Solution manual for graphical approach to college algebra 4th edition by john hornsby lial rockswold

Solution manual for Graphical Approach to College Algebra 4th Edition by John Hornsby Lial Rockswold Chapter 2: Analysis of Graphs of Functions 2.1: Graphs of Basic Functions and Relations; Symmetry 1 q, q2 q, q2; 0, q2 10, 02 0, q2; 0, q2 increases q, ; 0, q2 x-axis even odd 10 y-axis; origin 11 The domain can be all real numbers, therefore the function is continuous for the interval: q, q2 12 The domain can be all real numbers, therefore the function is continuous for the interval: q, q2 13 The domain can only be values where x 0, therefore the function is continuous for the interval: 0, q2 14 The domain can only be values where x 0, therefore The function is continuous for the interval: q, 15 The domain can be all real numbers except 3, therefore the function is continuous for the interval: q, 32; 3, q2 16 The domain can be all real numbers except 1, therefore the function is continuous for the interval: q, 12; 11, q2 17 (a) The function is increasing for the interval: 3, q2 (b) The function is decreasing for the interval: q, (c) The function is never constant, therefore: none (d) The domain can be all real numbers, therefore the interval: q, q2 (e) The range can only be values where y 0, therefore the interval: 0, q2 18 (a) The function is increasing for the interval: 4, q2 (b) The function is decreasing for the interval: q, (c) The function is constant for the interval: 1, 4 (d) The domain can be all real numbers, therefore the interval: q, q2 (e) The range can only be values where y 3, therefore the interval: 3, q2 19 (a) The function is increasing for the interval: q, (b) The function is decreasing for the interval: 4, q2 (c) The function is constant for the interval: 1, 4 (d) The domain can be all real numbers, therefore the interval: q, q2 (e) The range can only be values where y 3, therefore the interval: q, 58 CHAPTER Analysis of Graphs of Functions 20 (a) The function never is increasing, therefore: none (b) The function is always decreasing, therefore the interval: q, q2 (c) The function is never constant, therefore: none (d) The domain can be all real numbers, therefore the interval: q, q2 (e) The range can be all real numbers, therefore the interval: q, q2 21 (a) The function never is increasing, therefore: none (b) The function is decreasing for the intervals: q, ; 3, q2 (c) The function is constant for the interval: 2, 32 (d) The domain can be all real numbers, therefore the interval: q, q2 (e) The range can only be values where y 2, therefore the interval: q, 1.5 ´ 2, q2 1.5 or y 22 (a) The function is increasing for the interval: 13, q (b) The function is decreasing for the interval: q, 32 (c) The function is constant for the interval: 3, (d) The domain can be all real numbers except 3, therefore the interval: q, 32 ´ 3, q2 (e) The range can only be values where y 1, therefore the interval: 11, q2 x5, See Figure 23 As x increases for the interval: q, q2, y increases, therefore increasing 23 Graph f 1x2 24 Graph f 1x2x3, See Figure 24 As x increases for the interval: q, q2, y decreases, therefore decreasing x4, See Figure 25 As x increases for the interval: q, , y decreases, therefore decreasing 25 Graph f 1x2 x4, See Figure 26 As x increases for the interval: 0, q2, y increases, therefore increasing 26 Graph f 1x2 10, 10 by Xscl 10, 10 Yscl 10, 10 by Xscl 10, 10 Yscl 10, 10 by Xscl 10, 10 Yscl 10, 10 by Xscl 10, 10 Yscl Figure 23 Figure 24 Figure 25 Figure 26 27 Graph f 1x210x See Figure 27 As x increases for the interval: q, , y increases, therefore increasing 28 Graph f 1x210x , See Figure 28 As x increases for the interval: 0, q2, y decreases, therefore decreasing 29 Graph f 1x23 x, See Figure 29 As x increases for the interval: q, q2, y decreases, therefore decreasing 30 Graph f 1x2x, See Figure 30 As x increases for the interval: 0, q2, y decreases, therefore decreasing 10, 10 by Xscl 10, 10 Yscl Figure 27 10, 10 by Xscl 10, 10 Yscl Figure 28 10, 10 by Xscl 10, 10 Yscl Figure 29 10, 10 by Xscl 10, 10 Figure 30 Yscl Graphs of Basic Functions and Relations; Symmetry SECTION 2.1 59 31 Graph f 1x2 x , See Figure 31 As x increases for the interval: q, q2, y decreases, therefore decreasing f 1x2 x q2 32 Graph As x increases for the interval: 1, 2x, See Figure 32 , y increases, therefore increasing 33 Graph f 1x2 34 Graph f 1x2 10, 10 Xscl See Figure 33 As x increases for the interval: q, , y increases, therefore increasing x x , See Figure 34 As x increases for the interval: q, , y decreases, therefore decreasing by 10, 10 Yscl 10, 10 by Xscl Figure 31 10, 10 Yscl 10, 10 by Xscl Figure 32 35 (a) No (b) Yes (c) No 36 (a) Yes (b) No (c) No 37 (a) Yes (b) No (c) No 38 (a) No (b) No (c) Yes 39 (a) Yes (b) Yes (c) Yes 40 (a) Yes (b) Yes (c) Yes 41 (a) No (b) No (c) Yes 10, 10 Yscl Figure 33 10, 10 by Xscl 10, 10 Yscl Figure 34 42 (a) No (b) Yes (c) No 43 If f is an even function then f x2 f 1x2 or opposite domains have the same range See Figure 43 44 If g is an odd function then g1 x2g1x2 or opposite domains have the opposite range See Figure 44 x g(x) –5 13 –3 –2 –5 0 –1 –13 x ƒ(x) –3 21 –2 –12 –1 –25 –25 –12 21 Figure 43 Figure 44 45 (a) Since f x2 f 1x2, this is an even function and is symmetric with respect to the y-axis See Figure 45a (b) Since f x2f 1x2, this is an odd function and is symmetric with respect to the origin See Figure 45b y y 2 x x Figure 45a Figure 45b 60 CHAPTER Analysis of Graphs of Functions 46 (a) Since this is an odd function the graph is symmetric with respect to the origin See Figure 46a (b) Since this is an even function the graph is symmetric with respect to the y-axis See Figure 46b y y 2 x x Figure 46a 47 Since f x2 f 1x2, it is even Figure 46b 48 Since f x2f 1x2, it is odd 49 If f 1x2 x4 7x2 6, then f x2 x24 71 x22 f x2 x4 7x2 Since f x2 f 1x2, the function is even 50 If f 1x2 2x6 8x2 , then f x2 21 x26 81 x22 f x2 2x6 8x2 Since f x2 f 1x2, the function is even 51 If f 1x2 x6 4x4 5, then f x2 x26 41 x24 f x2 x6 4x4 Since f x2 f 1x2, the function is even 52 If f 1x2 82,then f x2 Since2 f x2 f 1x2, the function2 is even 53 If f 1x20 5x , then f x20 51 x f x20 5x Since f x2f 1x2, the function is even 54 If f 1x2x2 1, then f x21 x221 f x2x2 Since f x2f 1x2, the function is even 55 If f 1x2 3x3 x, then f x2 31 x23 x2 f x2 3x3 x and f 1x213x3x2 f 1x23x3 56 If f 1x2 x5 f 1x21 x52x3 57 If f 1x2 3x5 f 1x213x5x3 58 If f 1x2 x3 3x, then f x2 2x3 3x2 f 1x2 x52x3 7x, then f x2 x3 7x2 f 1x23x5 x3 4x, then f x2 x23 f 1x21x34x2 f 1x2x3 x25 21 x23 31 x2 f x2 x5 2x , then f x2 31 x25 x23 71 x2 f x2 3x5 41 x2 f x2 21 x2 f x2 x3 2x and f 1x2a f 1x2a4x 1 x2 xb x3 7x and 7x Since f x2f 1x2, the function is odd 4xx, then f 3x and 3x Since f x2f 1x2, the function is odd 4x and 2x b 1f 1x2 f x2 f 1x2, the function is odd 60 If f 1x2 2x3 4x Since f x2f 1x2, the function is odd 59 If f 1x2 x Since f x2f 1x2, the function is odd 41 f 1x2 x2 4x 1 x f x2 4xx and x Since f x2 f 1x2, the function is odd 2x Since 61 If f 1x2 Graphs of Basic Functions and Relations; Symmetry 2x, then f x2 x23 21 x2 f x2 x3 2x and x3 f 1x21 x32x2 f 1x2 origin Graph f 1x2 62 If f 1x2 x3 x 2x The graph supports symmetry with respect to the origin f 1x21x52x32 f 1x2x5 the origin f 1x2 f 12x2 x x25 21 x23 f x2 x5 2x3 and 2x3 Since f x2f 1x2, the function is symmetric with respect to Graph 63 If x The graph supports symmetry with respect to the origin 2x x2.5 x2 f1 then 2x1, 21 x2 1 f x2 x 24x 2 Since f x2 f 1x2, the function is symmetric with respect to the y-axis Graph f 1x2.5 x supports symmetry with respect to the y-axis 2x 1, then f x2 75 x22 64 If f 1x2 75 x x x 01 f x2 f x2 f 1x2, the function is symmetric with respect to the y-axis Graph f 1x2.75 x supports symmetry with respect to the y-axis The graph 75 x2 x Since x The graph 3 x 3, then f x2 x and f 1x2 1x3x x231 x23 65 If f 1x2 x f x2 x f 1x2x3 x Since f 1x2f x2f 1x2, the function is not symmetric with respect to the y-axis or origin Graph f 1x2 66 If f 1x2 x4 f 1x2x4 5x 5x x3 f 1x2x6 32 The graph supports no symmetry with respect to the y-axis or origin x24 51 x2 21f1 x2 x4 5x and f 1x2 1x4 5x 22 Since f 1x2f x2f 1x2, the function is not symmetric with respect to the y-axis 5x 4x3, then f x2 x6 The graph supports no symmetry with respect to the y-axis or origin x26 41 x23 f x2 x6 4x3 and f 1x2 1x6 4x32 4x3 Since f 1x2f x2f 1x2, the function is not symmetric with respect to the y-axis or ori- gin Graph f 1x2 68 If f 1x2 x 2, then f x2 or origin Graph f 1x2 x4 67 If f 1x2 61 2x Since f x2f 1x2, the function is symmetric with respect to the 2x3, then f x2 x5 SECTION 2.1 x6 4x3 The graph supports no symmetry with respect to the y-axis or origin 3x, then f x2 x3 x23 31 x2 f x2 x3 3x and f 1x2 1x3 3x2 f 1x2 x3 3x Since f x2 f 1x2, the function is symmetric with respect to the origin Graph f 1x2 x3 3x The graph supports symmetry with respect to the origin 69 If f 1x2 6, then f x2 Since f x2 f 1x2, the function is symmetric with respect to the y-axis Graph f 1x2 The graph supports symmetry with respect to the y-axis 70 If f 1x2 x 0 x , then f x2 x2 0 x f x2 x Since f x2 f 1x2, the function is symmetric with respect to the y-axis Graph f 1x2 x The graph supports symmetry with respect to the y-axis 1 1 1 12 f x2 4x3 and f1x2a 71 If f 1x2 4x3 , then f x2 x 4x3 b f 1x2 4x3 Since f x2 f 1x2, the function is symmetric with respect to the origin Graph f 1x2 supports symmetry with respect to the origin 72 If f 1x2 x2 f 1x2 x, then f x2 x2 f x2 function is symmetric with respect to the y-axis Graph f 1x22x respect to the y-axis x2 f x2 x 4x3 The graph Since f x2 f 1x2, the The graph supports symmetry with 62 CHAPTER Analysis of Graphs of Functions 73 (a) Functions where f x2 f 1x2 are even, therefore exercises: 63, 64, 69, 70, and 72 are even (b) Functions where f x2f 1x2 are odd, therefore exercises: 61, 62, 68, and 71 are odd (c) Functions where f 1x2 f x2 f 1x2 are neither odd or even, therefore exercises: 65, 66, and 67 are neither odd or even 74 Answers may vary If a function f is even, then f 1x2 f x2 for all x in the domain Its graph is symmetric with respect to the y-axis If a function f is odd, then f x2 f 1x2 for all x in the domain Its graph is symmetric with respect to the origin 2.2: Vertical and Horizontal Shifts of Graphs The equation y The equation y shifted units upward is: y x2 shifted units downward is: y x3 x shifted units downward is: y The equation y The equation y 1x x shifted units upward is: y1x The equation y y The equation The equation y x x x shifted units to the right is: y0 x4 x x shifted units to the left is: y 30 shifted units to the left is: y x 1x 72 The equation y x shifted units to the right is: y 1x Shift the graph of f units upward to obtain the graph of g 10 Shift the graph of f units to the left to obtain the graph of g 11 The equation y 12 The equation y x2 1x is y x2 shifted units downward, therefore graph B 322 is y x2 shifted units to the right, therefore graph C 13 The equation y 1x 322 is y x2 shifted units to the left, therefore graph A 14 The equation y0 x 04 is y0 x shifted units upward, therefore graph A 15 The equation y0 x4 03 is y0 x shifted units to the left and units downward, therefore graph B 16 The equation y0 x4 03 is y0 x shifted units to the right and units downward, therefore graph C 17 The equation y 1x 323 is y x3 shifted units to the right, therefore graph C 18 The equation y 1x 223 is y x3 shifted units to the right and units downward, therefore graph A 19 The equation y 1x 223 is y x3 shifted units to the left and units downward, therefore graph B 20 If y x h k with h and k 0, then the graph of y x is shifted to the left h units and k units downward This would place the vertex or lowest point of the absolute value graph in the third quadrant 21 For the equation y x2 , the Domain is: q, q2 and the Range is: 0, q2 Shifting this units downward Domain: gives us: 22 For the equation y x gives us: (a) (b) Range: 1q, q2 33, q2 , the Domain is: q, q2 and the Range is: 0, q2 Shifting this units to the right (a) Domain: 1q, q2 (b) Range: 30, q2 Vertical and Horizontal Shifts of Graphs 23 For the equation y (a) Domain: q, q2 (b) Range: 3, q2 x , the Domain is: q, q2 and the Range is: 0, q2 Shifting this units to the right and units downward gives us: (a) Domain: q , q2 (b) Range: Domain: (a) , q q2 (b) Range: , q q2 and units downward gives us: (a) Domain: q, q2 27 Using Y2 Y1 k and x 0, we get 19 15 k k (b) Range: q, q2 28 Using Y2 Y1 k and x 0, we get 53 k k2 29 From the graphs 16, 22 is a point on Y1 and 16, 12 a point on Y2 Using Y2 3, q2 , the Domain is: q, q2 and the Range is: q, q2 Shifting this units to the right For the equation y x x , the Domain is: q, q2 and the Range is: q, q2 Shifting this units to the right 25 For the equation y 26 gives us: 63 x , the Domain is: q, q2 and the Range is: 0, q2 Shifting this units to the left and units downward gives us: 24 For the equation y SECTION 2.2 Y1 k and x 6, we get k k3 30 From the graphs 4, 32 k k is a point on Y1 and 4, 82 a point on Y2 Using Y2 122 is the graph of the equation y 31 The graph of yx x2 Y1 k and x4, we get shifted unit to the right See Figure 31 1x shifted units to the left See Figure 32 x3 shifted unit upward See Figure 33 32 The graph of y1x is the graph of the equation y 33 The graph of y x3 is the graph of the equation y y y y 1 x x –2 x 0 Figure 31 Figure 32 Figure 33 34 The graph of y0 x2 is the graph of the equation y0 x shifted units to the left See Figure 34 1x 35 The graph of y 123 is the graph of the equation y x3 shifted unit to the right See Figure 35 x is the graph of the equation y 36 The graph of y 36 y x shifted units downward y See Figure y x –2 0 –1 x –3 Figure 34 Figure 35 x –3 Figure 36 Analysis of Graphs of Functions 37 The graph of y 1x is the graph of the equation y 1x shifted units to the right and unit downward See Figure 37 38 The graph of yx is the graph of the equation yx shifted units to the left and 64CHAPTER2 units downward See Figure 38 39 The graph of y 1x 222 is the graph of the equation y x2 shifted units to the left and units upward See Figure 39 y y y x –1 x –3 –2 x –4 Figure 37 Figure 38 Figure 39 40 The graph of y 1x 422 is the graph of the equation y x2 shifted units to the right and units downward See Figure 40 41 The graph of y x is the graph of the equation y x shifted units to the left and units downward See Figure 41 42 The graph of y 1x 323 is the graph of the equation y x3 shifted units to the left and unit downward See Figure 42 y y y x x – – –2 –2 –4 x –2 Figure 40 43 Since h and k are positive, the equation y 1x Figure 41 Figure 42 h2 k is y x shifted to the right and down, therefore: B 44 Since h and k are positive, the equation y 1x h22 k is y x2 shifted to the left and down, therefore: D 45 Since h and k are positive, the equation y 1x h22 k is y x2 shifted to the left and up, therefore: A 46 Since h and k are positive, the equation y 1x h22 k is y x2 shifted to the right and up, therefore: C 47 The equation y f 1x2 is y f 1x2 shifted up units or add to the y-coordinate of each point as follows: 3, 22 1 3, 02; 1, 42 1 1, 62; 15, 02 15, 22 See Figure 47 Vertical and Horizontal Shifts of Graphs SECTION 2.2 48 The equation y f 1x2 is y f 1x2 shifted down units or subtract from the y-coordinate of each point as 65 follows: 3, 22 1 3, 2; 1, 42 1 1, 22; 15, 02 15, 22 See Figure 48 y y (– 1, 6) (–1, 2) (5, 2) x (– 3, 0) x 0 (5, –2) (–3, –4) Figure 47 Figure 48 49 The equation y f 1x 22 is y f 1x2 shifted left units or subtract from the x-coordinate of each point as follows: 3, 22 1 5, 22; 1, 42 1 3, 42; 15, 02 13, 02 See Figure 49 50 The equation y f 1x 22 is y f 1x2 shifted right units or add to the x-coordinate of each point as follows: 3, 22 1 1, 22; 1, 42 11, 42; 15, 02 17, 02 See Figure 50 y y (– 3, 4) (1, 4) (3, 0) (7, 0) x x 0 (– 5, – 2) (–1, –2) Figure 49 Figure 50 51 The graph is the basic function y is: y 1x 422 x2 translated units to the left and units up, therefore the new equation The equation is now increasing for the interval: (a) 4, q2 and decreasing for the interval: (b) q, 4 52 The graph is the basic function y x translated units to the left, therefore the new equation is: y1 The equation is now increasing for the3 interval: (a) 5, q2 and does not decrease, therefore: (b) 53 The graph is the basic function y x translated units down, therefore the new equation is: y 54 The graph is the basic function y x0 none translated 10 units to the left, therefore the new equation is: y0 x10 The equation is now increasing for the interval: (a) 10, q2 and decreasing for the interval: (b) q, 10 55 The graph is the basic function y x translated units to the right and unit up, therefore the new equation The equation is now increasing for the interval: (a) 2, q2 and does not decrease, is: y1 x therefore: (b) none 56 The graph is the basic function y x translated units to the right and units down, therefore the new equation is: y 1x 222 the interval: (b) q, none x The equation is now increasing for the interval: (a) q, q2 and does not decrease, therefore: (b) x The equation is now increasing for the interval: (a) 2, q2 and decreasing for 66 CHAPTER 57 (a) f 1x2 0: 53, 46 Analysis of Graphs of Functions (b) f 1x2 0: for1 the intervals q, 32 ´ 14, q2 (c) f 1x2 0: for the interval 13,142 58 (a) f 1x2 0: 261 (b) f 1x2 0: for the interval 2, q2 (c) f 1x2 0: for the interval q, 22 59 (a) f 1x2 0: 4, 56 (b) f 1x20: for the intervals q, 4 ´ 5, q2 (c) f 1x20: for the interval 4, 60 (a) f 1x2 0: never, therefore: (b) f 1x20: for the interval 1, q2 (c) f 1x20: never, therefore: 61 The translation is units to the left and unit up, therefore the new equation is: y0 x3 01 The form y x h0 k x when: h3 and k will equal y 0 62 The equation y has a Domain: q, q2 and a Range: 0, q2 After the translation the Domain is still: x q, q2, but now the Range is: 38, q2, a positive or upward shift of 38 units Therefore, the horizontal shift can be any number of units, but the vertical shift is up 38 This makes h any real number and k 38 63 (a) Since corresponds to 1998, our equation using exact years would be: y 895.512006 19982 14,709 y (b) y 14,709 y 7164 895.51x 19982 299.81x 19982 14,709 $21,873 64 (a) Since corresponds to 1998, our equation using exact years would be: y 5249.1 (b) y 299.812007 19982 5249.1 y 2698.2 5249.1 y $ 7947.3 billion 65 (a) Enter the year in L1 and enter tuition and fees in L The year 1991 corresponds to x and so on The regression equation is: y 190x 2071.3 (b) Since x corresponds to 1991, the equation when the exact year is entered is: y 1901x 19912 2071.3 (c) y 19012008 19912 2071.3 y 3230 2071.3 y $5300 66 (a) Enter the year in L1 and enter the percent of women in the workforce in L The year 1965 corresponds to x and so on The regression equation is: y 6162x 40.6167 (b) Since x corresponds to 1965, the equation when the exact year is entered is: y 61621x 19652 40.6167 (c) y 616212010 19652 40.6167 y 67 See Figure 67 22 m 69 Using slope-intercept form yields: y1 68 m 70 11, 40.6167 y 68.3% 42 71 m 27.729 11 and 3, 1m 2 21x 32 y1 1, and 3, 2 2x y1 2x 108 CHAPTER x 90 (a) If x 4s then s (b) If y s Analysis of Graphs of Functions x , then y1x2 and s (c) Sine x is perimeter and x 6, y 36 16 16 x2 13 (d) Show that the point 16, 2.252 is on the graph y 13 12x2 91 (a) A12x2 (b) A1x2 11622 (c) On the graph of y 16 162 13 x2 x a b y1x2 A square with perimeter will have area 2.25 square units 16 13 14x A12x213 x 12562 A1x264 2.25 square units x2, locate the point where x 16 to find y 92 (a) If A1r2 π r and r1t2 2t then 1A r2 1t2 A r 1t2 110.85, an approximation for 64 A 2t π 12t22 13 π t (b) 1A r2 1t2 is a composite function that expresses the area of the circular region covered by the pollutants as a function of time t (in hours) (c) Since t (d) Graph y1 1A r2 142 is A.M., noon would be t π 1422 64 π mi π x2 and show that for x 4, y 201 (an approximation for 64 π) 93 (a) The function h is the addition of functions f and g x 1999 2000 2001 2002 2003 h(x) 76 82 79 89 103 (b) The function h is the addition of functions f and g Therefore h1x2 94 (a) The domain is the years See Table Therefore: D f 1x2 g1x2 51998, 1999, 2000, 2001, 20026 x 1998 1999 2000 2001 2002 h(x) 94.2 95.3 99.3 106.4 93.2 (b) The function h computes the value of animals produced in the U.S and sold in billions of dollars 95 (a) f g2 119702 32.4 17.6 50.0 (b) The function f g2 1x2 computes the total SO2 emissions from burning coal and oil during year x (c) Add functions f and g x (f g)(x) 1860 2.4 1900 12.8 1940 26.5 1970 50.0 2000 78.0 96 (a) The function h is the addition of functions f and g x 1990 2000 2010 2020 2030 h(x) 32 35.5 39 42.5 46 (b) The function h is the addition of functions f and g Therefore h1x2 f 1x2 97 (a) The function h is the subtraction of function f from g Therefore h1x2 (b) h119962g119962f 119962841 694 147 h120062g120062f 1200621165 1012 153 g 1x2 (c) Using the points 11996, 1472 and 12006, 1532 from part b, the slope is: m Now using point slope form: y 147 61x 19962 y 61x 19962 g1x2 f 1x2 153 2006 147 147 1996 10 Operations and Composition 1900 1x 1982 22 619 98 (a) Graph h1x2 SECTION 2.6 109 , in the window 1982, 1994 by 0, See Figure 98a 3200 x 1982 1586 Approximately 59% of the people who contracted AIDS during this time period died (b) Divide number of deaths by number of cases for each year The results compare favorably with the graph See Figure 98b 10, 10 by Xscl 10, 10 Yscl 4 Year Ratio 1982 39 1984 51 1986 59 Figure 98a 1988 58 1990 61 1992 60 1994 61 Figure 98b Reviewing Basic Concepts (Sections 2.4—2.6) 1 1 (a) ` x `4 x 12x Therefore, the solution set is: 12, 46 (b) ` x ` 2x 274 1x 1 x x or Therefore, the solution interval is: q, 122 1 (c) ` x 41 42x Therefore, the solution interval is: ` or 2x 2x 12 x x6 2x 6 1x 12 x 12 h 14, q2 41 2x 12, 4 For the graph of y f 1x2 , we reflect the graph of y f 1x2 across the x-axis for all points for which y Where y 0, the graph remains unchanged See Figure y (–3, 2) (3, 2) x Figure 110 CHAPTER Analysis of Graphs of Functions (a) The range of values for RL 26.75 01.42 is: 1.42 range of values for RE 2251RE2 then the range for TL is: 225136.58 If TE (a) f 32 21 32 TE TL 40.922 (b) f 102 1022 1.42 25.33 26.75 2.17 is: 2.17 2251RL2 then the range for TL is: 225125.33 (b) If TL 38.75 RL RE 38.75 28.172 8230.5 2.17 36.58 5699.25 TE TL 28.17 The RE 40.92 6338.25 9207 (c) f 122 1222 4 RL (a) See Figure 5a x22 * 1x (b) Graph y1 42 * 1x 02 in the window 10, 10 by 10, 10 See Figure 5b 1x 02 y 10, 10 by 10, 10 Xscl Yscl 0, 1000 by 20, 100 Xscl 100 Yscl 10 x –2 –4 Figure 5a Figure 5b (a) f g2 1x2 (b) f 21 g x 13x (c) f g2 1x2 agb1 1x 42 3x x 23 x x 2 22 3x 4 Therefore, agb1 4x Therefore, f g2 2231 22 32 f 1x2g f 1x2 41 3x4 solutions for One of many possible 2x2 1g f g x 21 3x 9x f 1x2 h x 52 is: f 21x g2 41 22 46 422 24 16 x 244 x x 16 and g x 2xh h x Then 3x 3h h 2x 3x 3h 2x2 3x 54xh 2h2 3h4x 2h h h (a) At 4% simple interest the equation for interest earned is: y1 04 x 4xh 2h2 f g2 1x2 f g1x2 41x22 21x h231x h2 2x h f f 3x f 31 32 (d) x Therefore, 2 (e) f g2 1x2f g1x2 431x22 1 f g2 1x23x2 (f) 1g fg2 112 112 3112 Therefore, 3x 3x42 1x 3x x Figure 3x (b) If he invested x dollars in the first account, then he invested x 500 in the second account The equation for the amount of interest earned on this account is: y2 0251 x 5002 y2 025x 12.5 (c) It represents the total interest earned in both accounts for year (d) Graph y1 y2 04x 1.025x 12.52 y1 y2 04x 025x 0, 1000 by 0, 100 See Figure An input value of x (e) At x2 250, y1 y2 2.0412502 12.5 in the window 250, results in $28.75 earned interest .025122502 12.5 10 If the radius is r, then the height is 2r and the equation is S π r r 212r22π r r 24r π r 5r S π r 10 16.25 12.5 $28.75 Chapter Review SECTION 2.R 111 Chapter Review Exercises The graphs for exercises 1–10 can be found in the “Function Capsule” boxes located in section 2.1 in the text True Both f 1x2 and f 1x2 True Both f 1x x False The function f 1x2 False The function f 1x2 True The function f 1x2 x has a domain and range of: q, q2 False The function f 1x2 True All of the functions show increases on the interval: 0, q2 True Both f 1x2 x x2 and f x have the interval: 0, q2 as the range x0 x increase on the interval: 0, q 3 and f 1x21x the domain: q, q 13x has the domain: 0, q2 x increases on its entire domain 1x is not defined on q, 02, so certainly cannot be continuous 2and f 1x2 x have graphs that are symmetric with respect to the origin True Both f 1x2 x and f 1x2 x have graphs that are symmetric with respect to the y-axis 10 True No graphs are symmetric with respect to the x-axis can be input for x, therefore the domain of f 1x21 11 Only values where x x is: 0, q2 x is: 0, q2 12 Only positive solutions are possible in absolute value functions, therefore the range of f 1x2 range of is: 13 All solutions are possible in cube root functions, therefore the input for x, therefore the domain of 14 All values can be 15 The function f 1x2 1 16 The function f x 17 The equation y domain of y f 1x2 q, q2 x q, q2 increases for all inputs where x x is the equation y1x Only values where x is: x increases for all inputs for x, therefore the interval is: q, q2 x x f 1x2 0, therefore the interval is: 0, q can be input for x, therefore the x is: 0, q2 18 The equation y x is the equation y1 x Square root functions have both positive and negative solutions and all solution are possible, therefore the range of y 1x is: q, q2 19 The graph of f 1x2 1x 32 is the graph y x shifted units to the left and unit downward See Figure 19 20 The graph of f 1x2 of 2x is the graph y x reflected across the x-axis, vertically shrunk by a factor , and shifted unit upward See Figure 20 21 The graph of f 1x2 1x 122 is the graph y x2 shifted unit to the left and units downward See Figure 21 y y y –2 x x –1 x –2 Figure 19 Figure 20 Figure 21 112 CHAPTER Analysis of Graphs of Functions 22 The graph of f 1x2 2x is the graph y x2 reflected across the x-axis, vertically stretched by a factor of 2, and shifted units upward See Figure 22 23 The graph of f 1x2 x3 is the graph y x3 reflected across the x-axis and shifted units upward See Figure 23 24 The graph of f 1x2 323 is the graph y 1x x3 shifted units to the right See Figure 24 y y y 2 x x x –2 Figure 22 25 The graph of f 1x2 26 The graph of f 1x2 See Figure 26 Figure 23 B2 x 2x 1x horizontally stretched by a factor of See Figure 25 is the graph y 1x shifted units to the right and unit upward is the graph y 27 The graph of f 1x2 Figure 24 1x is the graph y 1x vertically stretched by a factor of See Figure 27 y y y 1 x x Figure 25 28 The graph of f 1x2 29 The graph of f 1x2 x Figure 26 1x is the graph y1x 01 is the graph y x Figure 27 shifted units downward See Figure 28 x shifted units right and unit upward See Figure 29 30 The graph of f 1x2 2x is the graph y0 x horizontally shrunk by a factor of units to the left, and reflected across the y-axis See Figure 30 2, shifted a b 13 or Chapter Review y SECTION 2.R y 113 y 3 x –1 x 0 x –2 Figure 28 Figure 29 Figure 30 31 (a) From the graph, the function is continuous for the intervals: q, 22, 2, , and 11, q2 (b) From the graph, the function is increasing for the interval: 2, (c) From the graph, the function is decreasing for the interval: q, 22 (d) From the graph, the function is constant for the interval: 11, q2 (e) From the graph, all values can be input1 for x, therefore 1the domain is: q, q2 (f) From the graph, the possible values of y or the range is: 26 h 1, h 12, q2 32 x y2 y2 x yx and yx 33 From the graph, the relation is symmetric with respect to the x-axis, y-axis, and origin The relation is not a function since some inputs x have two outputs y 6, then F x x3 34 If F x x 1 2 F1x2 1x 62 F1x2 x neither an even nor an odd function 35 If f 1x2 x f x2 f 1F x x Since F1x2 and F1 x2F1x2, the function has no symmetry and is , then f x2 0x2 04 f x20 x 04 and f 1x20 x 04 Since x2, the function is symmetric with respect to the y-axis and is an even function 36 If f 1x2 and f x21 1x 5, then f x2 1 x25 x Since f 1x2f x2f 1x2, the function has no symmetry and is neither an even nor an odd function x then y1 37 If y x Since f 1x21 x is the reflection of f 1x21 x across the x-axis, the relation has symmetry with respect to the x-axis Also, one x input can produce two y outputs The relation is not a function 2x2 1, then f x2 31 x24 21 x22 1 f x2 3x4 2x2 and 38 If f 1x2 3x f 1x2 3x4 2x2 Since f x2 f 1x2, the function is symmetric with respect to the y-axis and is an even function 39 True, a graph that is symmetrical with respect to the x-axis means that for every 1x, y2 there is also 1x, y2, which is not a function 40 True, since an even function and one that is symmetric with respect to the y-axis both contain the points 1x, y2 and x, y2 41 True, since an odd function and one that is symmetric with respect to the origin both contain the points 1x, y2 and x, y2 114 CHAPTER Analysis of Graphs of Functions function, if a, b is on the graph, then a, b is on the graph and not a, b 42 False, for an even For example, f 1x2 x 2 is even, and 12, 42 is on the graph, but 12, 42 is not 43 False, for an odd function, if a, b is on the graph, then 1a, b For example, f 1x2 x is on the graph and not a, b is odd, and 12, 82 is on the graph, but 2, 82 is not 44 True, if 1x, 02 is on the graph2 of f 1x2 0, then x, 02 is on the graph 45 The graph of y31x 42 is the graph of y x shifted units to the left, vertically stretched by a factor of 3, reflected across the x-axis, and shifted units downward 1x 46 The equation y reflected across the y-axis is: y x , then reflected across the x-axis is: y1 1x , and finally shifted units upward is: y now vertically shrunk by a factor of is: y 3 47 Shift the function f upward units See Figure 47 x 1x , 48 Shift the function f to the right units See Figure 48 49 Shift the function f to the left units and downward units See Figure 49 y y y (8, 3) (–4, 3) –4 (4, 1) –3 (10, 0) x x (6, –2) (–7, –2) –4 x (–2, 0) –4 (5, –2) (1, –4) Figure 47 Figure 48 Figure 49 50 For values where f 1x2 the graph remains the same For values where f 1x2 reflect the graph across the x-axis See Figure 50 51 Horizontally shrink the function f by a factor of See Figure 51 52 Horizontally stretch the function f by a factor of See Figure 52 y y y 2 (4, 2) (0, 0) (2, 0) x (–4, 0) (8, 0) (–1, 0) (0, 0) x (1, –2) (16, 0) x (–8, 0) –2 (8, –2) Figure 50 Figure 51 Figure 52 53 The function is shifted upward units, therefore the domain remains the same: 3, 4 and the range is increased by and is: 2, Chapter Review SECTION 2.R 115 54 The function is shifted left 10 units, therefore the domain is decreased by 10 and is: 13, ; and the function is stretched vertically by a factor of 5, therefore the range is multiplied by and is: 10, 25 55 The function is horizontally shrunk by a factor of 2, therefore the domain is divided by and is: c 2, d ; and the function is reflected across the x-axis, therefore the range is opposite of the original and is: 5, 56 The function is shifted right unit, therefore the domain is increased by and is: 2, ; and the function is also shifted upward units, therefore the range is increased by and is: 1, 57 We reflect the graph of y f 1x2 across the x-axis for all points for which y Where y 0, the graph remains unchanged See Figure 57 58 We reflect the graph of y f 1x2 across the x-axis for all points for which y Where y 0, the graph remains unchanged See Figure 58 y (–2, 2) y y (0, 2) (0, 2) (0, 2) x Figure 57 59 Since the range is 526, y x x (2, 0) (–2, 0) (2, 0) 0 Figure 58 0, so the graph remains unchanged Figure 60 60 Since the range is 26, y 0, so we reflect the graph across the x-axis See Figure 60 15 or 4x 312 4, 61 4x 012 4x 12 4x x 4x15 x therefore the solution set is: e 15, f 4 62 2x 1 2x Since an absolute value equation can not have a solution less than zero, the solution set is: 63 5x x 147 x 11 5x x , therefore the solution set is: e 11 4x 81x or 5x 1x 112 6x 7, f 3 64 2x 2x 2x x or 2x 2x 12 x 6, therefore the solution set is: 6, 16 65 2x5 07 2x 12 2x 21 x , therefore the interval is: 6, 66 2x 2x 2x x or 2x 2x 12 x 6, therefore the solution is the interval: q, h 1, q2 14 116 CHAPTER Analysis of Graphs of Functions 12 67 5x12 5x 12 5x 12 x 12 or 5x 12 12 12 5x 12 12 x , therefore the solution is the interval: a q, b h a ,qb or e x x 68 Since an absolute value equation can not have a solution less than zero, the solution set is: 69 3x 01 21 3x 110 3x9 70 2x 0 3x x2 x 3x 020 1x 1 2x 3x 10 3x 10 3x 11 11 or 11 x , therefore the solution set is: e 3, f 3x 1 5x 01x or 2x 1 3x 12 , therefore the solution set is: 50, 26 71 The x-coordinates of the points of intersection of the graphs are and Thus, 6, 16 is the solution set of y1 y2 The graph of y1 lies on or below the graph of y2 between and 1, so the solution set of y1 y2 is 6, The graph of y1 lies above the graph of y2 everywhere else, so the solution set of y1 y2 is q, h 1, q2 72 Graph y1 x 0 x and y2 See Figure 72 The intersections are x and x , therefore the solution set is: 3, 321 00 323 08 Check:0 152 00 1523 08 00 08 08 818 818 and y 10, 10 by Xscl y 4, 16 Yscl (30, 500) Gallons 500 400 300 200 100 (0, 500) (10, 0) x 10 20 30 x Minutes Figure 72 Figure 74 Figure 75 73 Initially, the car is at home After traveling 30 mph for hr, the car is 30 mi away from home During the second hour the car travels 20 mph until it is 50 mi away During the third hour the car travels toward home at 30 mph until it is 20 mi away During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home During the last hour, the car travels 60 mi at 60 mph until it arrives home 74 See Figure 74 75 See Figure 75 76 See Figure 76 77 Graph y1 13x 78 See Figure 78 12 * 1x 221 x42 * 1x 22 in the window 10, 10 by 10, 10 See Figure 77 Chapter Review SECTION 2.R 117 y 10, 10 by 10, 10 Xscl 5, by 5, Yscl Xscl 100 Yscl 10 x Figure 76 Figure 77 79 From the graphs f g2 112 80 From the graphs f g2 102 43 Figure 78 81 From the graphs f g2 12 102 1320 f 82 From the graphs a g b 122 83 From the graphs f g 122 f g122 f 122 84 From the graphs 1g f 122 g f 122 g 132 85 From the graphs 1g f 42 g f 42 g 122 86 From the graphs f g 22 f g1 22 f 122 87 From the table f g2 112 88 From the table f g2 132 9 89 From the table f g2 12132 226 f 90 From the table a g b 102 , which is undefined 91 From the tables 1g f 22 g f 22 4g 1122 92 From the graphs f g2 132 93 21x h2 94 1xh2 2xh 12x 922x f g132 4f 221 2h 2x h 92h h 1x 51x h23 5x 32 h x 5h h2 5x 2x h x2 5x solutions for 3x2 is: 1f f g2 1x2f g1x2 41x g2 1x2 h1x2 1 f g2 1x2 f g1x2 x and f 1x2 x Then g1x2 x 3x 96.One of many possible solutions for f g2 1x2 h1x2 5h h 95.One of many possible is: f 1x2 x and g1x2 x Then π r 3, then a inch increase would be: V1r24 π 1r V1r2 2xh h h2 h 97 If V1r2 4 π 1r423 3 πr 423, and the volume gained would be: Full file at https://fratstock.eu 118 98 CHAPTER (a) Since h d, r d V1d2π a 2b d (b) Since h d, r d Analysis of Graphs of Functions d , and the formula for the volume of a can is: V πd3 V1d2 π r 2h, the function is: , c 2π r , and the formula for the surface area of a can is: A 2πrh 2πr2, the d d πd2 3πd2 2b function is: S1d2 π a b d2 π a S1d2 π d 2 S1d2 99 The function for changing yards to inches is: f 1x2 36x and the function for changing miles to yards is: 1760x The composition of this which would change miles into inches is: f g 1x2 g1x2 ƒ g2 1x2 36 17601x2 63,360 x 100 If x width, then length2x A formula for Perimeter can now be written as: P the function is: P1x2 6x This is a linear function x 2x x 2x and Chapter Test (a) D, only values where x can be input into a square root function (b) D, only values where y can be the range of a square root function (c) C, all values can be input for x in a squaring function (d) B, only values where y can be the range of f 1x2 x2 (e) C, all values can be input for x in a cube root function (f) C, all values can be the range of a cube root function (g) C, all values can be input1 for x in an absolute value function (h) D, only values where y (i) D, if x can be the range to an absolute value function y2 then yx and only values where x can be input into a square root function (j) C, all values can be the range in this function (a) This is f 1x2 shifted units upward See Figure 2a (b) This is f 1x2 shifted units to the left See Figure 2b (c) This is f 1x2 reflected across the x-axis See Figure 2c (d) This is f 1x2 reflected across the y-axis See Figure 2d (e) This is f 1x2 vertically stretched by a factor of See Figure 2e (f) We reflect the graph of y f 1x2 across the x-axis for all points for which y Where y 0, the graph remains unchanged See Figure 2f Chapter Test y SECTION 2.T 119 y y (1, 3) (0, 2) (4, 2) (0, 0) (–2, 0) (4, 0) x x x (2, 0) (1, –1) (–1, –3) Figure 2a y Figure 2b Figure 2c y y (0, 0) (–4, 0) (1, 3) x (0, 0) (0, 0) (4, 0) x x (4, 0) (1, –6) (–1, –3) Figure 2d Figure 2e Figure 2f (a) Since y f 12x2 is y f 1x2 horizontally shrunk by a factor of , the point 2, 42 on y f 1x2 becomes the point 1, 42 on the graph of y f 12x2 f a xb is y f 1x2 horizontally stretched by a factor of , the point 2, 42 on y f 1x2 becomes (b) Since y the point 4, 42 on the graph of y f a xb is the basic graph f x2 (a) The graph of f 1x2 1x22 to the right, and shifted units upward See Figure 4a (b) The graph of f 1x2 x is the basic graph f 1x21x stretched by a factor of See Figure 4b x2 reflected across the x-axis, shifted units reflected across the y-axis and vertically y y 4 x Figure 4a –4 Figure 4b x 120 CHAPTER Analysis of Graphs of Functions (a) If the graph is symmetric with respect to the y-axis, then 1x, y2 1 x, y2, therefore 13, 62 1 3, 62 (b) If the graph is symmetric with respect to the x-axis, then 1x, y2 1 x, y2, therefore 13, 62 1 3, 62 (c) See Figure We give an actual screen here The drawing should resemble it y 4, 4 by 0, Xscl Yscl (3, 2) x 13 x to the left units, vertically stretch by a factor of 4, and shift units downward (a) Shift the graph of y Figure Figure x reflected across the x-axis, vertically shrunk by a factor of , shifted units to the right, and shifted up units See Figure From the graph the domain is: q, q2; and the range is: q, (a) From the graph, the function is increasing for the interval: q, 32 (b) From the graph, the function is decreasing for the interval: 14, q2 (c) From the graph, the function is constant for the interval: 3, 4 (d) From the graph, the function is continuous for the intervals: q, 32, 3, 4 , 14, q2 (e) From the graph, the domain is: q, q2 (f) From the graph, the range is: q, 22 (a) 4x 04 4x 4x x or 4x 4x 12 x , therefore the solution set is: 3, 16 From the graph, the x-coordinates of the points of intersection of the graphs of Y1 and Y2 are (b) 4x 4 4x and See Figure 8 12 4x 6 x 1, therefore the solution is: 3, 12 From the graph, See Figure 8, the graphs of Y1 lies below the graph of Y2 for x-values between and (c) 4x 4x 4x x or 4x 4x 12 x , therefore the solution is: q, 32 h 1, q2 From the graph, See Figure 8, the graph of Y1 lies above the graph of Y2 for x-values less than or for x-values greater than 10, 10 by Xscl 10, 10 Figure Yscl (a) f g2 1x2 2x2 (b) a f b 1x22x2 3x g2x 3x 2x Chapter Test 2x2 x 12 1 f g2 1x2 SECTION 2.T (c) The domain can be all values for x, except any that make g1x2 Therefore 2x 1 2x 1 x interval: a q 6x or the b h a , qb (d) f g2 1x2f g1x2 421 2x 1 8x8x 121 12 2 31 2x 12 214x 4x 6x 3x 8x2x (e) 21x h22 31x h2 2x2 4xh 2h2 12x2 3x 22 2x2 3x 21x2 h22 3x 2xh 2x2 2h 3h h h 3x 12 3h h 2 4xh 2h2 h 3h 4x 10 (a) See Figure 10a (b) Graph y1 x2 See Figure 10b 32 * 1x 1x22 * 1x 12 in the window 4.7, 4.7 by 5.1, 5.1 12 y 4.7, 4.7 by Xscl 3 5.1, 5.1 Yscl 0, 10 Xscl 31 by 0, 4 Yscl –2 x Figure 10a Figure 10b Figure 11 11 (a) See Figure 11 (b) Set x 5.5, then $2.75 is the cost of a 5.5-minute call See the display at the bottom of the screen 12 (a) With an initial set-up cost of $3300 and a production cost of $4.50 the function is: C1x2 3300 4.50 x (b) With a selling price of $10.50 the revenue function is: R1x2 (c) P1x2R1x2C1x2 P1x210.50x 13300 4.50x2 P1x2 10.50 x 6x 3300 (d) To make a profit P1x2 0, therefore 6x 3300 6x 3300 x 550 Tyler needs to sell 551before he earns a profit (e) Graph y1 6x 3300, See Figure 12 The first integer x-value for which P1x2 is 551 Yscl 500 30, 1000 by 4000, 4000 Xscl 50 Figure 12 122 CHAPTER Analysis of Graphs of Functions Chapter Project Since the front is moving at 40 mph for hr the front moved 160 miles with each unit representing 100 miles This is a shift of 1.6 units south or downward The function f 1x2 1 x2 y x2 shifted 1.6 units downward is: 20 1.6 (a) Because the front has moved 250 south and 210 miles east, graph the shifted equation: y 1x 2.122 2.5 Plot the point 15.5, 82 for Columbus, Ohio Here we see that the front has 20 reached the city See Figure 2a (b) Because the front has moved 250 south and 210 miles east, graph the shifted equation: y 1x 2.122 2.5 Plot the point 11.9, 4.32 for Memphis, Tennessee Here we see that the front has 20 not reached the city See Figure 2b (c) Because the front has moved 250 south and 210 miles east, graph the shifted equation: 1 y x 2.122 2.5 Plot the point 14.2, 2.32 for Louisville, Kentucky Although the graph is 20 difficult to read, repeated zooming will show that the front has not reached the city See Figure 2c 10, 10 by Xscl 10, 10 Yscl Figure 2a 10, 10 by Xscl 10, 10 Yscl Figure 2b 10, 10 by Xscl 10, 10 Yscl Figure 2c ... coordinate 14, 12, therefore the solution set is: 546 (b) From the graph, y1 y2 never, therefore the solution set is: (c) From the graph, y1 y2 for all values for x, except 4, therefore for the intervals:... y1 y2 never, therefore the solution set is: (b) From the graph, y1 y2 for all values for x, therefore for the interval: q, q2 (c) From the graph, y1 y2 never, therefore the solution set is: ... Therefore, the solution is: a 54 3x1 the solution is: 18, 222 2x or 31x 52 19 53.4.2 x 3.1 4.2 x 02.1 x 5 x or x1 x or x Therefore, the solution set is: 50, 16 56 74x or 2x12 Therefore, The solution