ANSWERS, Homework Problems, Spring 2014 Now You Try It, Supplemental problems in written homework, Even Answers R.6 8) 27, 30) 25

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1 ANSWERS, Homework Problems, Spring 2014, Supplemental problems in written homework, Even Answers Review Assignment: Precalculus Even Answers to Sections R1 R7 R.1 24) 4a 2 16ab + 16b 2 R.2 24) Prime 5x 2 + 4x 4 R.3 38) x(x 1)(x + 1) R.6 8) 27, 30) Lecture 1 6x 2 6 1) 2) x = 2 or x = 2 2x ) a) 2x2 2x 4 b) x = 1 or x = 2 4) ( 6, 3] [6, ) (2x 1) 4/3 5) x + 4 x +, excluded value: x = 3 6) y = 2 or y = 6 12 x 7) (x + 3) 2 + (y 2) 2 = 17 Supplemental Problems in Written Homework x 2 4 1) a. b. x = ±2 2) a. 2x2 2x + 2 x x + 1 3) (3, 5) and (3, 1) 4) x 2 + y 2 = 36 Lecture 2 Even Answers to Sec ) x = 8 36) (a) k = 1 2, (b) k = 7 2 1) (a) y = 4: slope is 0 (b) x = 2: undefined slope 2) Find the slope of each line by solving for y. Then since parallel lines have the same slope, you can set up an equation to show the desired result. 3) k = 1 3 4) (a) N = 3 x + 89 (b) approximately 103 deaths (c) mg/m3 50 5) V (t) = ( )t + 20, 000 In 5 years, trade-in value is $ The machine will have no trade-in value after approximately 9.14 years. 1

2 Lecture 3 1) V (t) = 5 t + 200; 90 days 3 2) 75 miles: Avis charges $63.75 and Enterprise charges $ She should use Enterprise. The charge will be the same for a trip of 120 miles. 3) For a linear function, marginal cost and profit are simply the slope of the corresponding functions. In this example, marginal cost is 19: cost increases by $19 per unit. Marginal profit is 6: profit increases by $6 per unit. 4) (a) The supplier will provide no items (b) Equilibrium quantity: 25 and price: $176 5) (a) y = 7.25x , correlation coefficient: r = (b) approximately $ billion Lecture 3 Text Homework: Even Answers Sec. 1.2: 4) 12 8) ) C(x) = 45x ) C(x) = 120x + 12, ) a. C(x) = 1140x + 486, 000 b. $1,056,000 c. 450 Lecture 4 : 2 1. (a) only 2. k = 3 3. (for h 0): a. 4x+2h 3 b. (x + h + 2)(x + 2) [ ) 1 4. domain of f(x): (, ), domain of g(s): (, 2] 2, 5. domain: (, 3) (3, ) and range: [ 1, ) 6. a. neither b. odd 7. (a) 42,000 (b) 2002 (c) difference is thousand which rounds to 133 people (d) population will approach 48,000 since as t increases without bound, 12 t + 2 approaches 0 8. (a) Area: A(x) = 600x 2x 2, domain: (0, 300) (b) C(x) = 4800 (constant function) (c) C(x) = x 2

3 Lecture 4 Text Homework: Even Answers Sec. 2.1: 4) Not a function 6) function 8) not a function 18) (, ) 20) (, ) 26) (, 6) ( 6, 6) (6, ) 28) (, ) 32) (, 3) 40) domain: [ 2, 4], range: [3], a c: 3, d) nowhere 42) a. 0 b c. (a + 3)(a 4) d. 2(2 + 3m)(1 2m) m 2 54) a. 4x 2 8hx 4h 2 + 3x + 3h + 2 b. 8hx 4h 2 + 3h c. 8x 4h ) odd 70) neither 74) a. $410 b. $ c. $855 Lecture 5: Additional Exercises 1. Translate the graph of y = x to the right 4 units, then reflect across the x-axis, and finally shift up 3 units. 2. a) f(g(x)) = x 3, domain: [ 1, ), g(f(x)) = x 2 3, domain (, 3] [ 3, ) 2(x 1) b) f(g(x)) =, domain: (, 0) (0, 1) (1, ), x g(f(x)) = 2, domain: (, 0) (0, 2) (2, ) 2 x Lecture 5 1. a) 3 b) 4 c) 3 4) (f g)(4) = f( 1) which is not defined 2. (fg)(x) = 2, domain: (, 0) (0, 1) (1, ) ( ) x 1 f 2(x 1) (x) =, domain: (, 0) (0, 1) (1, ) g x 2 3. a) f(g(x)) = x 4, domain: [5, ), g(f(x)) = x 2 4, domain: (, 2] [2, ) 2(x 1) b) f(g(x)) =, domain: (, 0) (0, 1) (1, ), x g(f(x)) = 2, domain: (, 0) (0, 2) (2, ) 2 x 4. a) c(x(t)) = 0.05t b) 4 years 3

4 5. f(x) is neither even nor odd. f(x) = { x x < 0 3x x (a) (b) (c) (d) 7. a. reflect y = x across y-axis, shift left 3, up 2; domain (, ), range (, 2] a. b. f(x) = (x 4): reflect y = x across the y-axis, then shift 4 units to the right and reflect across the x-axis. domain: (, 4], range (, 0] b. c. Always reflect before you shift! 4

5 Lecture 5 Text Homework: Even Answers Sec. 4.3: 6) 1000z 2 80z ) f[g(x)] = (2 x), domain: (, 2) (2, ), g[f(x)] = x, domain: 4 (, 0) (0, ) Sec. 2.2: 4) F 6) B 8) E 26) A 28) B 30) F 32) shifts graph right 2, up 2 36) shifts y = x left 2, down 3 48) a. b b. b c. b Lecture 6 1. f(x) = (x + 2) To graph: translate the graph of y = x 2 left two units, reflect over the x-axis, and then shift up 7 units. ( ) ( ) vertex: 3, 51, x-intercepts: ( 1, 0), 3 3, 0, y-intercept (0, 5); parabola opens up 3. (a) p = 0.1x + 24 (b) R(x) = 0.1x x; maximum revenue is $1440, when the price is p(120) = 12 (c) break even when x = 50 or x = 150 (d) will earn a profit for 50 < x < 150. Maximum profit when 100 are sold. 4. R(x) = 0.05x 2 + x ; he will maximize his harvest when x = 10, or September a is negative and the smallest value of n is 5. Lecture 6 Text Homework: Even Answers Sec. 2.2: 2) reflects across the x-axis, stretches graph of y = x 2 if a < 1 and compresses graph if 1 < x < 0. 58) a. $15,000, b. $40,000 c) reflects the graph of f(x) = x 2 across the x-axis, compresses by a factor of 4, shifts right 10 units and up 40 units. 60) a mm and mm, b. 8.48mm and 39.4 weeks Sec. 2.3: 48) a. $54 billion, c. $750 billion, d. $1104 billion Lecture 7 1. x 2 5n y 5n 2n2 2) x = 0 and x = 3 5

6 3) a) b) ) x 4) B = 80 and e k = 7 ( 7 4 so f(x) = ) P (t) = 500(2) t/4 ; in ten years the population should be about ) After 2 years you would have $ in the first account, and $ in the second. So you get a better return in the account with interest compounded monthly. 7) 100%; 34%; in the long run, 24% since e 0.5x 0 as x increases without bound. Lecture 7 Text Homework: Even Answers Sec. 2.3: 40) No asymptotes; graph is the graph of the line y = 3 x with a hole at (3, 0). 58) a. [0, ) c. As x, denominator A + x x, so Kx A + x Kx x = K. d. limiting value of the growth rate as x increases over time e. replace x in the formula with A and simplify. Sec. 2.4: 4) D 6) F 8) D 10) B 30) graph of y = e x is reflected across the x-axis and stretched by a factor of 2, then the graph is shifted down 3 50) a. yes, although a linear model seems reasonable as well b. f(x) = e t c. 805,800; 855,900 Lecture 8 1) (b) and (c) are one-to-one 2) f(f(x)) = x. The graph of g is a semicircle centered at the origin with radius 3: not one-to-one so does not have an inverse 3) h 1 (x) = 2 x 1. Range of h is the domain of h 1 : (, 1) (1, ) 4) f 1 (x) = x2 + 1, x 0 2 5) not equivalent: domain of f is (, 0) (0, ) but domain of g is (0, ) 6

7 6) domain: x2 2x x + 3 7) x = 2 only 8) x = 2 > 0, so the domain is ( 3, 0) (2, ) 9) f 1 (x) = ln(x 1) + 2 Lecture 8 Additional Exercises from the Lecture Outline and Homework 1) a) 4 b) 1 c) 4 (Note that if g 1 (4) = 2 its graph contains (4, 2). What does that tell you about g(2)?) d) f 1 (x) = x (1, ), domain: (, ) 2) (a) f 1 (x) = 2x + 1, domain (, 1) 1 x (b) g 1 (x) = 3 x 2, domain [0, ) (c) h 1 (x) = ln x 1, domain (0, ) 2 3) Lecture 8 Text Homework: Even Answers Sec.2.5: 6) log 5/4 (16/25) = 2 24) 0 34) a + 2c 48) x = 1 56) x = 1 70) (, 3) (3, ) e 1 Lecture 9 (a) W (1) = 49.9 years (b) t 2.4, year 1955 (c) W (6) 80.5 years 2) Solve for r in the equation 10, 800 = 10, 000e 2r to get an interest rate of 3.8% 3) P = 100, 000 $72, 730 (1.01) 32 4) (a) p $16.39; Revenue is (400)(p) = $

8 (b) x = 200 ln(42/62) 78 (c) as x, p (0) = 8 dollars 5700 ln(.76) 5) t = 2257 years ln ) P (t) =, P (6) eln(99/249)t Since ln(99/249) < 0, e ln(99/249)t 0 as t, so the number of cases has an upper limit of Lecture a) 4 b) 0 c) DNE d) e) + f) DNE 2. f(x) is the parabola y = x 2 reflected over the x-axis and shifted up 4 units. The graph of g(x) is the same graph, with a hole at the point (0, 4). lim f(x) = lim g(x) = 4. x 0 x x2 + 3x x + 1 = 3x, x 1. Its graph is the graph of y = 3x with a hole at the point ( 1, 3); lim x 1 3x 2 + 3x x + 1 = 3 4. (1) 2 (2) DNE (3) 2 5. a. 1 b. 2 c. 1 4 Lecture 10 Text Homework: Even Answers Sec. 3.1: Problem 6) a. 4, b. 4 8) a. 2, b. Does not exist (DNE) 10) a. i. 1, ii. 1, iii. 1, iv. 2; i. 0, ii. 0, iii. 0, iv. 0 56) a. DNE, b. 7 82) a. 3, b. DNE, c. 2, d. 16 months Lecture (a) 7 16 (b) 3 2 (c) 4 5 (d) 0 (e) 2a 2. (a) Might be true: for example, if f(x) = x + 1, lim f(x) = 2 = f(1). But if { x 1 x + 1 x 1 g(x) =, then lim g(x) = 2, but g(1) = 5. 5 x = 1 x 1 (b) must be false: if lim f(x) = 1, then lim f(x) must equal 1 as well. x 0 8 x 0

9 (c) might be true: Let f(x) = x 2 + x and g(x) = x 3 + x 2. Then lim x 0 f(x) = 1 lim g(x) = 0, but lim[f(x)/g(x)] = lim which does not exist as a finite limit. x 0 x 0 x 0 x But consider problem 1c above, and let f(x) be the numerator and g(x) be the denominator. Note that lim ( ) 4 as a finite number. 5 x 4 f(x) = lim x 4 g(x) = 0, and lim x 4 3. (a) limit is 0; HA: y = 0 (b) limit is 0; HA: y = 0 (c) limit is ; no HA (d) limit is 1 2 ; HA: y = 1 2 * *For (d), note that as x, the function f(x) = 5 x2 3 4x properties we see that the limit can be found by finding lim x f(x) = [f(x)][g(x)] does exist > 0, so from the limit VA: x = ln 4 for horizontal asymptotes lim f(x) = 0, but lim f(x) = 2 x x 4 = 1 2. HA: y = 0 and y = lim x 4 f(x) = lim x 4 1 x 4 x 2 4 x = Lecture 11 Text Homework: Even Answers Sec. 3.1: 2) a 24) 2 28) 100 Lecture (a) x = 1: removable, x = 3: removable, x = 5: infinite, x = 7: jump (b) can define f( 1) = 0 and redefine f(3) = 1 but cannot so define at the other discontinuities since the limits do not exist at those points (nonremovable discontinuities) (c) (, 1), ( 1, 3), (3, 5), (5, 7), [7, ) 2. (a) (b) (c) (d) 0 3. (a) 1) 2 and 2) (b) x = 1: removable, x = 1: infinite (c) can define f( 1) = 2 = lim f(x); cannot so define f(1) since the limit does x 1 not exist 4. (a) (1) + (2) 1 (3) 2 (4) does not exist 9

10 (b) x = 1: infinite, x = 0: removable, x = 1: jump 5. (a) x = 5 and x = 1 (b) 1 6 (c) + (d) x = 5: infinite, cannot define f( 5) to make f continuous there since it is a nonremovable discontinuity. x = 1: removable, define f(1) = 1 to make f continuous at x = 1. 6 x 2 2( x) = 1 x < 0 6. Note that f(x) = x 3 4x x 2 x 2 2x x 3 4x = 1 x > 0 x + 2 (a) (1) 1 2 (2) 1 2 (b) x = 0: jump discontinuity, x = ±2, removable discontinuities (c) Lecture 12 Text Homework: Even Answers Sec. 3.1: 4) b 20) does not exist 56) a. Does not exist, b. 7 62) a., b. x = 4 Sec. 3.2: 12) nowhere 34) x = 6: infinite, x = 4: jump, x = 0: removable, x = 3: infinite, x = 4: removable Lecture 13 Additional Exercises 1) CORRECTION: f(x) = x2 2x x 3 + x 2 6x discontinuities are x = 3: infinite, x = 0 and x = 2: removable define f(0) = 1 3 and f(2) = 1 ; cannot so define f( 3) 5 10

11 2) f(x) has one jump (nonremovable) discontinuity at x = 2 since lim f(x) = 2 and lim f(x) = 2 x 2 x 2 + 3) x = 1: jump discontinuity, x = 2: removable discontinuity can define f(2) = 1 to make f continuous at x = 2, cannot so define f( 1) 4) removable discontinuity at x = 1; note that lim x 1 f(x) = 5 4 5) Note that f is continuous on [0, 1] and compare f(0) and f(1) to use the IVT Lecture (a) vertical asymptote: x = 6, x = 0, horizontal asymptote: y = 0 (b) (1) (2) (3) 1 8 or f(0) (c) can define f(6) = 1 8 to make f continuous at x = 6; cannot so define f( 6) 2. f has a jump discontinuity at x = 3 and an infinite discontinuity at x = 4; both are nonremovable. f is continuous at x = (a) f has a jump discontinuity at x = 3 11

12 (b) f has a jump discontinuity at x = 0 and an infinite discontinuity at x = 2 4. (a) c = 4 and d = 6 (b) K = Note that f is continuous on [0, 1] and compare f(0) and f(1) to use the IVT 6. The function is discontinuous at x = 160, x = 220, and x = 280 (x in thousands) Lecture 13 Text Homework: Even Answers Sec. 3.2: 28) k = 4 Lecture (a) concentration is increasing by 6/5 mg/ml per minute (b) concentration is increasing by 3.6 mg/ml per minute (c) 2mg/ml per minute increase 2. (a) cost increases by $2.84 (b) cost increases by $2.80 per unit 3. (a) object ends up where it started, so average velocity is 0 in/sec (b) traveling at 1 in/sec (c) v(a) = 2a 3 in/sec; object first comes to a stop after 3/2 seconds 4. (a) $3000 increase in profit per $1000 increase in advertising dollars (b) 2 a + 8 thousand dollar change in profit per $1000 increase in adv dollars 3 (c) $2000 decrease in profit per $1000 increase in advertising dollars (d) Profit is maximized when $12,000 is spent on advertising. 5. (a) 270 (b) population is increasing by 5 per year (c) population is increasing by 3 per year 12

13 Lecture 14 Text Homework: Even Answers Sec. 3.3: 24) increasing 30) a. $3000 b. 0 c. $ ) a. 5 ft/sec b. 2 ft/sec c. 3 ft/sec. d. 5 ft/sec. e. i. and ii. 2.5 ft/sec., f. i. and ii. 4 ft/sec. Lecture x = 5: approximately 2.5, x = 0: approximately 1.5, x = 4: 0 (answers may vary) 2. 2 : yes, f(x) is linear so the derivative is the slope of the line at each point 3 3. f 3 (x) = 2 ; graph is the square root function stretched up and reflected 10 3x across the y-axis, and then shifted 10/3 units to the right. At x = 5 the slope is 3 10 and at x = 1 3 the slope is 1 2. The slopes ; vertical tangent line at x = y = 3 4 x slope of the tangent line at any point is given by f 2 (x) = (1 x). ( ) ( ) Points are 2, 4 and 2, 4 6. (a) v(t) = 2t 2: at t = 4, the velocity is 6 in/sec, and at t = 10, the velocity is 18 in/sec (b) at one second 7. (a) MR = 8 x + 10; when x = 12, revenue is decreasing by $9.20 per unit 5 (b) P (x) = 4 5 x2 + 8x 15 ; MP = 8 5 x + 8 profit is maximized when MP = 0, at x = 5. This is the vertex of the graph of the profit function. 13

14 Lecture 15 Text Homework: Even Answers Sec. 3.4: 4) tangent line is horizontal 10) undefined 24) a) y = 1 4 x 7 4, b) y = 3 4 x ) a) 13, 13.4, 13 b) 13, 13.04, 13 58) answers may vary, for example at h = 5000, T (h) = dt dh 2.5 or degrees/foot 2000 Lecture The graph will look somewhat like a parabola opening down, with two x-intercepts, one negative and one positive 2. (a) car A is farthest from the starting point after 8 minutes, B is farthest after 12 minutes (comparing function values at those points) (b) first, after 7 minutes (on interval [0, 7]); next after 10 minutes (on interval [0, 10], also [7, 10]) (c) after two minutes, car B; after five minutes, car A (compare slopes of tangent lines at those points) (d) twice when tangent lines are parallel, for example approximately at t = 3 3. (a) x = 3 and x = 2 (b) x = 0 and x = 5 (c) x = 5 The graph of the derivative of f(x) has a hole at x = 3, at x = 2 a jump or asymptote (depending on whether you visualize f having a vertical asymptote at x = 2), a nonremovable discontinuity at x = 0 and a vertical asymptote at x = 5 4. (a) does not exist (sharp corner at x = 0) (b) f (0) = lim x 0 f(x) f(0) x 0 = 0; graph has horizontal tangent line at x = 0 (c) f(x) is not continuous at x = 0 so cannot be differentiable at x = 0 (d) f(x) is continuous at x = 0, but f f(x) f(0) (0) = lim does not exist x 0 x 0 (one-sided limits are not the same) so f(x) is not differentiable at x = 0 14

15 Lecture 16 Text Homework: Even Answers Sec. 3.4: 2) does not exist 16) f (x) = 3 x 2, f ( 2) = 3 4, f (0) does not exist, f (3) = ) 6 38) 5, 3, 0, 2, 4 40) a) distance, b) velocity Sec. 3.5: 10) 22) graph is an increasing function on (0, ) with vertical asymptote v = 0 and crosses the v-axis between 7 and 8. The derivative values approach as v approaches 0. Lecture (a) f(3) = 2 (b) f (3) = (a) P (2) = 6.08 so spending is 6.08% of GDP (b) average rate is 1.94; spending increased by 1.94% per decade (c) in 2010: P (2) = 1.94 so spending increased by 1.94% per decade, but in 2020, P (2) = 2.48 so spending is increasing by 2.48% per decade 3. N(10) = 6.926: number of cases after 10 years of exposure N (10) = ; number of cases is increasing by per year 4. (a) y = 43x + 44 (b) x = 2 and x = 6 5. f (x) = 2 x 4 x 2. Horizontal tangent line at x = 4 with equation y = 0 6. a = 3 4 and b =

16 7. (a) v(t) = 80 32t (b) v(t) = 0 at t = 5 2 ( ) 5 (c) maximum height: h = 900 feet 2 8. This limit represents f f(a + h) f(a) (a) = lim h 0 h Its value: f (1) = 8 for a = 1 and f(x) = x 8. Lecture 18! 1. (a) 6 25 (b) (x 2 + 2x + 3) 1 (2x + 2) = (4x + 4)(x 2 + 2x + 3) (illustrates the Chain rule) 3. y = 1 2 x f (x) = 1 x2/3 x 2 5. f (x) = 1 3x Horizontal tangent line at x =. x(x 2 + 1) (a) f (x) = x2 4 x 2 (b) x = ± price is decreasing by $14.15 per thousand units and revenue is increasing by $41.66 per thousand items 16