Solution manual for calculus 4th edition by smith

Solution Manual for Calculus 4th Edition by Smith ≤ x + < 3≤x x > −1 −1− − 2 2− − ,∞ − − 2x < − 2x < x > −2 (−2, ∞) + | −3 − + ✲ (x + 3) | −1 + | −1 + ✲ (x + 1) ✲ (x + 3)(x + 1) Solution Manual for Calculus 4th Edition by Smith CHAPTER PRELIMINARIES −3 < x < −1 (−3, −1) 20 |3 + x| > + x < −1 or + x > x < −4 or x > −2 13 x2 − x − < (x − 3)(x + 2) < − − | −2 + 0+ ✲ (x − 3) | + | −2 − (−∞, −4) ∪ (−2, ∞) 21 |2x + 1| > 2x + < or 2x + > 2x < −3 2x > x 2 −∞, − ∪ ,∞ 2 ✲ (x + 2) 0+ ✲ (x − 3)(x + 2) | 22 |3x − 1| < −4 < 3x − < (−2, 3) −3 < 3x < 5 −1 < x < −1, 14 x2 + > x2 > −1 This is always true (−∞, ∞) 15 3x2 + > 3x2 > −4 x2 > − This is always true (−∞, ∞) 23 16 x2 + 3x + 10 > x2 + 3x + 10 = x= −3 ± x+2 >0 x−2 − | −2 √ − 4(1)(10) −3 ± −31 = 2(1) + − The value of x2 + 3x + 10 is never zero, and it is always positive (−∞, ∞) + 17 |x − 3| < | −2 − ✲ (x + 2) | + x | + ✲ (x − 2) ✲ x+2 x−2 −4 < x − < −1 < x < x < −2 or x > (−1, 7) (−∞, −2) ∪ (2, ∞) 18 |2x + 1| < −1 < 2x + < −2 < 2x < −1 < x < (−1, 0) 19 |3 − x| < −1 < − x < − < −x < −2 4>x>2 2 (−∞, −4) ∪ (−4, −1) ∪ (2, ∞) − 2x 26 − ✲ (−8x) 35 d{(1, 1), (3, 4)} = √ (3 − 1)2√ + (4 − 1)2 = + = 13 d{(1, 1), (0, 6)} = √ (0 − 1)2 + √(6 − 1) = + 25 = 26 d{(3, 4), (0, 6)} = √ (0 − 3)2√ + (6 − 4)2 + = 13 √ √ = 9√ ( 13)2 + ( 13)2 = ( 26)2 This statement is true, so yes, this is a right triangle + ✲ (x + 1)3 − ✲ −8x (x + 1)3 36 d{(0, 2), (4, 8)} = √ (4 − 0)2 +√ (8 − 2)2 = 16 + 36 = 52 d{(0, 2), (−2, 12)} = √ (−2 − 0)2 √ + (12 − 2)2 = + 100 = 104 d{(4, 8), (−2, 12)} = √ (−2 − 4)2 √ + (12 − 8)2 = 36 + 16 = 52 Solution Manual for Calculus 4th Edition by Smith CHAPTER PRELIMINARIES √ √ √ ( 52)2 + ( 52)2 = ( 104)2 This statement is true, so yes, this is a right triangle (−2)]2 The y-values are increasing by 90, 110, 130, The next population is 3490 + 150 = 3640 41 y 3)2 37 d{(−2, 3), (2, 9)} = √ [2 − √+ (9 − = 16 + 36 = 52 d{(−2, 3), (−4, 13)} 2 = √ [−4 − (−2)] √ + (13 − 3) = + 100 = 104 d{(2, 9), (−4, 13)} = √ (−4 − 2)2 √ + (13 − 9)2 = 36 + 16 = 52 √ √ √ ( 52) + ( 52) = ( 104)2 (0,4000) 4,000 (1,3990) 3,975 (2,3960) 3,950 3,925 (3,3910) 3,900 This statement is true, so yes, this is a right triangle 2 38 d{(−2, 3), (0, 6)} = √ [0 − (−2)] √ + (6 − 3) = + = 13 2 d{(−2, 3), (−3, 8)} = √ [−3 − (−2)] √ + (8 − 3) = + 25 = 26 d{(0, 6), (−3, 8)} = √ (−3 − 0)√2 + (8 − 6)2 = √ + = 13 √ √ ( 13)2 + ( 13)2 = ( 26)2 x The y-values are decreasing by 10, 30, 50, The next population is 3910 − 70 = 3840 42 102 y 22 (1,2200) 21 (0,2100) (2,2100) 20 This statement is true, so yes, this is a right triangle 19 18 39 y 17 3,200 (3,1700) (3,3200) 16 2,800 x (2,2450) 2,400 Possible answer: 2,000 (1,1800) The increments in y are +100, −100, −400 A reasonable next increment would be −900 for a quadratic pattern, giving a next population of 1700 − 900 = 800 1,600 (0,1250) 1,200 800 400 x The y-values are increasing by 550, 650, 750, The next population is 3200 + 850 = 4050 40 102 44 When data points “curve up,” the vertical distance gets larger for the same horizontal distance This corresponds to larger increases in consecutive y-values y 35 43 When a calculation is done using a finite number of digits, the result can only contain a finite number of digits Therefore, the result can be written as a quotient of two integers (where the denominator is a power of 10), so the result must be rational (3,3490) 34 (2,3360) 33 45 (1,3250) 32 (0,3160) 31 x − 2 12 ≈ 0.013 or 1.3% 46 If x is the ratio of consecutive string lengths, then x12 is the ratio of string lengths spaced 12 strings (or octave) apart Therefore, x12 = Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.2 A BRIEF PREVIEW OF CALCULUS It would √ be difficult to measure lengths with a ratio of 12 because irrational numbers can be difficult to locate on a number line 47 P 0.551 0.587 0.404 0.538 0.605 0.2 Win % 0.568 0.593 0.414 0.556 0.615 A Brief Preview of Calculus -3 -2 -1 3 x -2 -4 12 y = −2(x − 1) + = −2x + Yes The slope of the line joining the points (2, 1) and (0, 2) is − , which is also the slope of the line joining the points (0, 2) and (4, 0) y No The slope of the line joining the points (3, 1) and (4, 4) is 3, while the slope of the line joining the points (4, 4) and (5, 8) is -2 No The slope of the line joining the points (4, 1) and (3, 2) is −1, while the slope of the line joining the points (3, 2) and (1, 3) is − No The slope of the line joining the points (1,2) and (2,5) is 3, but the slope of the line joining the points (2, 5) and (4, 8) is -1 Slope is Slope is Slope is 3−2 = 3−1 13 y = (0, 1) is a second point on the line 1.5 y 0.5 -3 -2 -1 14 y = 3 1 (x − 2) + = x 2 10 Slope is 11 y = 2(x − 1) + = 2x + For x = 2, y = 2(2) + = (2, 5) is a second point on the line -1 −2 − (−3) = − (−1) 2.4 − 2.1 = ≈ 0.158 3.1 − 1.2 1.9 In exercises 11-16, the equation of the line is given along with the graph Any point on the given line will suffice for a second point on the line x -0.5 −1 − (−6) 5 = =− 1−3 −2 −0.4 − (−1.4) 1.0 = =− −1.1 − 0.3 −1.4 x 6−2 Slope is = = 3−1 Slope is 1 y -3 -2 -1 x -1 -2 -3 Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith CHAPTER PRELIMINARIES 15 y = 1.2(x − 2.3) + 1.1 = 1.2x − 1.66 For x = 3.3, y = 1.2(3.3 − 2.3) + 1.1 = 2.3 (3.3, 2.3) is a second point on the line 26 (a) y = −1 (b) x = 3−1 = = through the given points 2−1 One possibility: y = 2(x − 1) + = 2x − 27 Slope When x = 4, y = 28 Slope −2 through the given points One possibility: y = −2(x − 3) + y -3 -2 -1 x -1 When x = 4, y = -2 29 Yes, passes vertical line test -3 30 Yes, passes vertical line test 31 No The vertical line x = meets the curve twice; nearby vertical lines meet it three times 1 16 y = − (x + 2) + = x + 4 32 No, does not pass vertical line test 33 Both: This is clearly a cubic polynomial, and also a rational function because it can be writx3 − 4x + ten as f (x) = (This shows that all polynomials are rational.) y -3 -2 -1 x -1 34 Both 35 Rational; both the numerator and the denominator are polynomials -2 -3 36 Rational 17 Parallel Both have slope 37 Neither: Contains square root 18 Neither Slopes are and 38 Neither: Contains exponent 19 Perpendicular product is -1 Slopes are −2 and ;their The domain is {x ∈ R|x ≥ −2} = [−2, ∞) 20 Neither Slopes are and −2 21 Perpendicular product is -1 39 We need the function under the square root to be non-negative x + ≥ when x ≥ −2 Slopes are and − ;their 22 Parallel Both have slope − 23 (a) y = 2(x − 2) + 1 (b) y = − (x − 2) + 24 (a) y = 3x + (b) y = − x + 3 25 (a) y = 2(x − 3) + 1 (b) y = − (x − 3) + 40 The function is defined only if x2 − x − ≥ and x = (x − 3) (x + 2) ≥ and x = (x − ≤ and x + ≤ 0) or(x − ≥ and x + ≥ 0)and x = x ≤ −2 or x ≥ and x = (−∞, −2] ∪ [3, 5) ∪ (5, ∞) 41 The function under square root should be nonnegative and the denominator should be nonzero 3 − 2x > when x < The domain of f is −∞, √ 42 The function √ is defined for all vales of + x but, + x is well defined for x ≥ The domain of f is [0, ∞) Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.2 A BRIEF PREVIEW OF CALCULUS 43 Negatives are permitted inside the cube root There are no restrictions, so the domain is (−∞, ∞) or all real numbers 44 We need the numerator function under square root be non-negative x2 − ≥ 2, when |x| ≥ Also the denominator cannot be zero − x2 > 0, when |x| < The domain is (−3, −2] [2, 3) 45 The denominator cannot be zero x2 − = when x = ±1 The domain is {x ∈ R|x = ±1} = (−∞, −1) ∪ (−1, 1) ∪ (1, ∞) 46 The denominator cannot be zero.√ x2 + 2x − = when x = −1 ± The domain is √ √ {x ∈ R|x √ = −1 ±√ 7} = (−∞,√−1 − 7) ∪ (−1 − 7, −1 + 7) ∪ (−1 + 7, ∞) 47 f (0) = 02 − − = −1 f (2) = 22 − − = f (−3) = (−3)2 − (−3) − = 11 1 f = − −1=− 2 0+1 = −1 0−1 2+1 =3 f (2) = 2−1 −2 + 1 f (−2) = = −2 − +1 f = = −3 −1 √ 49 f (0) = √0 + = f (3) = √ 3+1=2 f (−1) = −1 + = √ 1 f = +1= = 2 2 48 f (0) = =3 f (10) = = 0.3 10 f (100) = = 0.03 100 f = =9 3 50 f (1) = 51 The only constraint we know is that the width should not be negative, so a reasonable domain would be {x|x > 0} 52 Width can be anywhere from to 200 feet A reasonable domain is {x|0 ≤ x ≤ 200} 53 We know that x should not be negative, that x must be an integer and that x cannot exceed the number of candy bars made, i.e., a reasonable domain would be {x|0 ≤ x ≤ number made, xan integer} 54 Cost could be any positive value A reasonable domain is {x|x > 0} 55 Answers vary There may well be a positive correlation (more study hours = better grade), but not necessarily a functional relation.You may study the same amount of time for two different exams and get different grades 56 Answers vary Evidence supports a relationship 57 Answers vary While not denying a negative correlation (more exercise = less weight), there are too many other factors (metabolic rate, diet) to be able to quantify a person’s weight as a function just of the amount of exercise.Two people may exercise the same amount of time but have different weights 58 Answers vary Objects of all weights fall at the same speed unless friction affects them differently 59 A flat interval corresponds to an interval of constant speed; going up means that the speed is increasing while the graph going down means that the speed is decreasing It is likely that the bicyclist is going uphill when the graph is going down and going downhill when the graph is going up 60 Influxes of immigrants occur where graph rises War and plague occur where graph falls 61 The x-intercept occurs where = x2 − 2x − = (x − 4)(x + 2), so x = or x = −2; y-intercept at y = 02 − 2(0) − = −8 62 The x-intercept occurs where = x2 + 4x + = (x + 2)2 , so x = −2; y-intercept at y = 02 + 4(0) + = 63 The x-intercept occurs where = x3 − = (x − 2)(x2 + 2x + 4), so x = (using the quadratic formula on the quadratic factor gives the solutions Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith CHAPTER PRELIMINARIES √ x = −1 ± −3, neither of which is real so neither contributes a solution); y-intercept at y = 03 − = −8 64 The x-intercept occurs where = x3 − 3x2 + 3x − = (x − 1)3 , so x = 1; y-intercept at y = 03 − 3(0)2 + 3(0) − = −1 65 The x-intercept occurs where the numerator is zero, at = x2 − = (x − 2)(x + 2), so x = ±2; y-intercept at 02 − = −4 y= 0+1 66 The x-intercept occurs where the numerator is zero, at x = ; y-intercept at 2(0) − 1 y= = (0) − 4 67 x2 − 4x + = (x − 3)(x − 1), so the zeros are x = and x = 68 x2 + x − 12 = (x + 4)(x − 3), so the zeros are x = −4 and x = 69 Quadratic formula gives √ ± 16 − 8) x= =2± 2 70 Quadratic formula gives x= −4 ± 42 − 4(2)(−1) −2 ± = 2(2) √ 71 x3 − 3x2 + 2x = x(x2 − 3x + 2) = x(x − 2)(x − 1), so the zeros are x = 0, 1, and 72 x3 − 2x2 − x + = (x − 2)(x − 1)(x + 1), so the zeros are x = −1, 1, and 73 With t = x3 , x6 + x3 − becomes t2 + t − and factors as (t + 2)(t − 1) The expression is zero only if one of the factors is zero, i.e., if t = or t = −2 With x = t1/3 , the first occurs only if x = (1)1/3 = The latter occurs only if x = (−2)1/3 , about −1.2599 74 x3 + x2 − 4x − = (x − 2)(x + 1)(x + 2), so the zeros are x = −2, −1, and 75 If B(h) = −1.8h + 212, then we can solve B(h) = 98.6 for h as follows: 98.6 = −1.8h + 212 1.8h = 113.4 113.4 = 63 1.8 This altitude (63,000 feet above sea-level, more than double the height of Mt Everest) would be the elevation at which we humans boil alive in our skins Of course the cold of space and the near-total lack of external pressure create additional complications which we shall not try to analyze h= 76 Let x represent compression and L(x) represent spin rate Given the points (120, 9100) and (60, 10,000), the linear function is y = −15(x − 60) + 10, 000 The spin rate of a 90-compression ball is 9550, and the spin rate of a 100-compression ball is 9400 77 This is a two-point line-fitting problem If a point is interpreted as (R, T ) =(chirp rate, temperature), then the two given points are (160, 79) and (100, 64) 15 79 − 64 = = , The slope being 160 − 100 60 we could write T = 14 (R − 100) + 64 or T = 14 R + 39 78 From problem 77 we know the temperature is a function of chirping rate, T (r) = r + 39, where r is measured in chirps per minute The number of chirps in 15 seconds will then be r, and the temperature may conveniently be found by adding 39 79 Her winning percentage is calculated by the 100w formula P = , where P is the winning t percentage, w is the number of games won and t is the total number of games Plugging in w = 415 and t = 415 + 120 = 535, we find her winning percentage is approximately P ≈ 77.57, so we see that the percentage displayed is rounded up from the actual percentage Let x be the number of games won in a row If she doesn’t lose any games, her new winning percentage will be given by the for100(415 + x) mula P = In order to have 535 + x her winning percentage displayed as 80%, she only needs a winning percentage of 79.5 or greater Thus, we must solve the inequality 100(415 + x) 79.5 ≤ : 535 + x Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 100(415 + x) 535 + x 79.5(535 + x) ≤ 41500 + 100x 42532.5 + 79.5x ≤ 41500 + 100x 1032.5 ≤ 20.5x 50.4 ≤ x 79.5 ≤ −5 −4 −3 (In the above, we are allowed to multiply both sides of the inequality by 535 + x because we assume x (the number of wins in a row) is positive.) Thus she must win at least 50.4 times in a row to get her winning percentage to display as 80% Since she can’t win a fraction of a game, she must win at least 51 games in a row −2 −1 −1 x −2 y −3 −4 −5 (b) Intercepts: x ≈ 0.566, 19.434, y = 11 Maximum at (10, 111) No asymptotes 0.3 Graphing Calculators and Computer Algebra Systems 150 100 50 (a) Intercepts: x = ±1, y = −1 Minimum occur at (0, −1) No asymptotes 0 −5 10 15 20 25 x 2.0 −50 1.6 −100 1.2 0.8 0.4 0.0 −2 −1 −0.4 x −0.8 y −1.2 (a) Intercepts: x = −1, y = No extrema or asymptotes −1.6 −2.0 20 16 (b) Intercepts: y = (No x-intercepts) Minimum at (−1, 7) No asymptotes 12 18 16 14 −5 −4 −3 −2 −1 −4 x 12 −8 y y 10 −12 −16 −20 −10 −8 −6 −4 −2 10 x √ (a) Intercepts: x = = ±1.73, y = Maximum at (0, 3) No asymptotes (b) Intercepts: x ≈ −4.066, −0.72 and 4.788, y = −14 Local minimum: Approximately at (2.58, −48.427) Local maximum: Approximately at (−2.58, 20.4225) No asymptotes Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 10 CHAPTER PRELIMINARIES 2.0 50 1.6 1.2 25 0.8 0.4 −5.0 0.0 −2.5 2.5 0.0 5.0 −2 −1 −0.4 x −0.8 −25 y −1.2 −1.6 −50 −2.0 √ (a) Intercepts: x = 10 ≈ 2.1544, y = 10 No extrema or asymptotes (b) Intercepts: x ≈ 0.475, −1.395, y = √−1 Minimum at (approximately) (−1/ 2, −2.191) No asymptotes 15 10 x −2 −1 −1 y −2 −2 −1 −3 −5 (b) Intercepts: x ≈ 0.0334, −5.494 and 5.46, y = −1 Local minimum: Approximately at (−3.16, −64.24) Local maximum: Approximately at (3.16, 62.245) No asymptotes √ (a) Intercepts: x = ± 2, y = Maximum at (0, 2) No asymptotes 50 25 −5.0 −2.5 0.0 2.5 5.0 −3 −2 −1 x −25 −1 y −50 −2 −3 (a) Intercepts: x = ±1, y = −1 Minimum at (0, −1) No asymptotes (b) Intercepts: x ≈ ±2.33, and ±0.74, y = 3.√Local maximum at (0, 3) Minima at (± 3, −6) No asymptotes Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 33 Graph of y = (x2 − 1) − (2x + 1)3 : x ≈ ±1.177 by calculator or spreadsheet 36 Graph of y = sin x − (x2 + 1) : x −5.0 0.0 −2.5 2.5 17 7.5 5.0 10.0 x −2 −1 −250 −1 −500 y −2 −750 y −3 −1,000 −4 After zooming out, the graph shows that there are two solutions: one near zero, and one around ten Algebraically, 2/3 = 2x + x2 − 2 Graph shows no intersections 37 Calculator shows zeros at approximately ⇒ x − = (2x + 1) ⇒ x4 − 2x2 + = 8x3 + 12x2 + 6x + ⇒ x4 − 8x3 − 14x2 − 6x = ⇒ x x3 − 8x2 − 14x − = We thus confirm the obvious solution x = 0, and by solver or spreadsheet, find the second solution x ≈ 9.534 34 Graph of y = (x + 1) − (2 − x)3 : −1.879, 0.347 and 1.532 38 Calculator shows zeros at approximately 3.87, 0.79 and −0.66 39 Calculator shows zeros at approximately 5637 and 3.0715 40 Calculator shows zeros at approximately and 0.54 41 Calculator shows zeros at approximately −5.248 and 10.006 10 42 Calculator shows zeros at approximately 2.02, − 0.26, − 1.10 and −2.04 −2 −1 −2 x −4 y −6 −8 −10 Graph shows one solution at approximately x = 0.62 35 Graph of y = cos x − (x2 − 1) : 43 The graph of y = x2 on the window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10 appears identical (except for labels) to the graph of y = 2(x − 1) + if the latter is drawn on a graphing window centered at the point (1, 3) with √ √ − ≤ x ≤ + 2, −7 ≤ y ≤ 13 44 The graph of y = x4 is below the graph of y = x2 when −1 ≤ x ≤ 1, and above it when x > Both graphs have roughly the same upward parabola shape, but y = x4 is flatter at the bottom 45 y is the distance from (x, y) to the x-axis −2 −1 −1 x −2 y −3 −4 −5 The graph shows that there are two solutions: x2 + (y − 2) is the distance from (x, y) to the point (0, 2) If we require that these be the same, and we square both quantities, we have y = x2 + (y − 2) y = x2 + y − 4y + 4y = x2 + y = x2 + Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 18 CHAPTER PRELIMINARIES In this relation, we see that y is a quadratic function of x The graph is commonly known as a parabola 46 The distance between (x, y) and the x-axis is y The distance between (x, y)and (1, 4) is 2 (x − 1) + (y − 4) Setting these equal and squaring both sides yields 2 y = (x − 1) + (y − 4) which simplifies to y = (x − 1) + 16 (a parabola) 0.4 Trigonometric Functions (a) π 180◦ π = 45◦ (b) π 180◦ π = 60◦ (c) π ◦ (d) 4π 180◦ π = 240◦ (a) 3π 180◦ π = 108◦ (b) 180 π = 30◦ 180◦ π π ≈ 25.71◦ (c) 180◦ π ≈ 114.59 (d) 180◦ π ≈ 171.89◦ ◦ π =π 180◦ π 3π (b) (270◦ ) = 180◦ π 2π (c) (120◦ ) = 180◦ π π ◦ (d) (30 ) = 180◦ (a) (180◦ ) π 2π = 180◦ 4π π (b) 80◦ = 180◦ π 5π (c) 450◦ = 180◦ π 13π (d) 390◦ = 180◦ (a) 40◦ cos (x) − = when cos (x) = 1/2 This ocπ π curs whenever x = + 2kπ or x = − + 2kπ 3 for any integer k sin x + = when sin x = − This occurs π 5π whenever x = − + 2kπ or x = − + 2kπ for 6 any integer k √ cos (x) − = when cos (x) = 1/ This π π occurs whenever x = +2kπ or x = − +2kπ 4 for any integer k √ √ √ This occurs sin x − = when sin x = 2π π + 2kπ or x = + 2kπ for whenever x = 3 any integer k sin2 x−4 sin x+3 = (sin x − 1) (sin x − 3) when sin x = (sin x = for any x) This occurs π whenever x = + 2kπ for any integer k 10 sin2 x−2 sin x−3 = (sin x − 3) (sin x + 1) when sin x = −1 (sin x = for any x) sin x = −1 3π whenever x = + 2kπ for any integer k 11 sin2 x + cos x − = − cos2 x + cos x − = (cos x) (cos x − 1) = when cos x = or cos x = This occurs π + kπ or x = 2kπ for any whenever x = integer k 12 Use the sine double angle formula to get sin x cos x − cos x = (2 sin x − 1) cos x = π then (2 sin x − 1) = whenever x = + 2kπ 5π or x = + 2kπ and cos x = whenever π x = + kπ for any integer k 13 cos2 x + cos x = (cos x) (cos x + 1) = when cos x = or cos x = −1 this occurs whenever π x = + kπ or x = π + 2kπ for any integer k 14 sin2 x − sin x = sin x (sin x − 1) = whenever π x = kπ or x = + 2kπ for any integer k 15 The graph of f (x) = sin 3x Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.4 TRIGONOMETRIC FUNCTIONS 19 1.0 0.8 0.6 y 0.4 0.2 0.0 −5 −6 −4 −3 −1 −0.2 −2 x −5.0 −2.5 0.0 2.5 5.0 x −0.4 y −2 −0.6 −0.8 −1.0 −4 16 The graph of f (x) = cos 3x 20 The graph of f (x) = cos (x + π) 4.0 3.2 2.4 1.6 0.5 0.8 x -3 -2 -1 0.0 −6 −5 −4 −3 −2 −1 −0.8 −1.6 -0.5 −2.4 −3.2 -1 −4.0 21 The graph of f (x) = sin 2x − cos 2x 17 The graph of f (x) = tan 2x 10 0 −3 −2 −1 −2 x −2 −4 x −1 −4 y y −6 −2 −8 −3 −10 22 The graph of f (x) = cos 3x − sin 3x 18 The graph of f (x) = sec 3x y 0.5 -3 -2 -1 0 -3 -2 -1 x -2 x -0.5 -1 -4 19 The graph of f (x) = cos (x − π/2) 23 The graph of f (x) = sin x sin 12x Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 20 CHAPTER PRELIMINARIES 2π 32 Amplitude is 2, period is , 3 frequency is 2π 1.5 0.5 y -4 33 sin(α − β) = sin (α + (−β)) = sin α cos (−β) + sin (−β) cos α -2 x -0.5 = sin α cos β − sin β cos α 34 cos(α − β) = cos (α + (−β)) -1 = cos α cos (−β) − sin α sin (−β) -1.5 = cos α cos β + sin α sin β 24 The graph of f (x) = sin x cos 12x 35 (a) cos (2θ) = cos(θ + θ) 1.0 = cos (θ) cos (θ) − sin (θ) sin (θ) 0.8 = cos2 θ − sin2 θ 0.6 0.4 = cos2 θ − − cos2 θ 0.2 = 2cos2 θ − 0.0 −3 −2 −1 −0.2 −0.4 −0.6 −0.8 −1.0 2π 25 Amplitude is 3, period is = π, frequency is π 2π 26 Amplitude is 2, period is , 3 frequency is 2π 2π 27 Amplitude is 5, period is , 3 frequency is 2π 2π 28 Amplitude is 3, period is , 5 frequency is 2π 2π = π, frequency is 29 Amplitude is 3, period is We are completely ignoring the presence of π −π/2 This has an influence on the so-called “phase shift”which will be studied in Chapter 2π 30 Amplitude is 4, period is , 3 frequency is 2π 31 Amplitude is (the graph oscillates between −4 and 4, so we may ignore the minus sign), period is 2π, frequency is 2π (b) Just continue on, writing cos (2θ) = 2cos2 θ − = − sin2 θ − = − 2sin2 θ 36 (a) Divide sin2 θ + cos2 θ = by cos2 θ to get sin2 θ +1= or tan2 θ + = sec2 θ cos2 θ cos2 θ (b) Dividing sin2 θ + cos2 θ = by sin2 θ yields cot2 θ + = csc2 θ 37 Use the formula cos(x+β) = cos x cos β−sin β sin x Now we see that cos β must equal 4/5 and sin β must equal 2 3/5 Since (4/5) + (3/5) = 1, this is possible We see that β = sin−1 (3/5) ≈ 0.6435 radians, or 36.87◦ 38 Use the formula sin(x + β) = sin x cos β + sin √β cos x Now we see that cos β must equal 2/ and sin β must √ √ √ equal 1/ Since (2/ 5) +(1/ 5) = 1, this is possible We see that √ β = sin−1 2/ ≈ 0.4636 radians, or 26.57◦ 2π = π and sin(πx) has 39 cos(2x) has period 2π period = There are no common inteπ ger multiples of the periods, so the function f (x) = cos (2x) + sin (πx) is not periodic √ 40 sin √ x has period 2π and cos 2x has period 2π There are no common integer multiples of the periods, so the function √ f (x) = sin x − cos 2x is not periodic Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.4 TRIGONOMETRIC FUNCTIONS 21 2π 41 sin(2x) has period = π and cos (5x) has pe2 2π riod The smallest integer multiple of both of these is the fundamental period, and it is 2π 2π 2π 42 cos 3x has period and sin 7x has period The smallest integer multiple of both of these is the fundamental period, and it is 2π 43 cos2 θ = − sin2 θ = − =1− = 9 Because θ s in the first quadrant, its cosine is nonnegative Hence √ 2 = = 0.9428 cos θ = 44 First quadrant, 3-4-5 right triangle, so sin θ = √ 45 Second quadrant, 1- 3-2 right triangle, so √ cos θ = − √ 46 Second quadrant, 1- 3-2 right triangle, so tan θ = − √ 47 From graph the three solutions are −1.868, −0.538, and 2.585 48 From graph the three solutions are and ±2.28 49 From graph the two solutions are ±1.455 50 From graph the two solutions are and 0.88 51 Let h be the height of the rocket Then h = tan 20◦ ⇒ h = tan 20◦ ≈ 0.73(miles) 54 If the steeple is 20 inside the building, the height is 100 tan 50◦ ≈ 119.18 feet If the steeple is 21 inside the building, the height is 101 tan 50◦ ≈ 120.37 feet The difference is 1.19 feet 55 The period is π/30 (seconds) So, apparently, the frequency is f = 30/π (cycles per second) 170 and the meter voltage is √ ≈ 120.2 56 Revolutions per minute measures frequency The period is the reciprocal The period of minutes per revolua 33 rpm record is 100 tion Similarly, the period of a 45 rpm record is minutes per revolution 45 57 There seems to be a certain slowly increasing base for sales (110 + 2t), and given that the 2π sine function has period = 12 months, the π/6 sine term apparently represents some sort of seasonally cyclic pattern If we assume that travel peaks at Thanksgiving, the effect is that time zero would correspond to a time one quarterperiod (3 months) prior to Thanksgiving, or very late August The annual increase for the year beginning at time t is given by s (t + 12) − s (t) and automatically ignores both the seasonal factor and the basic 110, and indeed it is the constant × 12 = 24 (in thousands of dollars per year and independent of the reference point t) 58 The graph of sin 8t + sin 8t looks like 52 The person and the shadow form a right triangle similar to the triangle formed by the lightpole and the distance from the base of the pole to the tip of the shadow If x represents the height of the pole, we have that x = and therefore x = 18 4+2 2 -60 -30 30 60 -1 -2 53 Let h be the height of the steeple Then h = tan 50◦ 80 + 20 ⇒ h = 100 tan 50◦ ≈ 119.2 (feet) -3 The graph of sin 8t + sin 8.1t looks like Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 22 CHAPTER PRELIMINARIES √ f (g(x)) = x + − with domain {x|x ≥ −1} g(f (x)) = (x − 2) + = with domain {x|x ≥ 1} -60 -30 30 60 -1 -2 x−1 (f ◦ g)(x) = f (g(x)) = f (x3 + 4) = x +4 √ with domain {x|x = − 4} 1 (g ◦ f )(x) = g(f (x)) = g = +4 x x -3 with domain {x|x = 0} The difference in frequency produces clearly audible beats (to the trained ear) √ (f ◦ g)(x) = f (g(x)) = f (tan x) = − tan x For the domain, − tan x ≥ or tan x ≤ This happens on the intervals π π − + kπ, + kπ where k is the integer √ √ (g◦f )(x) = g(f (x)) = g( − x) = tan − x √ π For the domain − x = + kπ where k is the integer and x ≤ The domain is π {x|x ≤ and x = − + kπ where k is the integer } 59 (f ◦ g)(x) = f (sin x) = sin2 x + with domain (−∞, ∞) or all real numbers (g ◦ f )(x) = g(x2 + 1) = sin(x2 + 1) with domain (−∞, ∞) or all real numbers 60 − 2)2 − 1 = x − 4x2 + = (x − 3)(x2 − 1) √ This is valid if x = ± and x = ±1 g(f (x)) = − x2 − This is valid if x = ±1 √ x4 + = f (g(x)) when f (x) = x and f (g(x)) = 0.8 0.6 y 0.4 0.2 –1 –0.8 –0.6 –0.4 –0.2 0.2 –0.2 0.4 0.6 0.8 x –0.4 –0.6 –0.8 –1 When zoomed in sufficiently, the graph appears to go to zero near x = 0.5 √ Transformations of Functions g(x) = x4 + 1, for example √ √ x + = f (g(x)) when f (x) = x and g(x) = x + 3, for example √ (f ◦ g)(x) = f (g(x)) = g(x) + = x − + with domain {x|x ≥ 3} (g ◦ f )(x) = g(f (x)) = f (x) − √ = (x + 1) − = x − with domain {x|x ≥ 2} (x2 10 = f (g(x)) when f (x) = 1/x and +1 g(x) = x2 + 1, for example x2 + = f (g(x)) when f (x) = x + and x2 g(x) = 1/x2 , for example Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.5 TRANSFORMATIONS OF FUNCTIONS 23 11 (4x + 1)2 + = f (g(x)) when f (x) = x2 + and g(x) = 4x + 1, for example 10 12 4(x + 1)2 + = f (g(x)) when f (x) = 4x2 + and g(x) = x + 1, for example y 3 13 sin x = f (g(x)) when f (x) = x and g(x) = sin x, for example -4 -2 14 sin(x3 ) = f (g(x)) when f (x) = sin x and x -2 -4 g(x) = x3 , for example = f (g(h(x)) when f (x) = 3/x, sin x + √2 g(x) = x, and h(x) = sin x + 2, 15 √ for example 16 x4 + = f (g(h(x)) when f (x) = √ 23 Graph of f (x − 3):The graph of f is translated to the right by units x, g(x) = x + 1, and h(x) = x4 , for example y 3 17 cos (4x − 2) = f (g(h(x))) when f (x) = x , g(x) = cos x, and h(x) = 4x − 2, for example 18 tan x2 +√1 = f (g(h(x))) when f (x) = tan x, g(x) = x, and h(x) = x2 + 1, for example −5 −4 −3 −2 −1 x −2 19 cos(x ) − = f (g(h(x))) when f (x) = 4x − 5, g(x) = cos x, and h(x) = x2 , for example 20 [tan−1 (3x + 1)]2 = f (g(h(x))) when f (x) = x2 , g(x) = tan−1 x, and h(x) = 3x + 1, for example 24 Graph of f (x) + 2: 10 21 Graph of f (x) − 3:The graph of f is translated down by units y 5.0 y 2.5 -4 -2 0.0 −5 −4 −3 −2 −1 -2 4 x x -4 −2.5 −5.0 22 Graph of f (x + 2): 25 Graph of f (2x): Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 24 CHAPTER PRELIMINARIES 30 y 20 −2 10 −1 x −1 y -4 −2 -2 x -10 −3 26 Graph of 3f (x): 30 29 Graph of f (x − 4): 10 y 20 10 y -2 0 -4 2 10 x x -5 -10 -10 27 The graph of y = −3f (x) + can be obtained by reflecting the graph of y = f (x) in the horizontal axis then vertically stretching it 3times and finally shifting it up by units 30 Graph of f (x + 3): 10 10.0 y 7.5 y 5.0 -8 -6 -4 -2 x 2.5 -5 0.0 −5 −4 −3 −2 −1 x -10 −2.5 −5.0 28 Graph of 3f (x + 2): 31 Graph of f (2x): The horizontal scale is divided by Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 0.5 TRANSFORMATIONS OF FUNCTIONS 25 10 10 y 0 -4 −5 −4 −3 −2 −1 -2 x x -5 −5 y -10 −10 35 Graph of 2f (x) − 4: The vertical scale is multiplied by and then the graph is shifted down by units 10 32 Graph of f (2x − 4): 10 y 5 −5 −4 -2 −2 −1 x -4 −3 y x −5 -5 −10 -10 36 Graph of 3f (x) + 3: 10 y 33 Graph of f (3x + 3): The graph of f is shifted units to the left and then the horizontal axis is divided by -4 -2 x -5 10 -10 -4 -2 y x -5 -10 34 Graph of 3f (x): 37 f (x) = x2 + 2x + = (x + 1)2 Shift y = x2 to the left unit 38 f (x) = x2 − 4x + = (x − 2)2 This is the graph of x2 shifted to the right 39 f (x) = x2 + 2x + = (x2 + 2x + 1) + − = (x + 1)2 + Shift y = x2 to the left unit and up units 40 f (x) = x2 − 4x + = x2 − 4x + − = (x − 2)2 − This is the graph of x2 shifted to the right and down Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 26 CHAPTER PRELIMINARIES 41 f (x) = 2x2 + 4x + = 2(x2 + 2x + 1) + − = 2(x + 1)2 + Shift y = x2 to the left unit, then multiply the scale on the y-axis by 2, then shift up units 42 f (x) = 3x2 − 6x + = 3(x − 1)2 − This is the graph of x2 with the y-scale multiplied by 3, shifted to the right and down 49 y = (x − 1) − = x2 − 2x 10.0 7.5 y 5.0 2.5 0.0 −5.0 −2.5 0.0 2.5 5.0 7.5 x −2.5 43 Graph is reflected across the x-axis and the scale on the y-axis is multiplied by 44 Graph is reflected across the x-axis, vertical scale tripled f (x) = (−x + 1) + (−x + 1) 10.0 45 Graph is reflected across the x-axis, the scale on the y-axis is multiplied by 3, and the graph is shifted up units 7.5 y 2.5 46 Graph is reflected across the x-axis, vertical scale doubled, and shifted down unit 0.0 −2.5 47 Graph is reflected across the y-axis 5.0 0.0 2.5 5.0 7.5 x −2.5 48 y = (x − 1) − = x2 − 2x 10.0 f (x) = (−x + 1) + (−x + 1) This graph is shifted unit to the right 7.5 y 5.0 50 Graph is reflected across the y-axis, horizontal scale tripled, and shifted down units 2.5 0.0 −5.0 −2.5 51 The graph is reflected across the x-axis and the scale on the y-axis is multiplied by |c| 52 For c < 0, the graph of f (cx) is the mirror image across the y-axis of f (x) with the horizontal scale multiplied by 1/|c| −2.5 0.0 2.5 5.0 7.5 x f (x) = −(−x) + (−x) 2.5 x −6 −5 −4 −3 −2 −1 0.0 −2.5 −5.0 y −7.5 −10.0 f (x) = − −x2 + (−x) This graph is shifted across x−axis and it has local maximum at x = −1 53 The graph of y = |x|3 is identical to that of y = x3 to the right of the y-axis because for x > we have |x|3 = x3 For y = |x|3 the graph to the left of the y-axis is the reflection through the y-axis of the graph to the right of the y-axis In general to graph y = f (|x|) based on the graph of y = f (x), the procedure is to discard the part of the graph to the left of the y-axis, and replace it by a reflection in the y-axis of the part to the right of the y-axis 54 If f (x) = x3 , then f (−x) = (−x)3 = −x3 = −f (x) If in general you have the right half of a graph satisfying f (−x) = −f (x), you can rotate 180◦ about the origin to see the left half Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith REVIEW EXERCISES 27 55 The rest of the first 10 iterates of f (x) = cos x with x0 = are: x4 = cos 65 ≈ 796 x5 = cos 796 ≈ 70 x6 = cos 70 ≈ 765 x7 = cos 765 ≈ 721 x8 = cos 721 ≈ 751 x9 = cos 751 ≈ 731 x10 = cos 731 ≈ 744 Continuing in this fashion and retaining more decimal places, one finds that x36 through x40 are all 0.739085 The same process is used with a different x0 56 We have x1 = f (x0 ) so x2 = f (x1 ) = f (f (x0 )) and x3 = f (x2 ) = f (f (f (x0 ))) and so on The graphs of cos x, cos cos x, cos cos cos x, and cos cos cos cos x: 2.0 x0 which is not in the second list), it must be true that L and f (L) are the same number If conditions are right (and they are in the two cases f (x) = cos(x) (#55) and f (x) = sin(x) (#57)), this “convergence” will indeed occur, and since there is in these cases only one solution (x about 0.739085 in #55 and x = in #57) it won’t matter where you started 60 The only fixed point is x = 0, since this is the only solution to sin x = x One can see that this is the only solution by graphing y = sin x and y = x on the same axes and looking for intersection points Review Exercises 1.5 m = 7−3 = = −2 0−2 −2 m = 4−1 =− 1−3 y 1.0 0.5 0.0 −3 −1 −2 x −0.5 −1.0 The limiting line is y = 0.739085 57 They converge to For x0 = 1, x1 = sin x0 ≈ 0.841, x2 = sin x1 ≈ 0.745, x3 = sin x2 ≈ 0.678 and so on 58 If you start with a number x with |x| < 1, the iterations converge to If you start with a number x with |x| > 1, the iterations diverge quickly If you start with x = ±1, the iterations all equal 59 If the iterates of a function f (starting from some point x0 ) are going to go toward (and remain arbitrarily close to) a certain number L, this number L must be a solution of the equation f (x) = x For the list of iterates x0 , x1 , x2 , x3 , is, apart from the first term, the same list as the list of numbers f (x0 ), f (x1 ), f (x2 ), f (x3 ), (Remember that xn+1 is f (xn ).) If any of the numbers in the first list are close to L, then the f -values (in the second list) are close to f (L) But since the lists are identical (apart from the first term These lines both have slope They are parallel unless they are coincident But the first line includes the point (0, 1) which does not satisfy the equation of the second line The lines are not coincident m1 = −1/m2 , so the lines are perpendicular Let P = (1, 2), Q = (2, 4), R = (0, 6) Then PQ has slope 4−2 =2 2−1 QR has slope 6−4 = −1 0−2 RP has slope 2−6 = −4 1−0 Since no two of these slopes are negative reciprocals, none of the angles are right angles The triangle is not a right triangle The slopes between points seem to be alternating between 950 and 1050 If the pattern continues, the next points will be (4, 6100), (5, 7050), and (6, 8100) Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 28 CHAPTER PRELIMINARIES 16 Intercepts at x ≈ 2.36, 0.17 and√ −2.53, and y = Local maximum √ at x = − Local minimum at x = No asymptotes 5000 4500 4000 80 3500 40 3000 2500 -4 -2 x 0.5 1.5 2.5 -40 The line apparently goes through (1, 1) and 2−1 (3, 2) If so the slope would be m = = 3−1 The equation would be 1 y = (x − 1) + or y = x + 2 Using the equation with x = 4, we find y = (4) + = 2 -80 17 Intercepts at x = −1 and 1, and y = Local minimum at x = and at x = −1 Local maximum at x = No asymptotes f (0) = −4, f (2) = −6, and f (4) = Using the point-slope method, we find y = − (x + 1) − 1 10 y = (x − 0) − = x − 4 11 The graph passes the vertical line test, so it is a function 10 y -3 -2 -1 x 12 Fails vertical line test: not a function 13 The radicand cannot be negative, hence we require − x2 ≥ ⇒ ≥ x2 Therefore the natural domain is {x| − ≤ x ≤ 2} or, in “interval-language”: [−2, 2] 14 The function is not defined where the denominator is zero, √ so the domain for f (x) is {x|x = ± 2} 18 Intercepts at x ≈ 1.97, −0.82, and −1.89, and y = −1 Local maximums at x ≈ −1.52 and 0.29 Local minimums at x ≈ −0.29 and 1.52 No asymptotes 10 15 Intercepts at x = −4 and 2, and y = −8 Local minimum at x = −1 No asymptotes x -2 -1 10 -5 -10 y -6 -4 -2 0 x -5 -10 19 Intercept at y = and at x = No extrema Horizontal asymptote y = Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith REVIEW EXERCISES 29 Vertical asymptote x = −2 20 y 15 10 -2 y -1 x -1 -10 -5 10 -2 x -5 -3 -10 20 Intercept at y = No x-intercept since the function is undefined at x = No extrema Horizontal asymptote y = Vertical asymptote x = −1 23 Intercept at y = and from the amplitude/phase shift form f (x) = √ √ sin x + sin−1 (2/ 5) , y -10 -5 10 x -1 we could write down all the intercepts only at considerable y takes √ inconvenience Extrema: √ maximum and minimum − with great predictability and regularity No asymptotes -2 kπ 21 Intercept at y = and x = for integers k Extrema: y takes maximum and minimum −1 with great predictability and regularity No asymptotes y -6 -4 -2 x -2 -4 1.5 0.5 y -3 -2 -1 0 x -0.5 -1 -1.5 kπ for inte22 Intercept at y = and x = gers k No extrema Vertical asymptotes at (2k + 1)π x= for integers k 24 f (0) = 1, so y-intercept at (0, 1) Solving the equation sin x + cos 2x = 0, π we find x-intercepts at x = + 2kπ, 11π 7π x= + 2kπ and x = + 2kπ 6 where k is an integer Local minima of occur π at x = +2kπ where k is an integer and of −2 3π occur at x = + 2kπ where k is an integer, and local maxima of occur where sin x = There are no vertical asymptotes Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith 30 CHAPTER PRELIMINARIES 29 Vertical asymptote x = −2 30 Vertical asymptote at x = −1 This is where the denominator is zero (and the numerator is not zero) Note that the function is not defined at x = −6 −5 −4 −3 −2 −1 31 x2 − 3x − 10 = (x − 5)(x + 2) x −1 The zeros are when x = and x = −2 y −2 32 x3 + 4x2 + 3x = x(x + 3)(x + 1) −3 Zeros are x = 0, −1 and −3 25 Intercept y = (2k + 1)π Local maximums at x = for integers k Local minimums at x = kπ for integers k (2k + 1)π for inteVertical asymptotes at x = gers k y -2 -2 -3 26 f (0) = 0, so y-intercept at (0, 0) Solving the equation tan 2x = 0, we find x-intercepts at kπ x = where k is an integer There are no extrema There are vertical asymptotes at π kπ x= + where k is an integer 2 y 0 −1 √ 22 − 4(1)(−2) = ± Complete list of three roots: x = 1, √ √ x = − ≈ −.732, x = + ≈ 2.732 x= 2± 35 There are solutions, one at x = and the other two negatives of one another The value in question is 928632 , found using the function “Goal Seek” in Excel The result can be checked, and a graphing calculator can find them by graphing y = x3 and y = sin x on the same axes and finding the intersection points 36 The graph shows two zeros Squaring both sides gives x2 +1 = x4 −2x√2 +1, or = x4 −3x2 The solutions are x = ± (x = is an extraneous solution.) 37 Let h be the height of the telephone pole Then h = tan 34◦ ⇒ h = 50 tan 34◦ ≈ 33.7 feet 50 −2 Solve the quadratic by formula: x -1 −3 Factor the left side: (x − 1)(x2 − 2x − 2) 34 Zeros are at x ≈ 1.618, and −0.618 √ Exact values are x = (1 ± 5)/2 -4 33 Guess a root: x = 1 x −1 −2 −3 27 Intercepts at x = −4 and where x2 + 2x − = (x + 4)(x − 2) = 0, and y = −8 where x = 28 Intercepts y = 1, and x = ±1 38 The triangle in the first quadrant with adjacent side and hypotenuse has opposite side √ √ 24 24, so sin θ = √ 1 −1/2 39 (a) = 1/2 = √ = 5 1 (b) 3−2 = = 2 40 (a) √ = 1/2 = 2x−1/2 x x (b) = 3x−2 x Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith REVIEW EXERCISES 31 41 The natural domain for f is the full real line The natural domain for g is {x|1 ≤ x} Because f has a universal domain, the natural domain for f ◦ g is the same as the domain for g, namely {x|1 ≤ x} Because g requires its inputs be not less than 1, the domain for g ◦ f is the set of x for which ≤ f (x), i.e., {x|1 ≤ x2 } = {x|1 ≤ |x|}, or in interval language (−∞, −1] ∪ [1, ∞) The formulae are easier: √ √ (f ◦ g)(x) = f ( x − 1) = ( x − 1)2 = x − √ (g ◦ f )(x) = g(x2 ) = x2 − Caution: the formula for f ◦ g is defined for any x, but the domain for f ◦ g is restricted as stated earlier The formula must be viewed as irrelevant outside the domain 42 (f ◦ g)(x) = x −1 and x4 − are both valid for x = ±1 (g ◦ f )(x) = 43 cos(3x2 + 2) = f (g(x)) for f (x) = cos x and g(x) = 3x2 + √ √ 44 sin x + = f (g(x)) for f (x) = x and g(x) = sin x + 45 x2 − 4x + = x2 − 4x + − + 1, so f (x) = (x − 2)2 − The graph of f (x) is the graph of x2 shifted two units to the right and three units down 46 x2 + 4x + = (x2 + 4x + 4) + 2, so f (x) = (x + 2)2 + The graph of f (x) is the graph of x2 shifted two units to the left and two units up 47 sin 2x = ⇒ π 2x = + 2kπ for any integer k so π x = + kπ for any integer k 48 cos 3x = whenever π 3x = ± + 2kπ for any integer k, or π 2kπ x=± + for any integer k Full file at https://TestbankDirect.eu/Solution-Manual-for-Calculus-4th-Edition-by-Smith ... https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith. .. https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith. .. https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith Solution Manual for Calculus 4th Edition by Smith Full file at https://TestbankDirect.eu /Solution- Manual- for- Calculus- 4th- Edition- by- Smith