4 #82. Find the vertical asymptote of the graph of f(x) = 2 − log3(x + 1). (a) x = −1 (d) y = 0 (b) x = 0 (e) None of these (c) x = 1 6.4 G. The equation logπ x = 1 2 can be written in exponential form as (a) x = ( 1 2 )π (d) π = x1/2 (b) x = π1/2 (e) π = ( 1 2 )x (c) xπ = 12 91 6.4 #89. Solve: log2(2x + 1) = 3 (a) x = 1 (d) x = 4 (b) x = 0 (e) None of these (c) x = 3 6.5 #14. log6 4 + log6 9 = (a) 2 (d) log6(4/9) (b) 13/6 (e) None of these (c) log6 13 6.5 A. (log2 6)(log6 8) = (a) 2 (d) log2(4/3) (b) 3 (e) None of these (c) log6 4 6.5 B. (log3 6)(log6 9) = (a) log6 3 (d) 3 (b) log3(3/2) (e) None of these (c) 2 6.5 #29. If ln 2 = a and ln 3 = b, then ln 5 √ 6 = (a) 15ab (d) 5(a + b) (b) 15(a + b) (e) None of these (c) 5ab 6.5 #46. log ( x3 √ x + 1 (x − 2)2 ) = (a) 3 log x + 12 log(x + 1) − 2 log(x − 2) (b) 3 log x + 12 log(x + 1) + 2 log(x − 2) (c) 3 log x + log(x + 1) − log(x − 2) (d) 3 log x + log(x + 1) + log(x − 2) (e) None of these 92 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.5 #49. ln ( 5x √ 1 + 3x (x − 4)3 ) = (a) 5 ln x + ln(1 + 3x) + ln(x − 4) (b) 5 ln x + ln(1 + 3x) − ln(x − 4) (c) ln 5 + ln x + 12 ln(1 + 3x) − 3 ln(x − 4) (d) ln 5 + ln x + 12 ln(1 + 3x) + 3 ln(x − 4) (e) None of these 6.5 #72. logπ √ 2 = (a) 1 2 ln π (d) ln 2 ln π (b) ln 2 ln π (e) None of these (c) ln 2 2 ln π 6.5 #83. Express y as a function of x: ln y = ln x + ln(x + 1) + ln C (a) y = 2x + 1 + C (d) y = eCx(x+1) (b) y = Cx(x + 1) (e) None of these (c) y = Cex(x+1) 6.5 #85. Express y as a function of x (the constant C is positive). ln y = 3x + ln C (a) y = ln(3x) + C (d) y = e3x + C (b) y = Ce3x (e) None of these (c) y = C3x 6.5 #87. Solve for y (the constant C is positive): ln(y − 3) = −4x + ln C (a) y = 3 − 4 ln x + C (d) y = 3 + e−4x + C (b) y = 3 + C−4x (e) None of these (c) y = 3 + Ce−4x 95 6.3 #55. (e) 6.3 D. (d) 6.3 #61. (d) 6.3 #63. (d) 6.3 E. (c) 6.3 #85. (c) 6.3 #86. (e) 6.4 A. (b) 6.4 B. (e) 6.4 C. (a) 6.4 D. (c) 6.4 E. (b) 6.4 #29. (e) 6.4 #33. (e) 6.4 #35. (b) 6.4 F. (d) 6.4 #43. (d) 6.4 #82. (a) 6.4 G. (b) 6.4 #89. (e) 6.5 #14. (a) 6.5 A. (b) 6.5 B. (c) 6.5 #29. (b) 6.5 #46. (a) 6.5 #49. (c) 6.5 #72. (c) 6.5 #83. (b) 6.5 #85. (b) 6.5 #87. (c) 6.6 A. (d) 96 CHAPTER 6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.6 B. (e) 6.6 C. (d) 6.6 D. (c) 6.6 E. (c) Solutions 6.1 A. Let f(x) = 2x2 + 1 and let g(x) = x + 3. Find the composite function (f ◦ g)(x). Solution: (d) Write the formula for f(x) in this way: f( ) = 2( )2 + 1. Then you have (f ◦g)(x) = f(g(x)) = 2g(x)2 +1 = 2(x+3)2 +1 = 2(x2 +6x+9)+1 = 2x2 +12x+19. 6.1 #36. Find the domain of f ◦ g if f(x) = 1x+3 and g(x) = −2x . Solution: (d) {x | x 6= 0 and x 6= 2/3} First, you must exclude x = 0, since it is not in the domain of the first function g(x). If you compute the composite function f(g(x)), you get f(g(x)) = 1 − 2x + 3 = x −2+3x , so you must also exclude x = 2/3 (found by setting the denominator −2 + 3x equal to 0). 6.1 B. If f(x) = 3x2 − 7 and g(x) = 2x + a, find a so that the graph of f ◦ g crosses the y-axis at 5. Solution: (a) One method of solution is to compute the composite function f(g(x)). You get f(g(x)) = 3(2x + a)2 − 7 = 3(4x2 + 4ax + a2) − 7 = 12x2 + 12ax + 3a2 − 7. The problem asks you to find a so that the y-intercept is 5. Since the y-intercept is 3a2 − 7, you need to solve 3a2 − 7 = 5. You get 3a2 = 12, so a = ±2. Another method of solution is to find the y-intercept in two steps. The first is by substituting x = 0 into g(x). You get g(0) = a, and then, as the second step, you get f(g(0)) = 3a2 − 7. The answer a = ±2 again comes from the solution of the equation 3a2 − 7 = 5. 6.2 A. If f(x) has an inverse, and (2,− 12) is on the graph of f(x), then what point is on the graph of f−1(x)? Solution: (d) (− 12 , 2) If (a, b) is on the graph of f(x), then (b, a) is on the graph of f−1(x). 6.2 B. The function f(x) = √ x − 2, for x ≥ 2, is a one-to-one function. Find the inverse function f−1. Solution: (a) f−1(x) = x2 + 2, for x ≥ 0. Step 1. Write the function in the form y = √ x − 2. Step 2. Interchange x and y to get x = √ y − 2. Step 3. Solve for y in terms of x. x = √ y − 2 x2 = (√y − 2)2 = y−2 y = x2 +2 To find the domain of f−1(x), it may be easiest to find the range of f(x). Since x ≥ 2, this includes the square root of every number ≥ 0, so the range of f(x) is {y | y ≥ 0}. This means that the domain of f−1(x) is {x | x ≥ 0}, so the solution is the one given above. 97 6.2 #60. The function f(x) = 42−x , for x 6= 2, is a one-to-one function. Find the inverse function f−1. Solution: (a) f−1(x) = 2 − 4x Write y = 42−x , then exchange x and y and solve for y. x = 42−y (2− y)x = 4 2x− yx = 4 −yx = 4− 2x y = 4−2x−x y = 2x−4x = 2− 4x 6.2 C. The function f(x) = 3x − 2 is a one-to-one function. Find the inverse function f−1. Solution: (d) f−1(x) = 13x + 2 3 y = 3x− 2 x = 3y − 2 x + 2 = 3y y = 13x + 23 6.2 #94. The period T of a simple pendulum is T = 2π √ x g , where x is its length and g is a constant (the acceleration due to gravity). Solve for x as a function of T . Solution: (d) T = 2π √ x g , T 2 = 4π2 ( √ x g )2 T 2 = 4π2 xg gT 2 = 4π2x Answer: x = gT 2 4π2 6.3 A. Which answer describes the graph of the exponential function f(x) = ex? (a) The graph goes through (0, e) and decreases as x increases. (b) The graph goes through (0, e) and increases as x increases. (c) The graph goes through (0, 1) and decreases as x increases. (d) The graph goes through (0, 1) and increases as x increases. (e) The graph is a straight line through (1, e). Solution: (d) When x = 0, we have f(0) = e0 = 1, so the answer must be (c) or (d). As the exponent x increases, the values of ex get larger and larger, so the y-values increase as x increases, and the answer must be (d). 6.3 B. Which of the following is the graph of y = 2x? (a) (b) (c) (d) (1, 0) x y −5 5 (0, 1) x y −5 5 (0, 0) x y −5 5 (0, 1) x y −5 5 Solution: (d) The graph must go through (0, 1), and must increase as x increases. 6.3 C. Which exponential function is represented by this graph? (see the original problem) Solution: The first choice (a) f(x) = −2x2 + x is not an exponential function. The graph is decreasing, not increasing, so it cannot be (d) f(x) = 1 + 2x or (e) f(x) = 1 + ex. We need to decide between (b) f(x) = 1 − 2−x and (c) f(x) = 1 − 2x. Note that (0, 0) and (1,−1) are on the graph. Both (b) and (c) have f(0) = 0, but only (c) has f(1) = −1.