Solucao impares Calculo 6ed CAP.4
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(2) F. 148. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. 15. i ({) = 8 3{, { 1. Absolute maximum i (1) = 5; no. local maximum. No absolute or local minimum.. 19. i ({) = {2 , 0 { ? 2. Absolute minimum i (0) = 0; no. local minimum. No absolute or local maximum.. 23. i ({) = ln {, 0 ? { 2. Absolute maximum. i (2) = ln 2 0=69; no local maximum. No absolute or. TX.10 17. i ({) = {2 , 0 ? { ? 2. No absolute or local maximum or. minimum value.. 21. i ({) = {2 , 3 { 2. Absolute maximum. i (3) = 9. No local maximum. Absolute and local minimum i (0) = 0.. 25. i ({) = 1 . {. Absolute maximum i (0) = 1; no local. maximum. No absolute or local minimum.. local minimum.. + 27. i ({) =. 1{. if 0 { ? 2. 2{ 4. if 2 { 3. Absolute maximum i(3) = 2; no local maximum. No absolute or local minimum..
(3) F. TX.10 29. i ({) = 5{2 + 4{. SECTION 4.1. MAXIMUM AND MINIMUM VALUES. ¤. 149. i 0 ({) = 10{ + 4. i 0 ({) = 0 { = 25 , so 25 is the only critical number.. 31. i ({) = {3 + 3{2 24{. i 0 ({) = 3{2 + 6{ 24 = 3({2 + 2{ 8).. i 0 ({) = 0 3({ + 4)({ 2) = 0 { = 4, 2. These are the only critical numbers. 33. v(w) = 3w4 + 4w3 6w2. v0 (w) = 12w3 + 12w2 12w. v0 (w) = 0 12w(w2 + w 1) . w = 0 or w2 + w 1 = 0. Using the quadratic formula to solve the latter equation gives us s 1 ± 12 4(1)(1) 1 ± 5 1 ± 5 = 0=618, 1=618. The three critical numbers are 0, . w= 2(1) 2 2 |1 |2 | + 1. 35. j(|) =. . (| 2 | + 1)(1) (| 1)(2| 1) | 2 | + 1 (2|2 3| + 1) | 2 + 2| |(2 |) = = 2 = 2 . 2 2 2 2 (| | + 1) (| | + 1) (| | + 1)2 (| | + 1)2. j 0 (|) =. j 0 (|) = 0 | = 0, 2. The expression | 2 | + 1 is never equal to 0, so j 0 (|) exists for all real numbers. The critical numbers are 0 and 2. 37. k(w) = w. 3@4. 1@4. 2w. k0 (w) = 0 3. 0. k (w) =. w=2. 39. I ({) = {4@5 ({ 4)2. . 3 1@4 w 4. w=. 2 3. . 2 3@4 w 4. =. 1 3@4 w (3w1@2 4. 2) =. 3. w2 . 4 4 w3. w = 49 . k0 (w) does not exist at w = 0, so the critical numbers are 0 and 49 .. . I 0 ({) = {4@5 · 2({ 4) + ({ 4)2 · 45 {1@5 = 15 {1@5 ({ 4)[5 · { · 2 + ({ 4) · 4] =. ({ 4)(14{ 16) 2({ 4)(7{ 8) = 5{1@5 5{1@5. I 0 ({) = 0 { = 4, 87 . I 0 (0) does not exist. Thus, the three critical numbers are 0, 87 , and 4. 41. i () = 2 cos + sin2 . i 0 () = 2 sin + 2 sin cos . i 0 () = 0 2 sin (cos 1) = 0 sin = 0. or cos = 1 = q [q an integer] or = 2q . The solutions = q include the solutions = 2q , so the critical numbers are = q . 43. i ({) = {2 h3{. i 0 ({) = {2 (3h3{ ) + h3{ (2{) = {h3{ (3{ + 2). i 0 ({) = 0 { = 0,. [h3{ is never equal to 0]. i 0 ({) always exists, so the critical numbers are 0 and 23 . 45. The graph of i 0 ({) = 5h0=1|{| sin { 1 has 10 zeros and exists. everywhere, so i has 10 critical numbers.. 2 3.
(4) F. 150. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. 47. i ({) = 3{2 12{ + 5, [0> 3].. TX.10. i 0 ({) = 6{ 12 = 0 { = 2. Applying the Closed Interval Method, we nd that. i (0) = 5, i (2) = 7, and i (3) = 4. So i (0) = 5 is the absolute maximum value and i (2) = 7 is the absolute minimum value. 49. i ({) = 2{3 3{2 12{ + 1, [2> 3].. i 0 ({) = 6{2 6{ 12 = 6({2 { 2) = 6({ 2)({ + 1) = 0 . { = 2> 1. i (2) = 3, i (1) = 8, i (2) = 19, and i (3) = 8. So i(1) = 8 is the absolute maximum value and i (2) = 19 is the absolute minimum value. 51. i ({) = {4 2{2 + 3, [2> 3]=. i 0 ({) = 4{3 4{ = 4{({2 1) = 4{({ + 1)({ 1) = 0 { = 1, 0, 1.. i (2) = 11, i(1) = 2, i (0) = 3, i(1) = 2, i (3) = 66. So i (3) = 66 is the absolute maximum value and i(±1) = 2 is the absolute minimum value. 53. i ({) =. { ({2 + 1) {(2{) 1 {2 0 , [0> 2]. i ({) = = = 0 { = ±1, but 1 is not in [0> 2]. i(0) = 0, {2 + 1 ({2 + 1)2 ({2 + 1)2. i (1) = 12 , i (2) = 25 . So i (1) = 55. i (w) = w. 1 2. is the absolute maximum value and i (0) = 0 is the absolute minimum value.. 4 w2 , [1> 2].. w2 + (4 w2 ) w2 4 2w2 i 0 (w) = w · 12 (4 w2 )1@2 (2w) + (4 w2 )1@2 · 1 = + 4 w2 = = . 2 2 4w 4w 4 w2 i 0 (w) = 0 4 2w2 = 0 w2 = 2 w = ± 2, but w = 2 is not in the given interval, [1> 2]. i 0 (w) does not exist if 4 w2 = 0 w = ±2, but 2 is not in the given interval. i (1) = 3, i 2 = 2, and i (2) = 0. So i. 2 = 2 is the absolute maximum value and i (1) = 3 is the absolute minimum value.. 57. i (w) = 2 cos w + sin 2w, [0, @2].. i 0 (w) = 2 sin w + cos 2w · 2 = 2 sin w + 2(1 2 sin2 w) = 2(2 sin2 w + sin w 1) = 2(2 sin w 1)(sin w + 1). i 0 (w) = 0 sin w = 12 or sin w = 1 w = 6 . i (0) = 2, i ( 6 ) = 3 + 12 3 = 32 3 2=60, and i ( 2 ) = 0. So i ( 6 ) =. 3 is the absolute maximum value and i ( 2 ) = 0 is the absolute minimum value.. 3 2. 2. 59. i ({) = {h{. @8. , [1> 4]. i 0 ({) = { · h{. 2. @8. · ( {4 ) + h{. 2. @8. 2. · 1 = h{. @8. 2. 2. ( {4 + 1). Since h{. @8. is never 0,. i 0 ({) = 0 {2 @4 + 1 = 0 1 = {2 @4 {2 = 4 { = ±2, but 2 is not in the given interval, [1> 4]. i (1) = h1@8 0=88, i(2) = 2h1@2 1=21, and i (4) = 4h2 0=54. So i (2) = 2h1@2 is the absolute maximum value and i(1) = h1@8 is the absolute minimum value. 1 · (2{ + 1) = 0 { = 12 . Since {2 + { + 1 A 0 for all {, the {2 + { + 1 domain of i and i 0 is R. i (1) = ln 1 = 0, i 12 = ln 34 0=29, and i(1) = ln 3 1=10. So i (1) = ln 3 1=10 is the absolute maximum value and i 12 = ln 34 0=29 is the absolute minimum value.. 61. i ({) = ln({2 + { + 1), [1> 1]. i 0 ({) =.
(5) F. TX.10. SECTION 4.1. MAXIMUM AND MINIMUM VALUES. ¤. 151. 63. i ({) = {d (1 {)e , 0 { 1, d A 0, e A 0.. i 0 ({) = {d · e(1 {)e1 (1) + (1 {)e · d{d1 = {d1 (1 {)e1 [{ · e(1) + (1 {) · d] = {d1 (1 {)e1 (d d{ e{) At the endpoints, we have i(0) = i (1) = 0 [the minimum value of i ]. In the interval (0> 1), i 0 ({) = 0 { = i. d d+e . . =. d So i d+e. d d+e. =. d 1. d d+e. e =. dd (d + e)d. . d+ed d+e. e =. d = d+e. dd ee dd ee · = . d e (d + e) (d + e) (d + e)d+e. dd ee is the absolute maximum value. (d + e)d+e. 65. (a). From the graph, it appears that the absolute maximum value is about i (0=77) = 2=19, and the absolute minimum value is about i (0=77) = 1=81.. t (b) i ({) = {5 {3 + 2 i 0 ({) = 5{4 3{2 = {2 (5{2 3). So i 0 ({) = 0 { = 0, ± 35 . t t t 5 t 3 2 t 3 3 3 3 i 35 = 35 35 + 2 = 35 5 + 5 5 +2 = 5 and similarly, i. t 3 5. 6 = 25. t. 3 5. 9 25. t3 5. +2 =. 6 25. t. 3 5. + 2 (maximum). + 2 (minimum).. 67. (a). From the graph, it appears that the absolute maximum value is about i (0=75) = 0=32, and the absolute minimum value is i (0) = i(1) = 0; that is, at both endpoints.. (b) i ({) = {. { {2. i 0 ({) = { ·. 1 2{ ({ 2{2 ) + (2{ 2{2 ) 3{ 4{2 + { {2 = = . 2 2 2 {{ 2 {{ 2 { {2. So i 0 ({) = 0 3{ 4{2 = 0 {(3 4{) = 0 { = 0 or 34 . t t 2 3 = 3163 (maximum). i (0) = i (1) = 0 (minimum), and i 34 = 34 34 34 = 34 16 69. The density is dened as =. [since. 1000 mass = (in g@cm3 ). But a critical point of will also be a critical point of Y volume Y (W ). g gY = 1000Y 2 and Y is never 0], and Y is easier to differentiate than . gW gW. Y (W ) = 999=87 0=06426W + 0=0085043W 2 0=0000679W 3. Y 0 (W ) = 0=06426 + 0=0170086W 0=0002037W 2 .. Setting this equal to 0 and using the quadratic formula to nd W , we get 0=0170086 ± 0=01700862 4 · 0=0002037 · 0=06426 3=9665 C or 79=5318 C. Since we are only interested W = 2(0=0002037).
(6) F. 152. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. TX.10. in the region 0 C W 30 C, we check the density at the endpoints and at 3=9665 C: (0) (30) . 1000 1=00013; 999=87. 1000 1000 0=99625; (3=9665) 1=000255. So water has its maximum density at 1003=7628 999=7447. about 3=9665 C. 71. Let d = 0=000 032 37, e = 0=000 903 7, f = 0=008 956, g = 0=03629, h = 0=04458, and i = 0=4074.. Then V(w) = dw5 + ew4 + fw3 + gw2 + hw + i and V 0 (w) = 5dw4 + 4ew3 + 3fw2 + 2gw + h. We now apply the Closed Interval Method to the continuous function V on the interval 0 w 10. Since V 0 exists for all w, the only critical numbers of V occur when V 0 (w) = 0. We use a rootnder on a CAS (or a graphing device) to nd that V 0 (w) = 0 when w1 0=855, w2 4=618, w3 7=292, and w4 9=570. The values of V at these critical numbers are V(w1 ) 0=39, V(w2 ) 0=43645, V(w3 ) 0=427, and V(w4 ) 0=43641. The values of V at the endpoints of the interval are V(0) 0=41 and V(10) 0=435. Comparing the six numbers, we see that sugar was most expensive at w2 4=618 (corresponding roughly to March 1998) and cheapest at w1 0=855 (June 1994). 73. (a) y(u) = n(u0 u)u2 = nu0 u2 nu3. y 0 (u) = 2nu0 u 3nu2 . y 0 (u) = 0 nu(2u0 3u) = 0 u = 0 or 23 u0 (but 0 is not in the interval). Evaluating y at 12 u0 , 23 u0 , and u0 , we get y 12 u0 = 18 nu03 , y 23 u0 =. and y(u0 ) = 0. Since. 4 27. 4 nu03 , 27. A 18 , y attains its maximum value at u = 23 u0 . This supports the statement in the text.. (b) From part (a), the maximum value of y is. 4 nu03 . 27. (c). 75. i ({) = {101 + {51 + { + 1. i 0 ({) = 101{100 + 51{50 + 1 1 for all {, so i 0 ({) = 0 has no solution. Thus, i ({). has no critical number, so i ({) can have no local maximum or minimum. 77. If i has a local minimum at f, then j({) = i ({) has a local maximum at f, so j 0 (f) = 0 by the case of Fermat’s Theorem. proved in the text. Thus, i 0 (f) = j 0 (f) = 0.. 4.2 The Mean Value Theorem 1. i ({) = 5 12{ + 3{2 , [1, 3].. Since i is a polynomial, it is continuous and differentiable on R, so it is continuous on [1> 3]. and differentiable on (1> 3). Also i (1) = 4 = i (3). i 0 (f) = 0 12 + 6f = 0 f = 2, which is in the open interval (1> 3), so f = 2 satises the conclusion of Rolle’s Theorem..
(7) F. TX.10 3. i ({) =. SECTION 4.2. THE MEAN VALUE THEOREM. ¤. 153. { 13 {, [0> 9]. i , being the difference of a root function and a polynomial, is continuous and differentiable. on [0> ), so it is continuous on [0> 9] and differentiable on (0> 9). Also, i (0) = 0 = i (9). i 0 (f) = 0 2. 1 3 1 =0 2 f=3 f= 3 2 f. f=. 9 9 , which is in the open interval (0> 9), so f = satises the 4 4. conclusion of Rolle’s Theorem. 5. i ({) = 1 {2@3 .. i (1) = 1 (1)2@3 = 1 1 = 0 = i (1). i 0 ({) = 23 {1@3 , so i 0 (f) = 0 has no solution. This. does not contradict Rolle’s Theorem, since i 0 (0) does not exist, and so i is not differentiable on (1> 1). 7.. i (8) i (0) 64 1 = = . The values of f which satisfy i 0 (f) = 80 8 4. 9. (a), (b) The equation of the secant line is. |5 =. 8=5 5 ({ 1) | = 12 { + 92 . 81. 11. i ({) = 3{2 + 2{ + 5, [1> 1].. and differentiable on R. i 0 (f) =. 1 4. seem to be about f = 0=8, 3=2, 4=4, and 6=1.. (c) i ({) = { + 4@{ i 0 ({) = 1 4@{2 . So i 0 (f) = 12 f2 = 8 f = 2 2, and 4 i (f) = 2 2 + 2 = 3 2. Thus, an equation of the 2 tangent line is | 3 2 = 12 { 2 2 | = 12 { + 2 2.. i is continuous on [1> 1] and differentiable on (1> 1) since polynomials are continuous i(e) i (d) ed. 6f + 2 =. i (1) i (1) 10 6 = = 2 6f = 0 1 (1) 2. f = 0, which is in (1> 1). 13. i ({) = h2{ , [0> 3].. i is continuous and differentiable on R, so it is continuous on [0> 3] and differentiable on (0> 3). 1 h6 i (e) i (d) h6 h0 1 h6 2h2f = h2f = 2f = ln i 0 (f) = ed 30 6 6 1 h6 1 0=897, which is in (0> 3). f = ln 2 6.
(8) F. 154. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. 2. 15. i ({) = ({ 3). 6 3 = 4 (f 3)3. i 0 ({) = 2 ({ 3)3 . i (4) i (1) = i 0 (f)(4 1) . 1 1 2 = ·3 12 (2)2 (f 3)3. (f 3)3 = 8 f 3 = 2 f = 1, which is not in the open interval (1> 4). This does not. contradict the Mean Value Theorem since i is not continuous at { = 3. 17. Let i ({) = 1 + 2{ + {3 + 4{5 . Then i(1) = 6 ? 0 and i (0) = 1 A 0= Since i is a polynomial, it is continuous, so the. Intermediate Value Theorem says that there is a number f between 1 and 0 such that i (f) = 0= Thus, the given equation has a real root. Suppose the equation has distinct real roots d and e with d ? e. Then i(d) = i (e) = 0. Since i is a polynomial, it is differentiable on (d> e) and continuous on [d> e]. By Rolle’s Theorem, there is a number u in (d> e) such that i 0 (u) = 0. But i 0 ({) = 2 + 3{2 + 20{4 2 for all {, so i 0 ({) can never be 0. This contradiction shows that the equation can’t have two distinct real roots. Hence, it has exactly one real root. 19. Let i ({) = {3 15{ + f for { in [2> 2]. If i has two real roots d and e in [2> 2], with d ? e, then i (d) = i (e) = 0. Since. the polynomial i is continuous on [d> e] and differentiable on (d> e), Rolle’s Theorem implies that there is a number u in (d> e) such that i 0 (u) = 0. Now i 0 (u) = 3u2 15. Since u is in (d> e), which is contained in [2> 2], we have |u| ? 2, so u2 ? 4. It follows that 3u2 15 ? 3 · 4 15 = 3 ? 0. This contradicts i 0 (u) = 0, so the given equation can’t have two real roots in [2> 2]. Hence, it has at most one real root in [2> 2]. 21. (a) Suppose that a cubic polynomial S ({) has roots d1 ? d2 ? d3 ? d4 , so S (d1 ) = S (d2 ) = S (d3 ) = S (d4 ).. By Rolle’s Theorem there are numbers f1 , f2 , f3 with d1 ? f1 ? d2 , d2 ? f2 ? d3 and d3 ? f3 ? d4 and S 0 (f1 ) = S 0 (f2 ) = S 0 (f3 ) = 0. Thus, the second-degree polynomial S 0 ({) has three distinct real roots, which is impossible. (b) We prove by induction that a polynomial of degree q has at most q real roots. This is certainly true for q = 1. Suppose that the result is true for all polynomials of degree q and let S ({) be a polynomial of degree q + 1. Suppose that S ({) has more than q + 1 real roots, say d1 ? d2 ? d3 ? · · · ? dq+1 ? dq+2 . Then S (d1 ) = S (d2 ) = · · · = S (dq+2 ) = 0. By Rolle’s Theorem there are real numbers f1 > = = = > fq+1 with d1 ? f1 ? d2 > = = = > dq+1 ? fq+1 ? dq+2 and S 0 (f1 ) = · · · = S 0 (fq+1 ) = 0. Thus, the qth degree polynomial S 0 ({) has at least q + 1 roots. This contradiction shows that S ({) has at most q + 1 real roots. 23. By the Mean Value Theorem, i (4) i (1) = i 0 (f)(4 1) for some f (1> 4). But for every f (1> 4) we have. i 0 (f) 2. Putting i 0 (f) 2 into the above equation and substituting i (1) = 10, we get i (4) = i (1) + i 0 (f)(4 1) = 10 + 3i 0 (f) 10 + 3 · 2 = 16. So the smallest possible value of i (4) is 16. 25. Suppose that such a function i exists. By the Mean Value Theorem there is a number 0 ? f ? 2 with. i 0 (f) =. 5 i (2) i (0) = . But this is impossible since i 0 ({) 2 ? 20 2. 5 2. for all {, so no such function can exist..
(9) F. TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH SECT 27. We use Exercise 26 with i ({) =. ¤. 155. 1 + {, j({) = 1 + 12 {, and d = 0. Notice that i (0) = 1 = j(0) and. 1 1 ? = j 0 ({) for { A 0. So by Exercise 26, i (e) ? j(e) 1 + e ? 1 + 12 e for e A 0. 2 2 1+{ Another method: Apply the Mean Value Theorem directly to either i ({) = 1 + 12 { 1 + { or j({) = 1 + { on [0> e].. i 0 ({) =. 29. Let i ({) = sin { and let e ? d. Then i ({) is continuous on [e> d] and differentiable on (e> d). By the Mean Value Theorem,. there is a number f (e> d) with sin d sin e = i (d) i (e) = i 0 (f)(d e) = (cos f)(d e). Thus, |sin d sin e| |cos f| |e d| |d e|. If d ? e, then |sin d sin e| = |sin e sin d| |e d| = |d e|. If d = e, both sides of the inequality are 0. 31. For { A 0, i ({) = j({), so i 0 ({) = j0 ({). For { ? 0, i 0 ({) = (1@{)0 = 1@{2 and j0 ({) = (1 + 1@{)0 = 1@{2 , so. again i 0 ({) = j 0 ({). However, the domain of j({) is not an interval [it is ( > 0) (0> )] so we cannot conclude that i j is constant (in fact it is not). 33. Let i ({) = arcsin. {1 {+1. 2 arctan. { + 2 . Note that the domain of i is [0> ). Thus,. ({ + 1) ({ 1) 1 1 2 1 1 · = = 0. 2 2 1 + { 2 { { ({ + 1) { ({ + 1) ({ + 1) {1 1 {+1. i 0 ({) = v. Then i ({) = F on (0> ) by Theorem 5. By continuity of i , i ({) = F on [0> ). To nd F, we let { = 0 arcsin(1) 2 arctan(0) + 2 = F 2 0 + {1 arcsin = 2 arctan { 2 . {+1. 2. = 0 = F. Thus, i ({) = 0 . 35. Let j(w) and k(w) be the position functions of the two runners and let i(w) = j(w) k(w). By hypothesis,. i (0) = j(0) k(0) = 0 and i (e) = j(e) k(e) = 0, where e is the nishing time. Then by the Mean Value Theorem, there is a time f, with 0 ? f ? e, such that i 0 (f) =. i (e) i (0) . But i(e) = i(0) = 0, so i 0 (f) = 0. Since e0. i 0 (f) = j 0 (f) k0 (f) = 0, we have j0 (f) = k0 (f). So at time f, both runners have the same speed j 0 (f) = k0 (f).. 4.3 How Derivatives Affect the Shape of a Graph 1. (a) i is increasing on (1> 3) and (4> 6).. (c) i is concave upward on (0> 2).. (b) i is decreasing on (0> 1) and (3> 4). (d) i is concave downward on (2> 4) and (4> 6).. (e) The point of inection is (2> 3). 3. (a) Use the Increasing/Decreasing (I/D) Test.. (b) Use the Concavity Test.. (c) At any value of { where the concavity changes, we have an inection point at ({> i({))..
(10) F. 156. ¤. CHAPTER 4. TX.10. APPLICATIONS OF DIFFERENTIATION. 5. (a) Since i 0 ({) A 0 on (1> 5), i is increasing on this interval. Since i 0 ({) ? 0 on (0> 1) and (5> 6), i is decreasing on these. intervals. (b) Since i 0 ({) = 0 at { = 1 and i 0 changes from negative to positive there, i changes from decreasing to increasing and has a local minimum at { = 1. Since i 0 ({) = 0 at { = 5 and i 0 changes from positive to negative there, i changes from increasing to decreasing and has a local maximum at { = 5. 7. There is an inection point at { = 1 because i 00 ({) changes from negative to positive there, and so the graph of i changes. from concave downward to concave upward. There is an inection point at { = 7 because i 00 ({) changes from positive to negative there, and so the graph of i changes from concave upward to concave downward. 9. (a) i ({) = 2{3 + 3{2 36{. i 0 ({) = 6{2 + 6{ 36 = 6({2 + { 6) = 6({ + 3)({ 2).. We don’t need to include the “6” in the chart to determine the sign of i 0 ({). Interval. {+3. {2. i 0 ({). i. { ? 3. . . +. increasing on ( > 3). 3 ? { ? 2. +. . . decreasing on (3> 2). +. +. +. increasing on (2> ). {A2. (b) i changes from increasing to decreasing at { = 3 and from decreasing to increasing at { = 2. Thus, i (3) = 81 is a local maximum value and i (2) = 44 is a local minimum value. (c) i 0 ({) = 6{2 + 6{ 36 i 00 ({) = 12{ + 6. i 00 ({) = 0 at { = 12 , i 00 ({) A 0 { A 12 , and i 00 ({) ? 0 { ? 12 . Thus, i is concave upward on 12 > and concave downward on > 12 . There is an inection point at 12 > i 12 = 12 > 37 . 2 11. (a) i ({) = {4 2{2 + 3. i 0 ({) = 4{3 4{ = 4{ {2 1 = 4{({ + 1)({ 1).. Interval. {+1. {. {1. i 0 ({). i. { ? 1. . . . . decreasing on ( > 1). 1 ? { ? 0. +. . . +. increasing on (1> 0). 0?{?1. +. +. . . decreasing on (0> 1). {A1. +. +. +. +. increasing on (1> ). (b) i changes from increasing to decreasing at { = 0 and from decreasing to increasing at { = 1 and { = 1. Thus, i (0) = 3 is a local maximum value and i (±1) = 2 are local minimum values. (c) i 00 ({) = 12{2 4 = 12 {2 13 = 12 { + 1@ 3 { 1@ 3 . i 00 ({) A 0 { ? 1@ 3 or { A 1@ 3 and 3@3> and concave i 00 ({) ? 0 1@ 3 ? { ? 1@ 3. Thus, i is concave upward on > 3@3 and . downward on 3@3> 3@3 . There are inection points at ± 3@3> 22 9.
(11) F. TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH SECT 13. (a) i ({) = sin { + cos {, 0 { 2 .. i 0 ({) = cos { sin { = 0 cos { = sin { 1 =. sin { cos {. ¤. 157. . Thus, i 0 ({) A 0 cos { sin { A 0 cos { A sin { 0 ? { ? 4 or 5 ? { ? 2 and i 0 ({) ? 0 cos { ? sin { 4 ? { ? 5 . So i is increasing on 0> 4 and 5 > 2 and i 4 4 4 . is decreasing on 4 > 5 4 tan { = 1 { =. 4. or. 5 . 4. . 2 is a. (b) i changes from increasing to decreasing at { = 4 and from decreasing to increasing at { = local maximum value and i 5 = 2 is a local minimum value. 4. 5 . 4. Thus, i. (c) i 00 ({) = sin { cos { = 0 sin { = cos { tan { = 1 { =. 7 . 4. Divide the interval. 3 4. or. 4. =. (0> 2 ) into subintervals with these numbers as endpoints and complete a second derivative chart. Interval 3 0> 4 3 7 > 4 4 7 > 2 4 There are inection points at 15. (a) i ({) = h2{ + h{. 3 4. i 00 ({) = sin { cos { i 00 2 = 1 ? 0 i 00 ( ) = 1 A 0 1 i 00 11 = 2 6. 1 2. 3?0. Concavity downward upward downward. > 0 and 7 >0 . 4. i 0 ({) = 2h2{ h{ . i 0 ({) A 0 2h2{ A h{. h3{ A. 1 2. 3{ A ln 12 { A 13 (ln 1 ln 2) { A 13 ln 2 [ 0=23] and i 0 ({) ? 0 if { ? 13 ln 2. So i is increasing on 13 ln 2> and i is decreasing on > 13 ln 2 .. (b) i changes from decreasing to increasing at { = 13 ln 2. Thus, s s 3 3 3 3 i 13 ln 2 = i ln 3 1@2 = h2 ln 1@2 + h ln 1@2 = hln 1@4 + hln 2 = 3 1@4 + 3 2 = 22@3 + 21@3 [ 1=89] is a local minimum value. (c) i 00 ({) = 4h2{ + h{ A 0 [the sum of two positive terms]. Thus, i is concave upward on ( > ) and there is no point of inection. ln {. 17. (a) | = i ({) = . (Note that i is only dened for { A 0.). { { (1@{) ln { 12 {1@2. 1 ln { 2 { 2 ln { { 2 { 0 = · = i ({) = A 0 2 ln { A 0 { { 2{3@2 2 { ln { ? 2 { ? h2 . Therefore i is increasing on 0> h2 and decreasing on h2 > . ln h2 2 (b) i changes from increasing to decreasing at { = h2 , so i (h2 ) = = is a local maximum value. 2 h h (c) i 00 ({) =. 2{3@2 (1@{) (2 ln {)(3{1@2 ) 2{1@2 + 3{1@2 (ln { 2) {1@2 (2 + 3 ln { 6) 3 ln { 8 = = = 2 3 3@2 4{ 4{3 4{5@2 (2{ ). i 00 ({) = 0 ln { =. { = h8@3 . i 00 ({) A 0 { A h8@3 , so i is concave upward on (h8@3 > ) and concave downward on (0> h8@3 ). There is an inection point at h8@3 > 83 h4@3 (14=39> 0=70). 8 3.
(12) F. 158. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. 19. i ({) = {5 5{ + 3. TX.10. i 0 ({) = 5{4 5 = 5({2 + 1)({ + 1)({ 1).. First Derivative Test: i 0 ({) ? 0 1 ? { ? 1 and i 0 ({) A 0 { A 1 or { ? 1. Since i 0 changes from positive to negative at { = 1, i (1) = 7 is a local maximum value; and since i 0 changes from negative to positive at { = 1, i (1) = 1 is a local minimum value. Second Derivative Test: i 00 ({) = 20{3 . i 0 ({) = 0 { = ±1. i 00 (1) = 20 ? 0 i (1) = 7 is a local maximum value. i 00 (1) = 20 A 0 i (1) = 1 is a local minimum value. Preference: For this function, the two tests are equally easy. 21. i ({) = { +. 1 { i 0 ({) = 1 + 12 (1 {)1@2 (1) = 1 . for { 1. i 0 ({) = 0 2. 1{= 1 1{=. 1 2. 2. 1 . Note that i is dened for 1 { 0; that is, 1{. 1{=. 1 4. { = 34 . i 0 does not exist at { = 1,. but we can’t have a local maximum or minimum at an endpoint. First Derivative Test: i 0 ({) A 0 { ? 34 and i 0 ({) ? 0 34 ? { ? 1. Since i 0 changes from positive to negative at { = 34 , i 34 = 54 is a local maximum value. 1 Second Derivative Test: i 00 ({) = 12 12 (1 {)3@2 (1) = 3 . 4 1{ i 00 34 = 2 ? 0 i 34 = 54 is a local maximum value. Preference: The First Derivative Test may be slightly easier to apply in this case. 23. (a) By the Second Derivative Test, if i 0 (2) = 0 and i 00 (2) = 5 ? 0, i has a local maximum at { = 2.. (b) If i 0 (6) = 0, we know that i has a horizontal tangent at { = 6. Knowing that i 00 (6) = 0 does not provide any additional information since the Second Derivative Test fails. For example, the rst and second derivatives of | = ({ 6)4 , | = ({ 6)4 , and | = ({ 6)3 all equal zero for { = 6, but the rst has a local minimum at { = 6, the second has a local maximum at { = 6, and the third has an inection point at { = 6. 25. i 0 (0) = i 0 (2) = i 0 (4) = 0. horizontal tangents at { = 0, 2, 4.. 0. i ({) A 0 if { ? 0 or 2 ? { ? 4 i is increasing on ( > 0) and (2> 4). i 0 ({) ? 0 if 0 ? { ? 2 or { A 4 i is decreasing on (0> 2) and (4> ). i 00 ({) A 0 if 1 ? { ? 3 i is concave upward on (1> 3). i 00 ({) ? 0 if { ? 1 or { A 3 i is concave downward on ( > 1) and (3> ). There are inection points when { = 1 and 3. 27. i 0 ({) A 0 if |{| ? 2. i is increasing on (2> 2).. 0. i ({) ? 0 if |{| A 2 i is decreasing on ( > 2) and (2> ). i 0 (2) = 0 horizontal tangent at { = 2. lim |i 0 ({)| = there is a vertical asymptote or. {2. vertical tangent (cusp) at { = 2. i 00 ({) A 0 if { 6= 2 i is concave upward on ( > 2) and (2> )..
(13) F. TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH SECT. ¤. 159. 29. The function must be always decreasing (since the rst derivative is always negative). and concave downward (since the second derivative is always negative).. 31. (a) i is increasing where i 0 is positive, that is, on (0> 2), (4> 6), and (8> ); and decreasing where i 0 is negative, that is,. on (2> 4) and (6> 8). (b) i has local maxima where i 0 changes from positive to negative, at { = 2 and at { = 6, and local minima where i 0 changes from negative to positive, at { = 4 and at { = 8. (c) i is concave upward (CU) where i 0 is increasing, that is, on (3> 6) and (6> ),. (e). and concave downward (CD) where i 0 is decreasing, that is, on (0> 3).. (d) There is a point of inection where i changes from being CD to being CU, that is, at { = 3. 33. (a) i ({) = 2{3 3{2 12{. i 0 ({) = 6{2 6{ 12 = 6({2 { 2) = 6({ 2)({ + 1).. i 0 ({) A 0 { ? 1 or { A 2 and i 0 ({) ? 0 1 ? { ? 2. So i is increasing on ( > 1) and (2> ), and i is decreasing on (1> 2). (b) Since i changes from increasing to decreasing at { = 1, i (1) = 7 is a local. (d). maximum value. Since i changes from decreasing to increasing at { = 2, i (2) = 20 is a local minimum value. (c) i 00 ({) = 6(2{ 1) i 00 ({) A 0 on 12 > and i 00 ({) ? 0 on > 12 . So i is concave upward on 12 > and concave downward on > 12 . There . is a change in concavity at { = 12 , and we have an inection point at 12 > 13 2 35. (a) i ({) = 2 + 2{2 {4. i 0 ({) = 4{ 4{3 = 4{(1 {2 ) = 4{(1 + {)(1 {). i 0 ({) A 0 { ? 1 or. 0 ? { ? 1 and i 0 ({) ? 0 1 ? { ? 0 or { A 1. So i is increasing on ( > 1) and (0> 1) and i is decreasing on (1> 0) and (1> ). (b) i changes from increasing to decreasing at { = 1 and { = 1, so i (1) = 3 and i (1) = 3 are local maximum values. i changes from decreasing to increasing at { = 0, so i (0) = 2 is a local minimum value..
(14) F. 160. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. TX.10. (c) i 00 ({) = 4 12{2 = 4(1 3{2 ). i 00 ({) = 0 1 3{2 = 0 {2 = 13 { = ±1@ 3. i 00 ({) A 0 on 1@ 3> 1@ 3 and i 00 ({) ? 0. (d). on > 1@ 3 and 1@ 3> . So i is concave upward on 1@ 3> 1@ 3 and i is concave downward on > 1@ 3 and 1@ 3> . i ±1@ 3 = 2 +. 2 3. . 1 9. =. 23 . 9. There are points of inection. . at ±1@ 3> 23 9 37. (a) k({) = ({ + 1)5 5{ 2. k0 ({) = 5({ + 1)4 5. k0 ({) = 0 5({ + 1)4 = 5 ({ + 1)4 = 1 . ({ + 1)2 = 1 { + 1 = 1 or { + 1 = 1 { = 0 or { = 2. k0 ({) A 0 { ? 2 or { A 0 and k0 ({) ? 0 2 ? { ? 0. So k is increasing on ( > 2) and (0> ) and k is decreasing on (2> 0). (b) k(2) = 7 is a local maximum value and k(0) = 1 is a local minimum value.. (d). (c) k00 ({) = 20({ + 1)3 = 0 { = 1. k00 ({) A 0 { A 1 and k00 ({) ? 0 { ? 1, so k is CU on (1> ) and k is CD on ( > 1). There is a point of inection at (1> k(1)) = (1> 3).. 39. (a) D({) = {. { + 3 D0 ({) = { · 12 ({ + 3)1@2 + { + 3 · 1 =. 3{ + 6 { { + 2({ + 3) + {+3= = . 2 {+3 2 {+3 2 {+3. The domain of D is [3> ). D0 ({) A 0 for { A 2 and D0 ({) ? 0 for 3 ? { ? 2, so D is increasing on (2> ) and decreasing on (3> 2). (b) D(2) = 2 is a local minimum value. 2 (c) D00 ({) = =. (d). 1 { + 3 · 3 (3{ + 6) · {+3 2 2 {+3. 6({ + 3) (3{ + 6) 3{ + 12 3({ + 4) = = 4({ + 3)3@2 4({ + 3)3@2 4({ + 3)3@2. D00 ({) A 0 for all { A 3, so D is concave upward on (3> ). There is no inection point. 41. (a) F({) = {1@3 ({ + 4) = {4@3 + 4{1@3. F 0 ({) = 43 {1@3 + 43 {2@3 = 43 {2@3 ({ + 1) =. 4({ + 1) . F 0 ({) A 0 if 3 3 {2. 1 ? { ? 0 or { A 0 and F 0 ({) ? 0 for { ? 1, so F is increasing on (1> ) and F is decreasing on ( > 1)..
(15) F. SECTION 4.3 SECT. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH. (b) F(1) = 3 is a local minimum value.. ¤. 161. (d). (c) F 00 ({) = 49 {2@3 89 {5@3 = 49 {5@3 ({ 2) =. 4({ 2) . 3 9 {5. F 00 ({) ? 0 for 0 ? { ? 2 and F 00 ({) A 0 for { ? 0 and { A 2, so F is concave downward on (0> 2) and concave upward on ( > 0) and (2> ). There are inection points at (0> 0) and 2> 6 3 2 (2> 7=56). 43. (a) i () = 2 cos + cos2 , 0 2 . i 0 () = 2 sin + 2 cos ( sin ) = 2 sin (1 + cos ).. i 0 () = 0 = 0> > and 2 . i 0 () A 0 ? ? 2 and i 0 () ? 0 0 ? ? . So i is increasing on ( > 2 ) and i is decreasing on (0> ). (b) i ( ) = 1 is a local minimum value. (c) i 0 () = 2 sin (1 + cos ) i 00 () = 2 sin ( sin ) + (1 + cos )(2 cos ) = 2 sin2 2 cos 2 cos2 = 2(1 cos2 ) 2 cos 2 cos2 = 4 cos2 2 cos + 2 = 2(2 cos2 + cos 1) = 2(2 cos 1)(cos + 1) Since 2(cos + 1) ? 0 [for 6= ], i 00 () A 0 2 cos 1 ? 0 cos ? 12 3 ? ? 5 3 and and i is CD on 0> 3 and ? ? 2 . So i is CU on 3 > 5 i 00 () ? 0 cos A 12 0 ? ? 3 or 5 3 3 5 5 5 > 2 . There are points of inection at 3 > i 3 = 3 > 54 and 5 > i 5 = 3>4 . 3 3 3 (d). 45. i ({) =. {2 {2 = has domain ( > 1) (1> 1) (1> ). 1 ({ + 1)({ 1). {2. {2@{2 1 1 = 1, so | = 1 is a HA. = lim = {± ({2 1)@{2 {± 1 1@{2 10. (a) lim i ({) = lim {±. lim. {1. lim. {2 = since {2
(16) 1 and ({2 1)
(17) 0+ as {
(18) 1 , so { = 1 is a VA. 1. {2. {2 = since {2
(19) 1 and ({2 1)
(20) 0+ as {
(21) 1+ , so { = 1 is a VA. 1. {1+ {2. (b) i ({) =. {2 1. {2. i 0 ({) =. ({2 1)(2{) {2 (2{) 2{[({2 1) {2 ] 2{ = = 2 . Since ({2 1)2 is 2 2 ({ 1) ({2 1)2 ({ 1)2. positive for all { in the domain of i , the sign of the derivative is determined by the sign of 2{. Thus, i 0 ({) A 0 if { ? 0 ({ 6= 1) and i 0 ({) ? 0 if { A 0 ({ 6= 1). So i is increasing on ( > 1) and (1> 0), and i is decreasing on (0> 1) and (1> ). (c) i 0 ({) = 0 { = 0 and i (0) = 0 is a local maximum value..
(22) F. 162. ¤. CHAPTER 4. (d) i 00 ({) = =. APPLICATIONS OF DIFFERENTIATION. TX.10. ({2 1)2 (2) (2{) · 2({2 1)(2{) [({2 1)2 ]2. (e). 2({2 1)[({2 1) + 4{2 ] 2(3{2 + 1) = . ({2 1)4 ({2 1)3. The sign of i 00 ({) is determined by the denominator; that is, i 00 ({) A 0 if |{| A 1 and i 00 ({) ? 0 if |{| ? 1. Thus, i is CU on ( > 1) and (1> ), and i is CD on (1> 1). There are no inection points. 47. (a) lim {2 + 1 { = and { {2 + 1 + { 1 2 2 = lim = 0, so | = 0 is a HA. { + 1 { = lim { +1{ lim { { {2 + 1 + { { {2 + 1 + { { { (b) i ({) = {2 + 1 { i 0 ({) = 1. Since ? 1 for all {, i 0 ({) ? 0, so i is decreasing on R. 2 2 { +1 { +1 (c) No minimum or maximum (d) i 00 ({) =. ({2 + 1)1@2 (1) { · 12 ({2 + 1)1@2 (2{) 2 {2 + 1 ({2 + 1)1@2 . =. {2. ({2. +1. {2 + 1)1@2. =. (e). ({2 + 1) {2 1 = 2 A 0, ({2 + 1)3@2 ({ + 1)3@2. so i is CU on R. No IP 49. i ({) = ln(1 ln {) is dened when { A 0 (so that ln { is dened) and 1 ln { A 0 [so that ln(1 ln {) is dened].. The second condition is equivalent to 1 A ln { { ? h> so i has domain (0> h). (a) As {
(23) 0+ > ln {
(24) > so 1 ln {
(25) and i ({)
(26) = As {
(27) h > ln {
(28) 1 > so 1 ln {
(29) 0+ and i ({)
(30) . Thus, { = 0 and { = h are vertical asymptotes. There is no horizontal asymptote. 1 1 1 ? 0 on (0> h) = Thus, i is decreasing on its domain, (0> h) = (b) i 0 ({) = = 1 ln { { {(1 ln {) (c) i 0 ({) 6= 0 on (0> h) > so i has no local maximum or minimum value.. (e). [{(1 ln {)]0 {(1@{) + (1 ln {) = {2 (1 ln {)2 [{(1 ln {)]2 ln { = 2 { (1 ln {)2. (d) i 00 ({) = . so i 00 ({) A 0 ln { ? 0 0 ? { ? 1= Thus, i is CU on (0> 1) and CD on (1> h) = There is an inection point at (1> 0) = 51. (a). lim h1@({+1) = 1 since 1@({ + 1)
(31) 0, so | = 1 is a HA.. {±. lim h1@({+1) = 0 since 1@({ + 1)
(32) ,. {1+. lim h1@({+1) = since 1@({ + 1)
(33) , so { = 1 is a VA. 1 [Reciprocal Rule] = h1@({+1) @({ + 1)2 (b) i ({) = h1@({+1) i 0 ({) = h1@({+1) (1) ({ + 1)2 {1. i 0 ({) A 0 for all { except 1, so i is increasing on ( > 1) and (1> ). (c) There is no local maximum or minimum.. .
(34) F. TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH SECT (d) i 00 ({) = =. ({ + 1)2 h1@({+1) 1@({ + 1)2 h1@({+1) [2({ + 1)]. (e). [({ + 1)2 ]2 h1@({+1) [1 (2{ + 2)] h1@({+1) (2{ + 1) = ({ + 1)4 ({ + 1)4. ¤. . i 00 ({) A 0 2{ + 1 ? 0 { ? 12 , so i is CU on ( > 1) and 1> 12 , and CD on 12 , . i has an IP at 12 > h2 . 53. The nonnegative factors ({ + 1)2 and ({ 6)4 do not affect the sign of i 0 ({) = ({ + 1)2 ({ 3)5 ({ 6)4 .. So i 0 ({) A 0 ({ 3)5 A 0 { 3 A 0 { A 3. Thus, i is increasing on the interval (3> ). From the graph, we get an estimate of i(1) 1=41 as a local maximum. 55. (a). value, and no local minimum value. {+1 i ({) = {2 + 1. 1{. i 0 ({) =. i 0 ({) = 0 { = 1. i (1) =. . + 1)3@2 = 2 is the exact value.. ({2 2 2. (b) From the graph in part (a), i increases most rapidly somewhere between { = 12 and { = 14 . To nd the exact value, we need to nd the maximum value of i 0 , which we can do by nding the critical numbers of i 0 . 3 + 17 2{2 3{ 1 3 ± 17 i 00 ({) = . { = corresponds to the minimum value of i 0 . = 0 { = 4 4 ({2 + 1)5@2 t The maximum value of i 0 is at 3 4 17 > 76 617 (0=28> 0=69). 57. i ({) = cos { +. (a). 1 2. cos 2{ i 0 ({) = sin { sin 2{ i 00 ({) = cos { 2 cos 2{ From the graph of i , it seems that i is CD on (0> 1), CU on (1> 2=5), CD on (2=5> 3=7), CU on (3=7> 5=3), and CD on (5=3> 2 ). The points of inection appear to be at (1> 0=4), (2=5> 0=6), (3=7> 0=6), and (5=3> 0=4).. (b). From the graph of i 00 (and zooming in near the zeros), it seems that i is CD on (0> 0=94), CU on (0=94> 2=57), CD on (2=57> 3=71), CU on (3=71> 5=35), and CD on (5=35> 2 ). Rened estimates of the inection points are (0=94> 0=44), (2=57> 0=63), (3=71> 0=63), and (5=35> 0=44).. 59. In Maple, we dene i and then use the command. plot(diff(diff(f,x),x),x=-2..2);. In Mathematica, we dene i and then use Plot[Dt[Dt[f,x],x],{x,-2,2}]. We see that i 00 A 0 for { ? 0=6 and { A 0=0 [ 0=03] and i 00 ? 0 for 0=6 ? { ? 0=0. So i is CU on ( > 0=6) and (0=0> ) and CD on (0=6> 0=0).. 163.
(35) F. 164. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. 61. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about. w = 8 hours, and decreases toward 0 as the population begins to level off. (b) The rate of increase has its maximum value at w = 8 hours. (c) The population function is concave upward on (0> 8) and concave downward on (8> 18). (d) At w = 8, the population is about 350, so the inection point is about (8> 350). 63. Most students learn more in the third hour of studying than in the eighth hour, so N(3) N(2) is larger than N(8) N(7).. In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so N 0 (w) decreases and the graph of N is concave downward. 65. V(w) = Dws hnw with D = 0=01, s = 4, and n = 0=07. We will nd the. zeros of i 00 for i(w) = ws hnw . i 0 (w) = ws (nhnw ) + hnw (sws1 ) = hnw (nws + sws1 ) i 00 (w) = hnw (nsws1 + s(s 1)ws2 ) + (nws + sws1 )(nhnw ) = ws2 hnw [nsw + s(s 1) + n2 w2 nsw] = ws2 hnw (n2 w2 2nsw + s2 s) Using the given values of s and n gives us i 00 (w) = w2 h0=07w (0=0049w2 0=56w + 12). So V 00 (w) = 0=01i 00 (w) and its zeros are w = 0 and the solutions of 0=0049w2 0=56w + 12 = 0, which are w1 =. 200 7. 28=57 and w2 =. 600 7. 85=71.. At w1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at w2 minutes, the rate of decrease is the greatest. 67. i ({) = d{3 + e{2 + f{ + g. i 0 ({) = 3d{2 + 2e{ + f.. We are given that i(1) = 0 and i (2) = 3, so i (1) = d + e + f + g = 0 and i (2) = 8d + 4e 2f + g = 3. Also i 0 (1) = 3d + 2e + f = 0 and i 0 (2) = 12d 4e + f = 0 by Fermat’s Theorem. Solving these four equations, we get d = 29 , e = 13 , f = 43 , g = 79 , so the function is i ({) = 19 2{3 + 3{2 12{ + 7 . 69. | =. 1+{ 1 + {2. | 00 = =. |0 =. (1 + {2 )(1) (1 + {)(2{) 1 2{ {2 = 2 2 (1 + { ) (1 + {2 )2. . (1 + {2 )2 (2 2{) (1 2{ {2 ) · 2(1 + {2 )(2{) 2(1 + {2 )[(1 + {2 )(1 {) (1 2{ {2 )(2{)] = 2 2 2 [(1 + { ) ] (1 + {2 )4 2(1 { {2 {3 2{ + 4{2 + 2{3 ) 2({3 + 3{2 3{ 1) 2({ 1)({2 + 4{ + 1) = = 2 3 2 3 (1 + { ) (1 + { ) (1 + {2 )3. So | 00 = 0 { = 1, 2 ±. 3. Let d = 2 3, e = 2 + 3, and f = 1= We can show that i (d) = 14 1 3 ,.
(36) F. TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH SECT i (e) =. 1 4. ¤. 165. 1 + 3 , and i (f) = 1. To show that these three points of inection lie on one straight line, we’ll show that the. slopes pdf and pef are equal. 3 1 14 1 3 + 14 3 i (f) i (d) 1 4 = = pdf = = fd 4 1 2 3 3+ 3 3 1 14 1 + 3 14 3 i (f) i (e) 1 = 4 = = pef = fe 4 1 2 + 3 3 3 71. Suppose that i is differentiable on an interval L and i 0 ({) A 0 for all { in L except { = f. To show that i is increasing on L,. let {1 , {2 be two numbers in L with {1 ? {2 . Case 1 {1 ? {2 ? f. Let M be the interval {{ L | { ? f}. By applying the Increasing/Decreasing Test to i on M, we see that i is increasing on M, so i ({1 ) ? i ({2 ). Case 2 f ? {1 ? {2 . Apply the Increasing/Decreasing Test to i on N = {{ L | { A f}. Case 3 {1 ? {2 = f. Apply the proof of the Increasing/Decreasing Test, using the Mean Value Theorem (MVT) on the interval [{1 > {2 ] and noting that the MVT does not require i to be differentiable at the endpoints of [{1 > {2 ]. Case 4 f = {1 ? {2 . Same proof as in Case 3. Case 5 {1 ? f ? {2 . By Cases 3 and 4, i is increasing on [{1 > f] and on [f> {2 ], so i ({1 ) ? i (f) ? i({2 ). In all cases, we have shown that i ({1 ) ? i({2 ). Since {1 , {2 were any numbers in L with {1 ? {2 , we have shown that i is increasing on L. 73. (a) Since i and j are positive, increasing, and CU on L with i 00 and j 00 never equal to 0, we have i A 0, i 0 0, i 00 A 0,. j A 0, j0 0, j 00 A 0 on L. Then (i j)0 = i 0 j + i j 0. (ij)00 = i 00 j + 2i 0 j 0 + i j 00 i 00 j + i j00 A 0 on L. . i j is CU on L. (b) In part (a), if i and j are both decreasing instead of increasing, then i 0 0 and j 0 0 on L, so we still have 2i 0 j 0 0 on L. Thus, (i j)00 = i 00 j + 2i 0 j0 + ij 00 i 00 j + i j 00 A 0 on L. i j is CU on L as in part (a).. (c) Suppose i is increasing and j is decreasing [with i and j positive and CU]. Then i 0 0 and j0 0 on L, so 2i 0 j0 0 on L and the argument in parts (a) and (b) fails. Example 1.. L = (0> ), i ({) = {3 , j({) = 1@{. Then (i j)({) = {2 , so (i j)0 ({) = 2{ and (ij)00 ({) = 2 A 0 on L. Thus, i j is CU on L.. Example 2.. Example 3.. L = (0> ), i ({) = 4{ {, j({) = 1@{. Then (i j)({) = 4 {, so (ij)0 ({) = 2@ { and (ij)00 ({) = 1@ {3 ? 0 on L. Thus, i j is CD on L. L = (0> ), i ({) = {2 , j({) = 1@{. Thus, (i j)({) = {, so i j is linear on L.. 75. i ({) = tan { {. i 0 ({) = sec2 { 1 A 0 for 0 ? { ?. on 0> 2 . Thus, i ({) A i (0) = 0 for 0 ? { ?. 2. 2. since sec2 { A 1 for 0 ? { ?. 2.. So i is increasing. tan { { A 0 tan { A { for 0 ? { ?. . 2.
(37) F. 166. ¤. CHAPTER 4. TX.10. APPLICATIONS OF DIFFERENTIATION. 77. Let the cubic function be i({) = d{3 + e{2 + f{ + g. i 0 ({) = 3d{2 + 2e{ + f i 00 ({) = 6d{ + 2e.. So i is CU when 6d{ + 2e A 0 { A e@(3d), CD when { ? e@(3d), and so the only point of inection occurs when { = e@(3d). If the graph has three {-intercepts {1 , {2 and {3 , then the expression for i ({) must factor as i ({) = d({ {1 )({ {2 )({ {3 ). Multiplying these factors together gives us i ({) = d[{3 ({1 + {2 + {3 ){2 + ({1 {2 + {1 {3 + {2 {3 ){ {1 {2 {3 ] Equating the coefcients of the {2 -terms for the two forms of i gives us e = d({1 + {2 + {3 ). Hence, the {-coordinate of the point of inection is . d({1 + {2 + {3 ) {1 + {2 + {3 e = = . 3d 3d 3. 79. By hypothesis j = i 0 is differentiable on an open interval containing f. Since (f> i (f)) is a point of inection, the concavity. changes at { = f, so i 00 ({) changes signs at { = f. Hence, by the First Derivative Test, i 0 has a local extremum at { = f. Thus, by Fermat’s Theorem i 00 (f) = 0. 81. Using the fact that |{| =. {2 , we have that j({) = { {2. j 0 ({) =. {2 + {2 = 2 {2 = 2 |{| . 1@2 2{ ? 0 for { ? 0 and j00 ({) A 0 for { A 0, so (0> 0) is an inection point. But j 00 (0) does not = j 00 ({) = 2{ {2 |{| exist. 83. (a) i ({) = {4 sin. 1 {. i 0 ({) = {4 cos. 1 {. 1 1 1 1 2 + sin (4{3 ) = 4{3 sin {2 cos . { { { {. 1 j({) = {4 2 + sin = 2{4 + i({) j 0 ({) = 8{3 + i 0 ({). { 1 4 k({) = { 2 + sin = 2{4 + i ({) k0 ({) = 8{3 + i 0 ({). {. 1 {4 sin 0 1 i ({) i (0) { = lim = lim {3 sin . Since It is given that i (0) = 0, so i (0) = lim {0 {0 {0 {0 { { 1 {3 {3 sin {3 and lim {3 = 0, we see that i 0 (0) = 0 by the Squeeze Theorem. Also, {0 { 0. j 0 (0) = 8(0)3 + i 0 (0) = 0 and k0 (0) = 8(0)3 + i 0 (0) = 0, so 0 is a critical number of i , j, and k. For {2q =. 1 1 1 [q a nonzero integer], sin = sin 2q = 0 and cos = cos 2q = 1, so i 0 ({2q ) = {22q ? 0. 2q {2q {2q. For {2q+1 =. 1 1 1 , sin = sin(2q + 1) = 0 and cos = cos(2q + 1) = 1, so (2q + 1) {2q+1 {2q+1. i 0 ({2q+1 ) = {22q+1 A 0. Thus, i 0 changes sign innitely often on both sides of 0. Next, j 0 ({2q ) = 8{32q + i 0 ({2q ) = 8{32q {22q = {22q (8{2q 1) ? 0 for {2q ? 18 , but j 0 ({2q+1 ) = 8{32q+1 + {22q+1 = {22q+1 (8{2q+1 + 1) A 0 for {2q+1 A 18 , so j 0 changes sign innitely often on both sides of 0. Last, k0 ({2q ) = 8{32q + i 0 ({2q ) = 8{32q {22q = {22q (8{2q + 1) ? 0 for {2q A 18 and k0 ({2q+1 ) = 8{32q+1 + {22q+1 = {22q+1 (8{2q+1 + 1) A 0 for {2q+1 ? 18 , so k0 changes sign innitely often on both sides of 0..
(38) F. TX.10SSECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE (b) i (0) = 0 and since sin. ¤. 167. 1 1 and hence {4 sin is both positive and negative ininitely often on both sides of 0, and { {. arbitrarily close to 0, i has neither a local maximum nor a local minimum at 0. 1 1 4 Since 2 + sin 1, j({) = { 2 + sin A 0 for { 6= 0, so j(0) = 0 is a local minimum. { { 1 1 ? 0 for { 6= 0, so k(0) = 0 is a local maximum. Since 2 + sin 1, k({) = {4 2 + sin { {. 4.4 Indeterminate Forms and L'Hospital's Rule H. Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign: = 1. (a) lim. {d. i ({) 0 is an indeterminate form of type . j({) 0. (b) lim. i ({) = 0 because the numerator approaches 0 while the denominator becomes large. s({). (c) lim. k({) = 0 because the numerator approaches a nite number while the denominator becomes large. s({). {d. {d. (d) If lim s({) = and i ({)
(39) 0 through positive values, then lim {d. {d. and i({) = {2 .] If i ({)
(40) 0 through negative values, then lim. {d. s({) = . [For example, take d = 0, s({) = 1@{2 , i ({). s({) = . [For example, take d = 0, s({) = 1@{2 , i({). and i ({) = {2 .] If i ({)
(41) 0 through both positive and negative values, then the limit might not exist. [For example, take d = 0, s({) = 1@{2 , and i ({) = {.] (e) lim. {d. s({). is an indeterminate form of type . t({). 3. (a) When { is near d, i({) is near 0 and s({) is large, so i ({) s({) is large negative. Thus, lim [i ({) s({)] = . {d. (b) lim [ s({) t({)] is an indeterminate form of type . {d. (c) When { is near d, s({) and t({) are both large, so s({) + t({) is large. Thus, lim [ s({) + t({)] = . {d. 5. This limit has the form 00 . We can simply factor and simplify to evaluate the limit.. lim. {1. {2 1 ({ + 1)({ 1) {+1 1+1 = lim = lim = =2 {1 {2 { {1 {({ 1) { 1. 7. This limit has the form 00 . lim. {1. 9. This limit has the form 00 .. {9 1 H 9{8 9 9 9 = lim = lim {4 = (1) = 5 { 1 {1 5{4 5 {1 5 5. lim. {( @2)+. 11. This limit has the form 00 . lim. w0. cos { H sin { = lim = lim tan { = . 1 sin { {( @2)+ cos { {( @2)+. hw 1 H hw = lim 2 = since hw
(42) 1 and 3w2
(43) 0+ as w
(44) 0. w0 3w w3.
(45) F. 168. ¤. CHAPTER 4. TX.10. APPLICATIONS OF DIFFERENTIATION. tan s{ H s sec2 s{ s(1)2 s = = lim = 2 {0 tan t{ t sec t{ t(1)2 t. 13. This limit has the form 00 . lim. {0. 15. This limit has the form. lim . {. ln { H 1@{ 2 = lim 1 1@2 = lim = 0 { { { { { 2. 17. lim [(ln {)@{] = since ln {
(46) as {
(47) 0+ and dividing by small values of { just increases the magnitude of the {0+. quotient (ln {)@{. L’Hospital’s Rule does not apply. 19. This limit has the form. lim . {. h{ H h{ H h{ H h{ = = lim = lim = lim 3 2 { 3{ { 6{ { 6 {. h{ 1 { H h{ 1 H h{ 1 = = lim = lim {0 {0 2 {2 2{ 2. 21. This limit has the form 00 . lim. {0. sech2 0 1 tanh { H sech 2 { = = =1 = lim {0 tan { {0 sec2 { sec2 0 1. 23. This limit has the form 00 . lim 25. This limit has the form 00 . lim. 5w 3w H 5w ln 5 3w ln 3 = ln 5 ln 3 = ln 53 = lim w0 w 1. 27. This limit has the form 00 . lim. 1 sin1 { H 1@ 1 {2 1 = lim = lim = =1 {0 {0 { 1 1 1 {2. 29. This limit has the form 00 . lim. 1 cos { H sin { H cos { 1 = = lim = lim {0 2{ {0 {2 2 2. w0. {0. {0. 31. lim. {0. 0+0 0 { + sin { = = = 0. L’Hospital’s Rule does not apply. { + cos { 0+1 1. 33. This limit has the form 00 . lim. {1. 1 { + ln { H 1 + 1@{ H 1@{2 1 1 = = 2 = lim = lim {1 {1 1 + cos { sin { 2 cos { 2 (1) . d(d 1) d(d 1){d2 {d d{ + d 1 H d{d1 d H = = lim = lim 2 {1 {1 2({ 1) {1 ({ 1) 2 2. 35. This limit has the form 00 . lim. cos { 1 + 12 {2 H sin { + { H cos { + 1 H sin { H cos { 1 = = lim = lim = lim = lim {0 {0 {0 {0 24{ {0 24 {4 4{3 12{2 24. 37. This limit has the form 00 . lim 39. This limit has the form · 0.. sin( @{) H cos( @{)( @{2 ) = lim = lim cos( @{) = (1) = { { { 1@{ 1@{2. lim { sin( @{) = lim. {. 41. This limit has the form · 0. We’ll change it to the form. lim cot 2{ sin 6{ = lim. {0. {0. 0 . 0. 6(1) sin 6{ H 6 cos 6{ = = lim =3 tan 2{ {0 2 sec2 2{ 2(1)2 2. {3 H 3{2 3{ H 3 = lim 2 2 = lim 2 = lim 2 = 0 { h{ { 2{h{ { 2h{ { 4{h{. 43. This limit has the form · 0. lim {3 h{ = lim {.
(48) F. TX.10SSECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE. ¤. 169. 45. This limit has the form 0 · ( ).. lim ln { tan( {@2) = lim. {1+. {1+. 1 2 ln { 1@{ H = = lim = cot( {@2) {1+ ( @2) csc2 ( {@2) ( @2)(1)2 . 47. This limit has the form .. . lim. {1. { 1 {1 ln {. . = lim. {1. H. = lim. {1. { ln { ({ 1) H {(1@{) + ln { 1 ln { = lim = lim {1 ({ 1)(1@{) + ln { {1 1 (1@{) + ln { ({ 1) ln { 1@{ { {2 1 1 · = = = lim {1 1 + { 1@{2 + 1@{ {2 1+1 2. 49. We will multiply and divide by the conjugate of the expression to change the form of the expression.. 2 2 { + { {2 { +{{ {2 + { + { · = lim { { { 1 {2 + { + { {2 + { + { { 1 1 1 = = lim = lim s = { 2 {2 + { + { { 1 + 1@{ + 1 1+1 s As an alternate solution, write {2 + { { as {2 + { {2 , factor out {2 , rewrite as ( 1 + 1@{ 1)@(1@{), and lim. {2 + { { = lim. apply l’Hospital’s Rule. 51. The limit has the form and we will change the form to a product by factoring out {.. ln { H 1@{ ln { lim ({ ln {) = lim { 1 = 0. = since lim = lim { { { { { 1 {. 53. | = {{. 2. . ln | = {2 ln {, so lim ln | = lim {2 ln { = lim {0+. {0+. {0+. ln { H 1@{ 1 2 { = lim = lim =0 1@{2 2 {0+ 2@{3 {0+. 2. lim {{ = lim hln | = h0 = 1.. {0+. {0+. 55. | = (1 2{)1@{. ln | =. ln(1 2{) H 2@(1 2{) 1 ln(1 2{), so lim ln | = lim = 2 = lim {0 {0 {0 { { 1. lim (1 2{)1@{ = lim hln | = h2 .. {0. {0. { 5 5 3 3 1+ + 2 ln | = { ln 1 + + 2 { { { { ! 3 3 5 5 3 10 2 3 1+ + 2 ln 1 + + 2 { { { { { { H lim ln | = lim = lim = lim 2 { { { { 1@{ 1@{. 57. | =. { 5 3 = lim hln | = h3 . 1+ + 2 { { { {. so lim. 59. | = {1@{. ln | = (1@{) ln { . lim {1@{ = lim hln | = h0 = 1. {. {. lim ln | = lim. {. {. ln { H 1@{ =0 = lim { 1 {. 3+ 1+. 10 {. 5 3 + { {2. = 3,. .
(49) F. 170. ¤. CHAPTER 4. cot {. 61. | = (4{ + 1). TX.10. APPLICATIONS OF DIFFERENTIATION. 4 ln(4{ + 1) H 4{ +1 = 4 ln | = cot { ln(4{ + 1), so lim ln | = lim = lim tan { {0+ {0+ {0+ sec2 {. lim (4{ + 1)cot { = lim hln | = h4 .. {0+. {0+. 63. | = (cos {)1@{. . 2. ln | =. 1 ln cos { {2. . lim ln | = lim. {0+. {0+. ln cos { H tan { H sec2 { 1 = = lim = lim 2 + + { 2{ 2 2 {0 {0. 2 lim (cos {)1@{ = lim hln | = h1@2 = 1@ h. {0+. {0+. From the graph, if { = 500, | 7=36. The limit has the form 1 . { 2 2 Now | = 1 + ln | = { ln 1 + { { 1 2 2 1 + 2@{ { ln(1 + 2@{) H = lim lim ln | = lim { { { 1@{ 1@{2. 65.. = 2 lim. {. 1 = 2(1) = 2 1 + 2@{. . { 2 lim 1 + = lim hln | = h2 [ 7=39] { { { i ({) i 0 ({) = lim 0 = 0=25= {0 j({) {0 j ({). From the graph, it appears that lim. 67.. We calculate lim. {0. 69. lim. {. h{ H h{ h{ h{ H H H = = lim = lim = · · · = lim q q1 q2 { q{ { q(q 1){ { q! { { 1 H = lim = lim { {2 + 1 { 12 ({2 + 1)1@2 (2{). 71. lim {. i ({) h{ 1 H h{ 1 = lim 3 = . = lim 2 {0 { + 4{ {0 3{ + 4 j({) 4. {2 + 1 . Repeated applications of l’Hospital’s Rule result in the {. original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator { {@{ 1 1 by {: lim = lim s = =1 = lim s { { 1 {2 + 1 { {2 @{2 + 1@{2 1 + 1@{2 u qw u qw u 1+ , so , which is of the form 1 . | = 1 + ln | = qw ln 1 + q q q q u@q2 ln(1 + u@q) H u u = w lim = w lim = wu = w lim lim ln | = lim qw ln 1 + q q q q (1 + u@q)(1@q2 ) q 1 + l@q q 1@q u qw lim | = huw . Thus, as q
(50) , D = D0 1 +
(51) D0 huw . q q. 73. First we will nd lim. .
(52) F. TX.10SSECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE . 75.. ¤. 171. . hH + hH 1 hH hH H H hH + hH 1 hH hH HhH + HhH hH + hH = lim = lim H hH ) H + + (h HhH HhH H0 H0 HhH + hH · 1 + H hH + hH · 1 hH + hH H = lim HhH + hH · 1 [H(hH ) + hH · 1] H0+. lim S (H) = lim. H0+. H0+. = lim. H0+. =. HhH. HhH HhH = lim + hH + HhH hH H0+. hH hH 0 , where O = lim 2+O H H0+. Thus, lim S (H) = H0+. hH hH hH hH + hH hH + H H. form is 00. H. = lim. H0+. form is 00. [divide by H]. hH + hH 1+1 = =2 1 1. 0 = 0. 2+2. 77. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:. 1 (2d3 { {4 )1@2 (2d3 4{3 ) d 13 (dd{)2@3 d2 2d3 { {4 d 3 dd{ H 2 = lim lim 4 {d {d 14 (d{3 )3@4 (3d{2 ) d d{3 =. =. 3 1 2 (2d d. d4 )1@2 (2d3 4d3 ) 13 d3 (d2 d)2@3 14 (dd3 )3@4 (3dd2 ). d 13 d (d4 )1@2 (d3 ) 13 d3 (d3 )2@3 = = 43 43 d = 3 3(d4 )3@4 3 4d 4. 16 d 9. 79. Since i (2) = 0, the given limit has the form 00 .. lim. {0. i (2 + 3{) + i (2 + 5{) H i 0 (2 + 3{) · 3 + i 0 (2 + 5{) · 5 = i 0 (2) · 3 + i 0 (2) · 5 = 8i 0 (2) = 8 · 7 = 56 = lim {0 { 1. 81. Since lim [i ({ + k) i({ k)] = i ({) i ({) = 0 (i is differentiable and hence continuous) and lim 2k = 0, we use k0. k0. l’Hospital’s Rule: i ({ + k) i({ k) H i 0 ({ + k)(1) i 0 ({ k)(1) i 0 ({) + i 0 ({) 2i 0 ({) = = = i 0 ({) = lim k0 k0 2k 2 2 2 lim. i ({ + k) i ({ k) is the slope of the secant line between 2k ({ k> i ({ k)) and ({ + k> i ({ + k)). As k
(53) 0, this line gets closer to the tangent line and its slope approaches i 0 ({).. 83. (a) We show that lim. {0. i ({) 1 = 0 for every integer q 0. Let | = 2 . Then {q { 2. i ({) h1@{ |q H q| q1 H q! H = lim = lim = · · · = lim | = 0 q = lim 2q 2 | {0 { {0 ({ ) | h | | h h| lim. lim. {0. i ({) i ({) i ({) i ({) i (0) i ({) = lim = 0. = lim {q 2q = lim {q lim 2q = 0. Thus, i 0 (0) = lim {0 {0 {0 { {0 {0 {q { {0 {.
(54) F. 172. ¤. CHAPTER 4. TX.10. APPLICATIONS OF DIFFERENTIATION. (b) Using the Chain Rule and the Quotient Rule we see that i (q) ({) exists for { 6= 0. In fact, we prove by induction that for each q 0, there is a polynomial sq and a non-negative integer nq with i (q) ({) = sq ({)i({)@{nq for { 6= 0. This is true for q = 0; suppose it is true for the qth derivative. Then i 0 ({) = i ({)(2@{3 ), so i (q+1) ({) = {nq [s0q ({) i ({) + sq ({)i 0 ({)] nq {nq 1 sq ({) i ({) {2nq = {nq s0q ({) + sq ({) 2@{3 nq {nq 1 sq ({) i ({){2nq = {nq +3 s0q ({) + 2sq ({) nq {nq +2 sq ({) i ({){(2nq +3) which has the desired form. Now we show by induction that i (q) (0) = 0 for all q. By part (a), i 0 (0) = 0. Suppose that i (q) (0) = 0. Then i (q) ({) i (q) (0) i (q) ({) sq ({) i ({)@{nq sq ({) i ({) = lim = lim = lim {0 {0 {0 {0 {0 { { {nq +1. i (q+1) (0) = lim. = lim sq ({) lim {0. {0. i ({) = sq (0) · 0 = 0 {nq +1. 4.5 Summary of Curve Sketching 1. | = i ({) = {3 + { = {({2 + 1). A. i is a polynomial, so G = R.. H.. B. {-intercept = 0, |-intercept = i (0) = 0 C. i ({) = i ({), so i is odd; the curve is symmetric about the origin. D. i is a polynomial, so there is no asymptote. E. i 0 ({) = 3{2 + 1 A 0, so i is increasing on ( > ). F. There is no critical number and hence, no local maximum or minimum value. G. i 00 ({) = 6{ A 0 on (0> ) and i 00 ({) ? 0 on ( > 0), so i is CU on (0> ) and CD on ( > 0). Since the concavity changes at { = 0, there is an inection point at (0> 0). . . 3. | = i ({) = 2 15{ + 9{2 {3 = ({ 2) {2 7{ + 1. i ({) = 0 { = 2 or (by the quadratic formula) { =. A. G = R B. |-intercept: i(0) = 2; {-intercepts:. 7 ± 45 2. 0=15, 6=85 C. No symmetry D. No asymptote. E. i 0 ({) = 15 + 18{ 3{2 = 3({2 6{ + 5) = 3({ 1)({ 5) A 0 1 ? { ? 5 so i is increasing on (1> 5) and decreasing on ( > 1) and (5> ). F. Local maximum value i (5) = 27, local minimum value i (1) = 5 G. i 00 ({) = 18 6{ = 6({ 3) A 0 { ? 3, so i is CU on ( > 3) and CD on (3> ). IP at (3> 11). H..
(55) F. TX.10 5. | = i ({) = {4 + 4{3 = {3 ({ + 4). SECTION 4.5 SUMMARY OF CURVE SKETCHING. A. G = R B. |-intercept: i (0) = 0;. ¤. 173. H.. {-intercepts: i ({) = 0 { = 4> 0 C. No symmetry D. No asymptote E. i 0 ({) = 4{3 + 12{2 = 4{2 ({ + 3) A 0 { A 3, so i is increasing on (3> ) and decreasing on ( > 3). F. Local minimum value i(3) = 27, no local maximum G. i 00 ({) = 12{2 + 24{ = 12{({ + 2) ? 0 2 ? { ? 0, so i is CD on (2> 0) and CU on ( > 2) and (0> ). IP at (0> 0) and (2> 16) 7. | = i ({) = 2{5 5{2 + 1. A. G = R B. |-intercept: i (0) = 1 C. No symmetry D. No asymptote. E. i 0 ({) = 10{4 10{ = 10{({3 1) = 10{({ 1)({2 + { + 1), so i 0 ({) ? 0 0 ? { ? 1 and i 0 ({) A 0 { ? 0 or { A 1. Thus, i is increasing on ( > 0) and (1> ) and decreasing on (0> 1). F. Local maximum value i (0) = 1, local minimum value i (1) = 2 G. i 00 ({) = 40{3 10 = 10(4{3 1) so i 00 ({) = 0 { = 1@ 3 4. i 00 ({) A 0 { A 1@ 3 4 and i 00 ({) ? 0 { ? 1@ 3 4, so i is CD on > 1@ 3 4 and CU # $ 1 9 3 on 1@ 4> . IP at > 1 2 (0=630> 0=786) 3 4 2 34. 9. | = i ({) = {@({ 1). C. No symmetry D. E. i 0 ({) =. H.. A. G = {{ | { 6= 1} = ( > 1) (1> ) B. {-intercept = 0, |-intercept = i (0) = 0 lim. {±. { { { = 1, so | = 1 is a HA. lim = , lim = , so { = 1 is a VA. {1 {1 { 1 {1+ { 1. ({ 1) { 1 = ? 0 for { 6= 1, so i is ({ 1)2 ({ 1)2. H.. decreasing on ( > 1) and (1> ) = F. No extreme values G. i 00 ({) =. 2 A 0 { A 1, so i is CU on (1> ) and ({ 1)3. CD on ( > 1). No IP. 11. | = i({) = 1@({2 9). A. G = {{ | { 6= ±3} = ( > 3) (3> 3) (3> ) B. |-intercept = i (0) = 19 , no. {-intercept C. i ({) = i ({) i is even; the curve is symmetric about the |-axis. D. is a HA. lim. {3. lim. {±. 1 = 0, so | = 0 {2 9. 1 1 1 1 = , lim 2 = , lim = , lim = , so { = 3 and { = 3 {2 9 {3+ { 9 {3 {2 9 {3+ {2 9. are VA. E. i 0 ({) = . ({2. 2{ A 0 { ? 0 ({ 6= 3) so i is increasing on ( > 3) and (3> 0) and 9)2.
(56) F. 174. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. TX.10. decreasing on (0> 3) and (3> ). F. Local maximum value i (0) = 19 . G. | 00 =. H.. 2({2 9)2 + (2{)2({2 9)(2{) 6({2 + 3) = 2 A0 2 4 ({ 9) ({ 9)3. {2 A 9 { A 3 or { ? 3, so i is CU on ( > 3) and (3> ) and CD on (3> 3). No IP 13. | = i ({) = {@({2 + 9). A. G = R B. |-intercept: i (0) = 0; {-intercept: i({) = 0 { = 0. C. i ({) = i ({), so i is odd and the curve is symmetric about the origin. D. HA; no VA E. i 0 ({) =. lim [{@({2 + 9)] = 0, so | = 0 is a. {±. ({2 + 9)(1) {(2{) 9 {2 (3 + {)(3 {) = 2 = A 0 3 ? { ? 3, so i is increasing ({2 + 9)2 ({ + 9)2 ({2 + 9)2. on (3> 3) and decreasing on ( > 3) and (3> ). F. Local minimum value i (3) = 16 , local maximum value i (3) =. 1 6. ({2 + 9)2 (2{) (9 {2 ) · 2({2 + 9)(2{) (2{)({2 + 9)[({2 + 9) 2(9 {2 )] 2{({2 27) = = [({2 + 9)2 ]2 ({2 + 9)4 ({2 + 9)3 = 0 { = 0> ± 27 = ±3 3 (2{)({2 + 9) ({2 + 9) 2(9 {2 ) ({2 + 9)2 (2{) (9 {2 ) · 2({2 + 9)(2{) = G. i 00 ({) = [({2 + 9)2 ]2 ({2 + 9)4 2{({2 27) = = 0 { = 0, ± 27 = ±3 3 H. 2 3 ({ + 9) i 00 ({) A 0 3 3 ? { ? 0 or { A 3 3, so i is CU on 3 3> 0 and 3 3> , and CD on > 3 3 and 0> 3 3 . There are three 1 inection points: (0> 0) and ±3 3> ± 12 3 . i 00 ({) =. 15. | = i ({) =. {1 {2. C. No symmetry D. E. i 0 ({) =. A. G = {{ | { 6= 0} = ( > 0) (0> ) B. No |-intercept; {-intercept: i ({) = 0 { = 1 lim. {±. {1 {1 = 0, so | = 0 is a HA. lim = , so { = 0 is a VA. {0 {2 {2. {2 · 1 ({ 1) · 2{ {2 + 2{ ({ 2) = = , so i 0 ({) A 0 0 ? { ? 2 and i 0 ({) ? 0 2 2 ({ ) {4 {3. { ? 0 or { A 2. Thus, i is increasing on (0> 2) and decreasing on ( > 0) and (2> ). F. No local minimum, local maximum value i(2) = 14 . G. i 00 ({) =. {3 · (1) [({ 2)] · 3{2 2{3 6{2 2({ 3) = = . 3 2 ({ ) {6 {4. i 00 ({) is negative on ( > 0) and (0> 3) and positive on (3> ), so i is CD on ( > 0) and (0> 3) and CU on (3> ). IP at 3> 29. H..
(57) F. TX.10 17. | = i ({) =. ({2 + 3) 3 3 {2 = =1 2 +3 {2 + 3 { +3. {2. SECTION 4.5 SUMMARY OF CURVE SKETCHING. ¤. 175. A. G = R B. |-intercept: i(0) = 0;. {-intercepts: i ({) = 0 { = 0 C. i ({) = i ({), so i is even; the graph is symmetric about the |-axis. D.. lim. {±. 2{ 6{ {2 = 1, so | = 1 is a HA. No VA. E. Using the Reciprocal Rule, i 0 ({) = 3 · 2 = 2 . {2 + 3 ({ + 3)2 ({ + 3)2. i 0 ({) A 0 { A 0 and i 0 ({) ? 0 { ? 0, so i is decreasing on ( > 0) and increasing on (0> ). F. Local minimum value i(0) = 0, no local maximum. G. i 00 ({) = =. ({2 + 3)2 · 6 6{ · 2({2 + 3) · 2{ [({2 + 3)2 ]2. H.. 6({2 + 3)[({2 + 3) 4{2 ] 6(3 3{2 ) 18({ + 1)({ 1) = = ({2 + 3)4 ({2 + 3)3 ({2 + 3)3. i 00 ({) is negative on ( > 1) and (1> ) and positive on (1> 1), so i is CD on ( > 1) and (1> ) and CU on (1> 1). IP at ±1> 14 19. | = i ({) = {. 5 { A. The domain is {{ | 5 { 0} = ( > 5] B. |-intercept: i (0) = 0;. {-intercepts: i ({) = 0 { = 0, 5 C. No symmetry. D. No asymptote. E. i 0 ({) = { · 12 (5 {)1@2 (1) + (5 {)1@2 · 1 = 12 (5 {)1@2 [{ + 2(5 {)] = and decreasing on 10 so i is increasing on > 10 >5 . 3 3 10 = 9 15 4=3; no local minimum F. Local maximum value i 10 3 2(5 {)1@2 (3) (10 3{) · 2 12 (5 {)1@2 (1) G. i 00 ({) = 2 2 5{. {?. 10 3{ A0 2 5{. 10 , 3. =. H.. (5 {)1@2 [6(5 {) + (10 3{)] 3{ 20 = 4(5 {) 4(5 {)3@2. i 00 ({) ? 0 for { ? 5, so i is CD on ( > 5). No IP. 21. | = i ({) =. s {2 + { 2 = ({ + 2)({ 1) A. G = {{ | ({ + 2)({ 1) 0} = ( > 2] [1> ). B. |-intercept: none; {-intercepts: 2 and 1 C. No symmetry D. No asymptote E. i 0 ({) = 12 ({2 + { 2)1@2 (2{ + 1) =. 2. 2{ + 1 , i 0 ({) = 0 if { = 12 , but 12 is not in the domain. {2 + { 2. i 0 ({) A 0 { A 12 and i 0 ({) ? 0 { ? 12 , so (considering the domain) i is increasing on (1> ) and i is decreasing on ( > 2). F. No local extrema 2({2 + { 2)1@2 (2) (2{ + 1) · 2 · 12 ({2 + { 2)1@2 (2{ + 1) 2 2 {2 + { 2 ({2 + { 2)1@2 4({2 + { 2) (4{2 + 4{ + 1) = 4({2 + { 2) 9 ?0 = 4({2 + { 2)3@2. G. i 00 ({) =. so i is CD on ( > 2) and (1> ). No IP. H..
(58) F. 176. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. . 23. | = i ({) = {@ {2 + 1. TX.10. A. G = R B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 { = 0. C. i ({) = i ({), so i is odd; the graph is symmetric about the origin. { {@{ {@{ 1 1 = lim s =1 = D. lim i ({) = lim = lim = lim { { { 1+0 {2 + 1 { {2 + 1@{ { {2 + 1@ {2 1 + 1@{2 and { {@{ {@{ 1 = lim s = lim = lim lim i({) = lim 2 2 { { { { { +1 { + 1@{ 1 + 1@{2 {2 + 1@ {2. {. =. 1 = 1 so | = ±1 are HA. 1+0. No VA. E. i 0 ({) =. {2 + 1 { ·. 2{ {2 + 1 {2 1 2 {2 + 1 = = A 0 for all {, so i is increasing on R. 3@2 2 2 [({2 + 1)1@2 ]2 ({ + 1) ({ + 1)3@2 H.. F. No extreme values G. i 00 ({) = 32 ({2 + 1)5@2 · 2{ =. 3{ , so i 00 ({) A 0 for { ? 0 ({2 + 1)5@2. and i 00 ({) ? 0 for { A 0. Thus, i is CU on ( > 0) and CD on (0> ). IP at (0> 0) 1 {2 @{ A. G = {{ | |{| 1, { 6= 0} = [1> 0) (0> 1] B. {-intercepts ±1, no |-intercept 1 {2 1 {2 = , lim = , C. i ({) = i ({), so the curve is symmetric about (0> 0) = D. lim { { {0+ {0 2 { @ 1 {2 1 {2 1 0 so { = 0 is a VA. E. i ({) = = ? 0, so i is decreasing {2 {2 1 {2. 25. | = i({) =. on (1> 0) and (0> 1). F. No extreme values G. i 00 ({) =. 2 3{2. H.. t t A 0 1 ? { ? 23 or 0 ? { ? 23 , so. {3 (1 {2 )3@2 t t t t 2 >1 . i is CU on 1> 23 and 0> 23 and CD on 23 > 0 and 3. t IP at ± 23 > ± 12 27. | = i ({) = { 3{1@3. A. G = R B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 { = 3{1@3 {3 = 27{ {3 27{ = 0 {({2 27) = 0 { = 0, ±3 3 C. i ({) = i ({), so i is odd;. the graph is symmetric about the origin. D. No asymptote E. i 0 ({) = 1 {2@3 = 1 . 1 {2@3 1 = . {2@3 {2@3. i 0 ({) A 0 when |{| A 1 and i 0 ({) ? 0 when 0 ? |{| ? 1, so i is increasing on ( > 1) and (1> ), and decreasing on (1> 0) and (0> 1) [hence decreasing on (1> 1) since i is continuous on (1> 1)]. F. Local maximum value i (1) = 2, local minimum value i (1) = 2 G. i 00 ({) = 23 {5@3 ? 0 when { ? 0 and i 00 ({) A 0 when { A 0, so i is CD on ( > 0) and CU on (0> ). IP at (0> 0). H..
(59) F. TX.10 29. | = i ({) =. SECTION 4.5 SUMMARY OF CURVE SKETCHING. ¤. 3 {2 1 A. G = R B. |-intercept: i (0) = 1; {-intercepts: i ({) = 0 {2 1 = 0 . { = ±1 C. i ({) = i ({), so the curve is symmetric about the |-axis= D. No asymptote E. i 0 ({) = 13 ({2 1)2@3 (2{) =. 2{ s . i 0 ({) A 0 { A 0 and i 0 ({) ? 0 { ? 0, so i is 3 3 ({2 1)2. increasing on (0> ) and decreasing on ( > 0). F. Local minimum value i (0) = 1 G. i 00 ({) = =. 2 2@3 2 1@3 (2{) 2 ({ 1) (1) { · 23 ({ 1) · 3 [({2 1)2@3 ]2. H.. 2 ({2 1)1@3 [3({2 1) 4{2 ] 2({2 + 3) · = 2 4@3 9 ({ 1) 9({2 1)5@3. i 00 ({) A 0 1 ? { ? 1 and i 00 ({) ? 0 { ? 1 or { A 1, so i is CU on (1> 1) and i is CD on ( > 1) and (1> ). IP at (±1> 0) 31. | = i ({) = 3 sin { sin3 {. A. G = R B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 . sin { (3 sin2 {) = 0 sin { = 0 [since sin2 { 1 ? 3] { = q , q an integer. C. i ({) = i ({), so i is odd; the graph (shown for 2 { 2 ) is symmetric about the origin and periodic with period 2 . D. No asymptote E. i 0 ({) = 3 cos { 3 sin2 { cos { = 3 cos { (1 sin2 {) = 3 cos3 {. i 0 ({) A 0 cos { A 0 { 2q 2 > 2q + 2 for each integer q, and i 0 ({) ? 0 cos { ? 0 { 2q + 2 > 2q + 3 for each integer q. Thus, i is increasing on 2q 2 > 2q + 2 for each integer q, 2 for each integer q. and i is decreasing on 2q + 2 > 2q + 3 2 F. i has local maximum values i(2q + 2 ) = 2 and local minimum values i (2q +. 3 ) 2. = 2.. G. i 00 ({) = 9 sin { cos2 { = 9 sin { (1 sin2 {) = 9 sin { (1 sin {)(1 + sin {). i 00 ({) ? 0 sin { A 0 and sin { 6= ±1 { 2q > 2q + 2 2q + 2 > 2q + for some integer q. i 00 ({) A 0 sin { ? 0 and sin { 6= ±1 { (2q 1) > (2q 1) + 2 (2q 1) + 2 > 2q H. for some integer q. Thus, i is CD on the intervals 2q > 2q + 12 and 2q + 12 > (2q + 1) [hence CD on the intervals (2q > (2q + 1) )] for each integer q, and i is CU on the intervals (2q 1) > 2q 12 and 2q 12 > 2q [hence CU on the intervals ((2q 1) > 2q )] for each integer q. i has inection points at (q > 0) for each integer q. 33. | = i ({) = { tan {, 2 ? { ?. symmetric about the |-axis. D.. 2. A. G = 2 > 2 B. Intercepts are 0 C. i({) = i ({), so the curve is lim. {( @2). { tan { = and. lim. {( @2)+. { tan { = , so { =. E. i 0 ({) = tan { + { sec2 { A 0 0 ? { ? 2 , so i increases on 0> 2 and decreases on 2 > 0 . F. Absolute and local minimum value i (0) = 0. G. | 00 = 2 sec2 { + 2{ tan { sec2 { A 0 for 2 ? { ? CU on 2 > 2 . No IP. , 2. so i is. H.. 2. and { = 2 are VA.. 177.
(60) F. ¤. 178. CHAPTER 4. 35. | = i ({) =. 1 2{. TX.10. APPLICATIONS OF DIFFERENTIATION. sin {, 0 ? { ? 3 . A. G = (0> 3 ) B. No |-intercept. The {-intercept, approximately 1=9, can be. found using Newton’s Method. C. No symmetry D. No asymptote E. i 0 ({) = 12 cos { A 0 cos { ? and 7 . ? { ? 5 or 7 ? { ? 3 , so i is increasing on 3 > 5 > 3 and decreasing on 0> 3 and 5 > 7 3 3 3 3 3 3 3 3 F. Local minimum value i 3 = 6 2 , local maximum value H. i. 5 3. =. 5 6. +. 3 , 2. local minimum value i. 7 3. =. 7 6. . 1 2. . . 3 2. G. i 00 ({) = sin { A 0 0 ? { ? or 2 ? { ? 3 , so i is CU on (0> ) and (2 > 3 ) and CD on ( > 2 ). IPs at > 2 and (2 > ) 5 37. | = i ({) =. sin { 1 + cos {. when. 6. 9cos { 6= 1 7 =. 1 cos { sin { (1 cos {) sin { 1 cos { : · = = csc { cot {8 = 1 + cos { 1 cos { sin { sin2 {. A. The domain of i is the set of all real numbers except odd integer multiples of . B. |-intercept: i (0) = 0; {-intercepts: { = q , q an even integer. C. i ({) = i ({), so i is an odd function; the graph is symmetric about the origin and has period 2 . D. When q is an odd integer, integer q. No HA. E. i 0 ({) =. lim. {(q ). i({) = and. lim. {(q )+. i ({) = , so { = q is a VA for each odd. (1 + cos {) · cos { sin {( sin {) 1 + cos { 1 . i 0 ({) A 0 for all { = = (1 + cos {)2 (1 + cos {)2 1 + cos {. except odd multiples of , so i is increasing on ((2n 1) > (2n + 1) ) for each integer n. F. No extreme values G. i 00 ({) =. sin { A 0 sin { A 0 (1 + cos {)2. H=. { (2n > (2n + 1) ) and i 00 ({) ? 0 on ((2n 1) > 2n ) for each integer n. i is CU on (2n > (2n + 1) ) and CD on ((2n 1) > 2n ) for each integer n. i has IPs at (2n > 0) for each integer n. 39. | = i ({) = hsin {. A. G = R B. |-intercept: i (0) = h0 = 1; {-intercepts: none, since hsin { A 0 C. i is periodic. with period 2 , so we determine E–G for 0 { 2 . D. No asymptote E. i 0 ({) = hsin { cos {. i 0 ({) A 0 3 0 [ i is decreasing]. cos { A 0 { is in 0> 2 or 3 2 > 2 [ i is increasing] and i ({) ? 0 { is in 2 > 2 3 1 F. Local maximum value i 2 = h and local minimum value i 2 = h G. i 00 ({) = hsin { ( sin {) + cos { (hsin { cos {) = hsin { (cos2 { sin {). i 00 ({) = 0 cos2 { sin { = 0 5 = sin1 1 +2 5 0=67 and 1 sin2 { sin { = 0 sin2 { + sin { 1 = 0 sin { = 1 ± 2 = 2=48. i 00 ({) ? 0 on (> ) [ i is CD] and i 00 ({) A 0 on (0> ) and (> 2 ) [ i is CU]. The inection points occur when { = , . H..
(61) F. TX.10 41. | = 1@(1 + h{ ). SECTION 4.5 SUMMARY OF CURVE SKETCHING. ¤. A. G = R B. No {-intercept; |-intercept = i (0) = 12 = C. No symmetry. D. lim 1@(1 + h{ ) =. 1 1+0. {. = 1 and lim 1@(1 + h{ ) = 0 since lim h{ = ], so i has horizontal asymptotes {. 0. {. { 2. | = 0 and | = 1. E. i ({) = (1 + h F. No extreme values G. i 00 ({) =. ). (h. { 2. (1 + h. {. {. )=h. {. ) (h. { 2. @(1 + h. ) . This is positive for all {, so i is increasing on R.. {. ) h (2)(1 + h{ )(h{ ) h{ (h{ 1) = { 4 (1 + h ) (1 + h{ )3. The second factor in the numerator is negative for { A 0 and positive for { ? 0,. H.. and the other factors are always positive, so i is CU on ( , 0) and CD on (0> ). IP at 0, 12. 43. | = i({) = { ln {. A. G = (0> ) B. |-intercept: none (0 is not in the domain); {-intercept: i({) = 0 . { = ln {, which has no solution, so there is no {-intercept. C. No symmetry D. lim ({ ln {) = > so { = 0 {0+. 0. is a VA. E. i ({) = 1 1@{ A 0 1 A 1@{ { A 1 and. H.. i 0 ({) ? 0 0 ? { ? 1, so i is increasing on (1> ) and i is decreasing on (0> 1). F. Local minimum value i(1) = 1; no local maximum value G. i 00 ({) =. 1 A 0 for all {, so i is CU on (0> ). No IP {2. 45. | = i ({) = (1 + h{ )2 =. 1 (1 + h{ )2. A. G = R B. |-intercept: i (0) = 14 . {-intercepts: none [since i ({) A 0]. C. No symmetry D. lim i ({) = 0 and lim i ({) = 1, so | = 0 and | = 1 are HA; no VA {. E. i 0 ({) = 2(1 + h{ )3 h{ =. {. {. 2h ? 0, so i is decreasing on R F. No local extrema (1 + h{ )3. G. i 00 ({) = (1 + h{ )3 (2h{ ) + (2h{ )(3)(1 + h{ )4 h{ = 2h{ (1 + h{ )4 [(1 + h{ ) 3h{ ] =. H.. 2h{ (1 2h{ ) = (1 + h{ )4. i 00 ({) A 0 1 2h{ ? 0 h{ A 12 { A ln 12 and i 00 ({) ? 0 { ? ln 12 , so i is CU on ln 12 > and CD on > ln 12 . IP at ln 12 > 49 47. | = i ({) = ln(sin {). A. G = {{ in R | sin { A 0} =. V q=. (2q > (2q + 1) ) = · · · (4 > 3 ) (2 > ) (0> ) (2 > 3 ) · · ·. B. No |-intercept; {-intercepts: i ({) = 0 ln(sin {) = 0 sin { = h0 = 1 integer q.. 179. C. i is periodic with period 2 . D.. 2. i({) = and. lim. {[(2q+1) ]. 2. for each. i ({) = , so the lines. cos { = cot {, so i 0 ({) A 0 when 2q ? { ? 2q + 2 for each sin { ? { ? (2q + 1) . Thus, i is increasing on 2q > 2q + 2 and. { = q are VAs for all integers q. E. i 0 ({) = integer q, and i 0 ({) ? 0 when 2q +. lim. {(2q )+. { = 2q +.
(62) F. 180. ¤. CHAPTER 4. APPLICATIONS OF DIFFERENTIATION. TX.10. decreasing on 2q + 2 > (2q + 1) for each integer q. F. Local maximum values i 2q + 2 = 0, no local minimum.. H.. G. i 00 ({) = csc2 { ? 0, so i is CD on (2q > (2q + 1) ) for each integer q= No IP. 49. | = i ({) = {h{. 2. about the origin. D.. A. G = R B. Intercepts are 0 C. i ({) = i ({), so the curve is symmetric 2. lim {h{ = lim. {±. {±. { H 1 = lim 2 = 0, so | = 0 is a HA. {± 2{h{ h{2. 2 2 2 E. i 0 ({) = h{ 2{2 h{ = h{ (1 2{2 ) A 0 {2 ? 12 |{| ? 12 , so i is increasing on 12 > 12 and decreasing on > 12 and 12 > . F. Local maximum value i 12 = 1@ 2h, local minimum 2 2 2 value i 12 = 1@ 2h G. i 00 ({) = 2{h{ (1 2{2 ) 4{h{ = 2{h{ (2{2 3) A 0 t t t 3 H. > and { A 32 or 32 ? { ? 0, so i is CU on 2 t t t 32 > 0 and CD on > 32 and 0> 32 . t t IP are (0, 0) and ± 32 > ± 32 h3@2 . 51. | = i ({) = h3{ + h2{ 0. 3{. E. i ({) = 3h h5{ A. 2 3. 2{. 2h. A. G = R B. |-intercept = i (0) = 2; no {-intercept C. No symmetry D. No asymptote , so i 0 ({) A 0 3h3{ A 2h2{ [multiply by h2{ ] . 5{ A ln 23. {A { ? 15 ln 23 . i is decreasing on > F. Local minimum value i. 1 5. H.. ln 23 0=081. Similarly, i 0 ({) ? 0 1 ln 23 and increasing on 15 ln 23 > . 5 1 5. 3@5 2 2@5 ln 23 = 23 + 3 1=96; no local maximum.. G. i 00 ({) = 9h3{ + 4h2{ , so i 00 ({) A 0 for all {, and i is CU on ( > ). No IP. p0 . The m-intercept is i (0) = p0 . There are no y-intercepts. lim i (y) = , so y = f is a VA. yf 1 y 2 @f2 p0 y p0 y p0 fy = 2 2 = 2 A 0, so i is i 0 (y) = 12 p0 (1 y 2 @f2 )3@2 (2y@f2 ) = 2 f (1 y 2 @f2 )3@2 (f y 2 )3@2 f (f y 2 )3@2 f3 increasing on (0> f). There are no local extreme values.. 53. p = i(y) = s. i 00 (y) = =. (f2 y 2 )3@2 (p0 f) p0 fy · 32 (f2 y 2 )1@2 (2y) [(f2 y 2 )3@2 ]2 p0 f(f2 y 2 )1@2 [(f2 y 2 ) + 3y 2 ] p0 f(f2 + 2y 2 ) = A 0, 2 2 3 (f y ) (f2 y 2 )5@2. so i is CU on (0> f). There are no inection points..
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