Business Calc Homework w answers_Part_16

Business Calc Homework w answers_Part_16 - 76 Section 3.2...

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Unformatted text preview: 76 Section 3.2 21. continued Check x 27. 3: Since g(3) lim g(x) x→3 3)2 (4 lim (2x x→3 1 and 1) 2(3) 1 7, the function is not continuous (and hence not differentiable) at x 3. The function is differentiable for all reals except x 2 22. Note that C(x) x, x 2, xx x x lim C(0 h→0 3. y NDER (x) actually has two asymptotes for each asymptote of y C(0) lim h→0 hh 0 h tan x. The asymptotes of y occur at x 0. 0: h) h tan x Note: Due to the way NDER is defined, the graph of 0 , so it is 0 differentiable for all x except possibly at x Check x [ 2 , 2 ] by [ 4, 4] dy dx k 2 NDER (x) 0.001, where k is an integer. A good window for viewing this behavior is [1.566, 1.576] by lim h h→0 0 [ 1000, 1000]. The function is differentiable for all reals. 28. 23. (a) x 0 is not in their domains, or, they are both discontinuous at x 0. 1 1 ,0 x x 1 1 For 2 : NDER 2 , 0 x x (b) For : NDER 1,000,000 [ 2 , 2 ] by [ 20, 20] 0 (c) It returns an incorrect response because even though these functions are not defined at x defined at x 0, they are 0.001. The responses differ from each 1 other because 2 is even (which automatically makes x 1 1 0) and is odd. NDER 2 , 0 x x 24. The graph of NDER (x) does not look like the graph of any basic function. 29. (a) lim f (x) x→1 f (1) x) a(1)2 2 lim (3 x→1 a The relationship is a b(1) b b 2. (b) Since the function needs to be continuous, we may assume that a b 2 and f (1) 2. lim f (1 h→0 [ 5, 5] by [ 10, 10] dy x3 dx f (1) f (1 h) h f (1) lim h→0 lim h→0 lim h→0 dy dx sin x 26. 3 (1 h) a(1 1 h)2 b(1 h) 2 h a 2ah ah 2 b bh h 2ah ah 2 bh (a b h lim (2a h→0 2a 2 h lim ( 1) h→0 [ 2 , 2 ] by [ 1.5, 1.5] lim h→0 h→0 lim 25. h) h ah 2 2) b) b Therefore, 2a b 1. Substituting 2 a for b gives 2a (2 a) 1, so a 3. Then b 2 a 2 ( 3) 5. The values are a 3 and b 5. [ 6, 6] by [ 4, 4] dy dx abs (x) or x 30. The function f (x) does not have the intermediate value property. Choose some a in ( 1, 0) and b in (0, 1). Then f (a) 0 and f (b) 1, but f does not take on any value between 0 and 1. Therefore, by the Intermediate Value Theorem for Derivatives, f cannot be the derivative of any function on [ 1, 1]. 77 Section 3.3 31. (a) Note that x 1 x so lim x sin x→0 1 x sin 0 by the Sandwich Theorem. Therefore, f is continuous at x f (0 (b) h) h 1 h sin h f (0) x 10. (a) f (x) 0. 0 h) h 1 h h 2 sin g(0) 0 1. 1 h h sin h As noted in part (a), the limit of this as x approaches zero is 0, so g (0) lim h→0 x(x h lim 1 h→0 f (x) lim x h h lim h→0 hx(x h) 2 x x2 h) x h h→0 Section 3.3 Exercises (d) No, because the limit in part (c) does not exist. g(0 lim 1 x h h→0 f (x lim lim interval containing 0). (e) hx h f (x) h→0 h→0 0 (that is, for h in any open h) h h) h x (x h) hx(x h) (b) f (x) 1 and 1 an infinite number of times arbitrarily close to h lim x h→0 1 oscillates h (c) The limit does not exist because sin f (x h→0 lim 1 h sin h between 9. These are all constant functions, so the graph of each function is a horizontal line and the derivative of each function is 0. x, for all x, 2. 0. 3. s Section 3.3 Rules for Differentiation (pp. 112–121) dy dx d 2y dx 2 d ( x 2) dx d ( 2x) dx dy dx d 2y dx 2 d 13 x dx 3 d2 (x ) dx dy dx d 2y dx 2 d (2x) dx d (2) dx dy dx d 2y dx 2 d2 (x ) dx d (2x) dx dy dx d 13 x dx 3 d (3) dx 2x 0 2x 2 d (x) dx d (1) dx x2 2x d (1) dx 1 0 2 2x 0 2 0 Quick Review 3.3 1. (x 2 x 2. 2)(x 1 1) x 2 2x 1 1 2 x 2x 3 2x 2 3x 4 1 5. (x 6. 1 x x 2x 1 3x 4 2x 2 4 2 2)(x x 3 x x 5 x2 1) 2x 1 x2 1 x2 x 3x 2 1 3. 3x 2 4. x2 1 x x x2 x 2x 2 1 x 1 x 1 x 1 4. 21 1 x 5. 2 5x x2 x 2 2x 3 2x 2 2 2x 32 x 2 4 2x 2 x 2 x 1 x 1 2x 2x 2 dy dx 2 2 21 6. 2 dy dx 3 x (x 3 1 x 2 ) x 2 x 1 7. dy dx At x 1.173, 500x 6 1305. At x 2.394, 500x 6 94,212 After rounding, we have: At x 1, 500x 6 1305. At x 2, 500x 6 94,212. 8. (a) f (10) 8. 7 (b) f (0) 7 (c) f (x h) f (x ) (d) lim x x→a 9. 7 f (a) a 7 a lim 0 x→a 0 2x 3x 2 1 2 6x 2x d (4x 3) dx 4x 1 0 2x 1 42x 4 d3 (x ) dx 3x 2 2x 6x d (2x 2) dx 0 4x 3 d (21x 2) dx d (15) dx 21x 2 d (4x) dx 4x 12x 2 dy dx d 2y dx 2 d d (5x 3) (3x 5) 15x 2 15x 4 dx dx d d (15x 2) (15x 4) 30x 60x 3 dx dx dy dx d (4x 2) dx 2 dy dx 2 1 d (3x 2) dx d (7x 3) dx 21x 2 2x d2 (x ) dx 2 8x 7 lim x→a x d (1) dx d (x) dx d4 (x ) dx d 2y dx 2 0 d (x) dx d (2x) dx 4x 3 [0, 5] by [ 6, 6] d (x) dx d ( 1) dx 0 1 1 d (1) dx d 2y dx 2 7. d 12 x dx 2 d2 (x ) dx 0 2 d d (x) (1) 2x dx dx d (1) 2 0 2 dx 3 8 d (8x) dx 0 d ( 8x 3) dx 8x d (8) dx d (1) dx 3 8 24x 4 0 24x 4 78 10. Section 3.3 dy dx d1 4 x dx 4 d (3) dx x 5 x 5 d 2y dx 2 x 4 x 6 dy dx d1 2 x dx 2 d (x 1) dx (x 18. x 4 3 x 3 d ( x 5) dx 5x 11. (a) d1 3 x dx 3 5 4x x 2 x 0 4 d (x 3 ) dx 1)(2x) (x 2x 2 2x x2 3x 2 2x 1) (x 19. d (x dx (x dy dx d [(x dx 1)(x d3 (x dx 3x (x 2 2 x2 d 1) (x dx 1) (2x 3 3x 2 dy dx 2 1)] x 1) h lim f (x 3x 3x x( 1)2 x 2 2 2)(2x h) h h→0 lim (1) 3) 5x 6) f (x) lim h→0 1 h→0 d dx (x 2 3) d ( u) dx u(x) is a function of x. x 2 u(x lim u(x 3 x2 d x2 3 x dx 3 1 x2 d (x dx 3x 1 ) 1 2 3x f (x ) 0 dy 14. dx d dx 0 5x x2 5x x2 2 2x d (x dx d (1 dx (3x 5 2 x 1 ) 0 1) 3x d1x dx 1 x 2 x 2 2x 1 (1 x 2)2 (1 d x2 dx 1 x 3 (1 (2x 2)2 23. (a) At x 5)(3) 4 2)2 0, d f (x) dx 1 d (uv) dx u(0)v (0) ( 1)( 3) f (x) [ f (x)]2 v(0)u (0) 13 x ) 0, (c) At x 0, (5)(2) x)(2x) (d) At x 7(2) x 3)(2x) x 2( 3x 2) (1 x 3)2 d (c) dx v(0)u (0) u(0)v (0) du [v(0)]2 dx v ( 1)( 3) (5)(2) 7 ( 1)2 (b) At x 2 d x3 1 dx x3 3 x4 x 2)( 1) (1 (1 x 2)2 du dx 19 (3x d (1 5x 1 dx 5 2 x2 x3 3 1)(x 2 x x3 3 2)(2) (3x u(x) [ f (x)]2 (5)(2) d 2x dx 3x u(x) d d (c f (x)) c f (x) f (x) dx dx d d c f (x) 0 c f (x) dx dx d 1 22. dx f (x) This is equivalent to the answer in part (a). dy 13. dx h) h h→0 3) h) [ u(x)] h h) h h→0 lim (x 2 u(x lim h→0 3) (x) 21. dy dx d x2 dx x 2 2) 2) f (x) h→0 h 1 x2 17. 1 1)2 x. d (x) dx 3 dx x dx x(2x) dy dx x 12 6x 2 (x 2 3x 2)2 1 x2 16. 1) 2 d dx dy dx x( 5x 6) (2x 3 3x 2 (x 2 3x 2)2 1)(1) lim 2x x (x 2 15. x 2 2 2 2)(2x 3) (x 2 3x (x 2 3x 2)2 3x (b) Note that u dy dx x 1) 1)2 1)(x 1)(x 20. (a) Let f (x) (b) 2 1 dy (b) dx 12. (a) ( 1 1 1) d (x 2) dx 3 d [(x 1)(x 2 1)] dx d (x 1) (x 2 1) (x 2 dx (x (x x (x x 2 1 2 x( x 2x 2 x d dx 2 d (x 4) dx 3x dy dx 1 1) x 4 2x (1 x 3)2 0, u(0)v (0) v(0)u (0) [u(0)]2 dv dx u ( 1)( 3) (5)2 d (7v dx 2u) 2( 3) 20 7 25 7v (0) 2u (0) (x h) h x Section 3.3 24. (a) At x 2, d (uv) dx (3)(2) (1)( 4) u(2)v (2) v(2)u (2) 2 v(2)u (2) u(2)v (2) du [v(2)]2 dx v (1)( 4) (3)(2) 10 (1)2 (b) At x 2, 29. y (x) 2, [ 3, 3] by [ 10, 30] d (d) Use the result from part (a) for (uv). dx d At x 2, (3u 2v 2uv) dx d 3u (2) 2v (2) 2 (uv) dx 3( 4) 2(2) 30. y (x) 3x 2 y ( 2) 12 The tangent line has slope 12 and passes through ( 2, so its equation is y 2(2) x-intercept is 3 x 2 3x 3 3(2)2 3 9 y 1 and passes through (2, 3). 9 1 (x 9 1 x 9 2) 12x 3 29 9 31. y (x) (x 2 1)(4) 4x(2x) (x 2 1)2 At the origin: y (0) The tangent is y At (1, 2): y (1) 4x 2 4 (x 2 1)2 4 4x. 0 The tangent is y 2. Graphical support: Graphical support: [ 4.7, 4.7] by [ 3.1, 3.1] [ 4.7, 4.7] by [ 2.1, 4.1] 28. y (x) 3x 2 1 The slope is 4 when 3x 2 1 4, at x 1. The tangent at x 1 has slope 4 and passes through ( 1, 2), so its equation is y 4(x 1) 2, or y 4x 2. The tangent at x 1 has slope 4 and passes through (1, 2), so its equation is y 4(x 1) 2, or y 4x 2. The smallest slope occurs when 3x 2 1 is minimized, so the smallest slope is 1 and occurs at x 0. 32. y (x) y (2) x 2)(0) 8(2x) (4 x 2)2 (4 16x x 2)2 1 2 The tangent has slope equation is y 1 (x 2 Graphical support: Graphical support: [ 3, 5] by [ 2, 4] [ 4.7, 4.7] by [ 3.1, 3.1] (4 8), 16. The [ 3, 3] by [ 20, 20] 2 The tangent line has slope 9, so the perpendicular line has y 8, or y 6, so the 3 slope is . (iii) 2 slope 2) Graphical support: 25. y (x) 2x 5 y (3) 2(3) 5 11 The slope is 11. (iii) 26. The given equation is equivalent to y 12(x 4 and the y-intercept is 16. 3 12 27. y (x) y (2) 1 Graphical support: u(2)v (2) v(2)u (2) dv [u(2)]2 dx u (3)(2) (1)( 4) 10 (3)2 9 (c) At x 6x 2 6x 12 6(x 2 x 2) 6(x 1)(x 2) The tangent is parallel to the x-axis when y 0, at x and at x 2. Since y( 1) 27 and y(2) 0, the two points where this occurs are ( 1, 27) and (2, 0). 79 1 and passes through (2, 1). Its 2 1 2) 1, or y x 2. 2 80 33. Section 3.4 an 2 d nRT V2 dV V nb d (V nb) (nRT) dP dV (c) (nRT) dV (V 0 (V nR T nb)2 d (V dV nb) nb)2 2an 2V d (an 2V 2) dV 3 d (uv) dx (V 34. 35. ds dt d 2s dt 2 d (4.9t 2) dt d (9.8t) dt dR dM 2an V3 nRT nb)2 CM Exploration 1 M2 37. If the radius of a sphere is changed by a very small amount r, the change in the volume can be thought of as a very thin layer with an area given by the surface area, 4 r 2, and a thickness given by r. Therefore, the change in the volume can be thought of as (4 r 2)( r), which means that the change in the volume divided by the change in the radius is just 4 r 2. 38. Let t(x) be the number of trees and y(x) be the yield per tree x years from now. Then t(0) d (ty) dx 156, y(0) 12, t (0) 13, 1.5. The rate of increase of production is t(0)y (0) y(0)t (0) (156)(1.5) (12)(13) 39. Let m(x) be the number of members and c(x) be the pavillion cost x years from now. Then m(0) 250, m (0) 6, and c (0) of each member’s share is (65)(10) (250)(6) (65)2 dc dx m Growth Rings on a Tree 1. Figure 3.22 is a better model, as it shows rings of equal area as opposed to rings of equal width. It is not likely that a tree could sustain increased growth year after year, although climate conditions do produce some years of greater growth than others. 2. Rings of equal area suggest that the tree adds approximately the same amount of wood to its girth each year. With access to approximately the same raw materials from which to make the wood each year, this is how most trees actually grow. 3. Since change in area is constant, so also is change in area . If we denote this latter constant by k, we 2 k have r, which means that r varies change in radius inversely as the change in the radius. In other words, the change in radius must get smaller when r gets bigger, and vice-versa. Exploration 2 390 bushels of annual production per year. c(0) u du u v dv v (u du)(v) (u)(v dv) (v dv)(v) uv vdu uv udv v 2 vdv vdu udv v2 s Section 3.4 Velocity and Other Rates of Change (pp. 122–133) 36. If the radius of a circle is changed by a very small amount r, the change in the area can be thought of as a very thin strip with length given by the circumference, 2 r, and width r. Therefore, the change in the area can be thought of as (2 r)( r), which means that the change in the area divided by the change in the radius is just 2 r. and y (0) u v 9.8 M 3 13 M 3 dv dx u . This is equivalent to the product (d) Because dx is ‘infinitely small,” and this could be thought of as dividing by zero. 9.8t C d M2 2 dM dC2 M dM 2 du dx rule given in the text. (e) d 2 v Modeling Horizontal Motion 1. The particle reverses direction at about t t 2.06. 0.61 and 65, 10. The rate of change m(0)c (0) c(0)m (0) [m(0)]2 0.201 dollars per year. Each member’s share of the cost is decreasing by approximately 20 cents per year. 40. (a) It is insignificant in the limiting case and can be treated as zero (and removed from the expression). (b) It was “rejected” because it is incomparably smaller than the other terms: v du and u dv. 2. When the trace cursor is moving to the right the particle is moving to the right, and when the cursor is moving to the left the particle is moving to the left. Again we find the particle reverses direction at about t 0.61 and t 2.06. ...
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