{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solnshw2

# solnshw2 - 142 Chapter 4 Exponential and Logarirhmic...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 142 Chapter 4 Exponential and Logarirhmic Functions Section 4.3 Derivatives of Exponential Functions 2. y’ = 261‘ N10) = 2 6. y’ = —e"" g’(~l)=3e‘l x— =;(.x+ l) 3 3 l 3 4 v=—x -— —=—x+—- e e e e e 22. y = (e4-" - 2)2 (0, 1) y’ = 2(6’4" — 2)(4e4-") _\"(0) = 2(-1)4 = —8 y— l = —8(x-0) y=—8x+l 4. v' = ~26‘1‘ )"(0) = ~ 2 —2 l —l .v'<1)= ,+—.=— e- e‘ e- 1 —l v—‘e—z FL’c—l) ve--l=-—x+1 yel+x—2=0 or y=2_,x e- 24. xzy —xe>" + 2 = 0 2xy+x3ﬂ—xe"—e‘=0 dx r3ﬂ=xe‘+e-‘—2xv dx Section 4.3 Derivatives of Exponential Functions 143 IO N a (”t + N I l I ll ' dx dx Efﬁe“ — 2\) = ﬁve)“ —- 2r EL" 2 —ye"“ — 2t : -y(l0 - x3 + yZ) - 2x _ xzy ' y" - 2r - IOy dx x0)“ - 2y x(10 — x3 + .vz) — 2y xyz — x3 + 10x - 2y 28. f’(x) = (l + 2x)(4e4’) + 264" = 264-"[(l + 210(2) + l] = 234-‘(4x + 3) f”(x) = 2e4"(4) + 8e4“‘(4x + 3) = 86""[1 + (4x + 3)] = 3264"(x + l) 30. f’(x) = (3 + 2x)(—3e‘3") + 26“" = e““’(—9 — 6x + 2) = -e““’(6x + 7) f” x) = —e‘3*"(6) + (6x + 7)(3e‘-"") = 3e_3~“(—2 + 6x + 7) = 36—3"(6X + 5) 32. f(x) = % e~" ~ w) 3- f“) = la(e"' + e“) > Ofor all x f ’(x) ¢ 0 for any values of x. Thus. there are no relative extrema. fit) = 56" — e“), " f ”(x) = 0 when x = O. We have a point of inﬂection at (0, 0). “HI!- -----n 34. f(x) = xe" V v f’(x) = -xe'-" + e“ = e‘-*(l — x) V 3‘ f’(x) = Owhenx = 1. . f”(x) = e“(— l) + (l — x)(—e"") = —e‘-"[l + (l - x)] = e”"’(x — 2) Since f”(l) < 0, we have a relative maximum at (l. e“). f”(x) = 0 when x = 2 and we have a point of inflection at (2. 2e‘3). Horizontal asymptote: y = 4 Vertical asymptote: x = 0 (from the left) 10 -10 10 144 Chapter 4 Exponemiul and Logarithmic Functions 38. (a) V = 500,0006‘”~333" (C) V15) = — 11 15009—022316] z —\$36,650.66 per year V’ = —]11,500e’”-333" (d) V10) = 500.000. V1.10) z 53.7105 600.000 V(10) — V(0) 53,7105 “ 500.000 ——— = a“ z — ’7 10 - 0 10 44‘6‘9 Linear model: V — 500.000 = ~44,629(t - 0) V = 500.000 - 44.629! 0 12 0 (e) Answers will vary. (b) V11) = “111,5009‘0'333'”) * —\$89.243.89 per year 95 , _ 96.9fm” _ _ 40. N — m. N — (1 + 8.5e_0“2’)2 42. (l - 20. b — 0.5 (a) When I = 5, N’ = 1.66 words/min/week. (b) Whent = 10. N’ = 2.30 words/min/week. % = «409’0-5’ Wh t=30,N’= 1.74 d/ 'n/ k. (C) 3“ wor 5”" “’66 (a) Whent = 1, dp/dt a —24.3%. (b) When! 2 31.01p/dtz -8.9%. 44. y = 115.46 + 1.5921 + 0.0552!2 — 0.000046’ (t = 5 46. p. = 64, 0' = 3.2 corresponds to 1995) 1 = ——___ *‘(A'r‘MF/ZUAR (a) 200 (a) flx) 3.2 —27Te (b) 0.2 12 (b) 1995: y’(5) *3 2.14 million per year 1998: )"(81 z 2.36 million per year (c) f’(x) : Talﬁg‘H-MWZO-‘W—ﬂx — 64)/20.48] 2002: y’(12) ~ —3.59 million per year ‘ "- (C) y’ = 1.592 + 0.1104! — 0.000046! y’(5) z 2.14 ‘\"(8) z 2.36 y'(12) z —3.59 11 — 1 . __ _ 64 ‘(X‘ 641720.414 32.768 J27“ ”3 (d) For x < p. = 64.f’(x) > 0, and forx > ,u = 64,f’(x) < 0. 1 “LT—p.172”: : 1 a 27re \/27r p. shifts the graph horizontally. e—h ~u13/2 48. f (x) = Section 4.4 Logarithmic Functions 145 W Section 4.4 Logarithmic Functions 2. e3-‘3"~' : 8.4 4. (2.3242... : 0.056 6. ln 7.389. . . : 2 8. ln 1.284 . . . = 0.25 10. The graph is a logarithmic curve that passes through the 12. The graph is a logarithmic curve that passes through the point (1, O) with a vertical asymptote at x = 0. Therefore, point (2» 0) with a vertical asymptote at x = 1. Therefore, it matches graph (d). it matches graph (3). IIIIII IIIIIm 20. f(g(x)) =f(ln(x + 1)) III : ean’l—l) —. l : (x + I) —] glfm) = g(e“' - l) ; =1n((e"’— 1) + l) = lne“ =x ll )6 22- flglx» =flln-x3) 24. In ez-‘Al = 2x - l :f(3 [n x) = (Jinx/3 : elnx : x g(f(x)) = 230””) = ln(e"'/3)3 = ln e*" = x 28. —8 + em” = —8 + x3 146 Chapter 4 Expunemial and Logarithmir Functions 30. (a) In 0.25 ‘ Inﬁ — In I In4 — In I In 22 — In I 2In2 — 0 2(0.693l) = - 1.3862 (b) In 24 = In(3 - 2‘) = 1n 3 + 3 In 2 = 10986 + 3(0.693I) = 3.1779 (c) In é/ﬁ = %In(3 - 21) = Hln 3 + 2 In 2] = %[I.0986 + 2(0.6931)] a 08283 (d) In7‘—2 = In I — In 72 = 0 - ln(2“ ~ 32) = —[3 In 2 + 2 In 3] = —[3I0.o931)+ 2(1.0986)] = —4.2765 32. In%=InI — In5 =0— In5 = —In5 34. Inf}:1nxy-Inz=lnx+lny—In: x3 36. In x+I _l .3_ -3 ._l . —2[In.\ In(x+1)]—2In.x 2In(.I+l) 38. In [x 57x2 + I] = Inx + In(x2 + I)”3 = Inx + lIn(.Ir2 +1) 3 40. “1% = In 2x — In \/x3 — I 42. In(2x + I) + In(2x * I) = In(2x + I)(,2x ~ I) x- _ I = In(4)c2 — I) = In2 + Inx — 3|n[(x + I)(x — |)] = In2 +1n.f—%[In(x + I) + In(x - I)] *12+l —il(+1)—lln(—I) — n nx 2 nx 2 x I 9 ' _ _ _ 2 + = .. 2 + = __._.._._.. 44 2In3 21n(x l) ln9 [mm 1 mm I 1 I (x + 3)3x .— -+3+.— -— =— ———=» 46 3[21n(2 ) Inr ln(x |)] 3In (x2 _ 1) In 48. 2[lnx +§1nIx +1)]= Zlnx + §In(x +1) = lnx2 + In(x +1)”2 = In[x3(.r + Um] 50. anon — I) + §In(x + 2) : In(x — l)'/3 + 1n(x + 2)” = In[(x ~ I)'/3(x + 2%] 52. e1" — 9 = 0 54. 2 In x = 4 ' 56. [05" = 0.075 x2 - 9 = 0 In x = 2 —O.5x = In 0.075 x2 = 9 x = e2 _ In 0.075 x ~05 x= t3 3 5.1805 Section 4.4 Logarithmic Functions 147 L W 58. 400e“’-‘“74’ = 1000 1000 5 41,0174, : _ = _ 2 5 e 400 2 2-- —0.0174; : In 2.5 In 2.5 :'_—_ x _S . [ —0.0174 ~266 50 62' W = 10.5 50 + ‘1).01‘ : . l 126 10.5 e—W-v z 0.3135 —0.02x a ln(0.3135) x z —501n(0.3135) z 57.9991 66. 400(1.06)' = 1300 13 .0 (1 6)'= 4 t1n1.06=lnl3—ln4 :ln13—In4 1n1.06 220.2279 70 3P=Pe” 3:en ln3=rt 1n. t=—— r 72. p = 250 — 0.8e0.005\- (a) p = 200 = 250 _ 0860.005;- 0.8e0.005x = 50 e1),(l()5.\' 2 625 0.005x = In 62.5 In 62.5 . x — 0005 ~ 827.03 ~ 827 umts ---- 2747 13.73 10.99 60. 2e“+l - 5 = 9 9““ =7 —x +1 =ln7 x = 1 ~1n7 ~ -O.9459 64. 2"":6 1n 2""'=1n 6 (1—x)1n2=1n6 1— len_6 ' ln2 1-1—n—6= 1c 1n2 ' x~ ~1,5850 121 68. 2000(1 + %) = 10.000 121 (1 + 0&) 10 000 12: _ 2000 .12t1n(1.005) = 1n 5 1n5 = 121n(1.005) 7 26'891 mm (b) p = 125 = 250 - 0.8420005. 0.8900051 : 125 90.0051- : 156.25 00051 = 1n 156.25 1n 156.25 . ‘ x — W ~ 1010.29 ~ 1010 umts ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

solnshw2 - 142 Chapter 4 Exponential and Logarirhmic...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online