Exponential Functions" Differentiation and lintegration

Transcription

1 5t)8 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.4 Exponential Functions" Differentiation and lintegration l. e lnx = x = e x/ e 4. e lnx = 50 x = 84 = e 4 X=6 3. e x = x = In,~ e x = 83 e x = 83 4 x= e x = 7 ln(8. ~] ~ e x = X In 84 = - x = n84 ~ e x e x 499 = e x In 499 = x x =-ln 499 ~ 3.9 e x = X= e x = 8 3e x = 4 e x = e -x = 30 e-x = ~_ 5 ~.540 ~ e -4"~ = 5 e-4x _ 5 _ ~ x =, ln x = 0 x = e ~ X = e o x = ±e 5 ~ ± o ln(x- 3) = 4. In 4x = 4x = e l x-3=e x = 3 +e ~ e X ~ in x/-~x + =.,/-~x + x+ x e =el=e = e 6o ln(x-) = (x- ) = e = x-=e 6 = e - = x = + e 6 ~ Brooks/Cole, Cengage Learning

2 Section 5.4 Exponential Functions." Differentiation and Integration 5)9 7o y = e -~". y = e -4z x 8. y = $e 3. (a) 4" (b) Horizontal shift units to the right 9o y=ex + t4 (c) -3 A reflection in the x-axis and a vertical shrink / 3 4. (a) Vertical shift 3 units upward and a reflection in the y-axis ", y = e -x Symmetric with respect to the y-axis Horizontal asymptote: y = 0 (b) ~ ~ ~ 0 - Horizontal asymptotes: y = 0 and y = 8 l0-8 ~ 0 - Horizontal asymptote: y = 4 5. y = Ce ~ Horizontal asymptote: y = 0 Matches (c) 00 Brooks/Cole, Cengage Learning

3 56 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 6. y = Ce -~ ~ 3. f(x) Horizontal asymptote: y = 0 Reflection in the y-axis Matches (d) g(x) 4- = ln(x + ) Vertical shift C units ~ Reflection in both the x- and y-axes Matches (a) C 8. y- l+e -ox C lim -- x~o~l + e -~ -C c lim x-~l + e -~ -0 Horizontal asymptotes: y = C and y = 0 Matches (b) 3 f(x) = e x- g(x) = l+lnx 9. f(x) : e x l lnx g(x) : ln-,/-~ = I 3 4 g - As x --~ 0% the graph off approaches the graph ofg. lim. + = e f(x) : e x/3 g(x) = lnx3 = 3nx y 34. Using the result from Exercise 33: lim + = e ~forr > O /,O0~,O00J ~ e ~ e > ( + Too,,oo6,ooo) ~ + -g + + ]~ + 7-T6 + = e ~ e>l+l+½+++ +-T6+7-~ -~4 +~ Brooks/Cole, Cengage Learning

4 Section 5.4 Exponential Functions." Differentiation and Integration (a) y : e 3x y = 3e 3x y (O) = 3 Tangent line: y - = 3(x - 0) y =3x+l y = e -3x y = _3e -3x y (O) = -3 Tangent line: y - = -3(x - 0) y = -3x (a) y = e x y = e x y (O) : Tangent line: y - = (x - 0) y = x+l (b) y = e -x y = _e -x y (o) -- Tangent line: y - = -(x - 0) S(x): f (x) : e x 40. y = e -Sx dy = _5e-SX y = -x y = e x In x 46. y = xe x y = xe x + ex() = ex(x + l) 47. y = x3e x, : x3e x + 3x~(e ~) : x~e~(x + 3): ex(x~ + 3x ~) 48. y = xe -x y = x(-e -x) + xe -x : xe- ~( - x) 49. g(t) = (e -t + e ) 3 g (,) : 3(e- + et)(e - e- ) 50. g(t) : e -3/t g,(t) = e_3/t(6t_3) = 6 t3 e3/t 5. y = ln( + e x) dy e x -- dx + e x 5. ha( l+ ex) = In(l+ ex) - In(l-e x) Y= ~,~) dy e x e x e x -- dx + e x - e x - e x 53. y - _ (ex +e_x) -~ ex + e -x 4. y = e "/~ dy e y = e -x dy = _xe-x 54. dy_dx (ex + e-x)-(ex -e-x) = (e :~ + e-:~) y ~~ e x _ e-x dy e x + e -x dx 43. y = e x-4 y = e x-4 -e x 44. f(x)= 3e l-x f (x) : 3ei-X (-x) : -6xe l-x 00 Brooks/Cole, Cengage Learning

5 5 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions -- e x 65. y = xe ~ -xe x +e x, (, e) y = xe x + xe x - xe x - e x + e x = xe x y () = e Tangent line: y - e = e(x - ) 57. y ~ ex(sin x + cos x) 58. y = ln ex = x dx e ~(cos x -sin x) + (sin x + cos x)(e*) ex( cos x) = e ~ cos x 59. F(x)= ::Xcose dt F (4 = ~os(e ~X). ± = ~os(~/ x x 60. F(x) =![ x ln(t + ) dt F (x) = ln(e x + )e ~" 6. :(x) = e -, (l, 0 f (x) =-e -x, f ()=-I Tangent line: y - = -l(x - ) 6, y = e -x+x, (, ) y=-x+ y = (x- )e -zr+xz, y () = Tangent line: y - = (x - ) y = x y= ln(e :3) = x ~, (-, 4) y = x y (-) = -4 Tangent line: y - 4 = -4(x + ) y = -4x - 4 e x + e 64. y : ~ -x (o, o) = e x _ e -x] 66. y = xe x -e x, (,0) y = xe x + e x _ e x = xe x y O) = e Tangent line: y - 0 = e(x - ) y=ex-e 67. :(x) = e -x ln x, (,0) f (x) = e-x(~;-e-xlnx = e-~(-~ - () = Tangent line: y - 0 = e-l(x - ) y = e e 68. f(x) = e ~lnx, (,0) :,(x) = :- x s () = e3 Tangent line: y - 0 = e3(x - ) y = e3(x -) 69. xe y - l Ox + 3 y = 0 xe y- : + e y : = 0 dx dx dy(xey + 3)= 0- e y dx ~ dy 0 - e y -- dx xe y ex~, + x _ y = 0 + x-ydy = 0 dx dy(xe~y - y)= -ye x, - x dx ~ dy ye "~ + x dx -- xe x* - y y (0) : 0 Tangent line: y = 0 00 Brooks/Cole, Cengage Learning

6 Section 5.4 Exponential Functions." Differentiation and Integration xe y + ye x =, (0, ) xeyy + ey + ye x + y e x = 0 At(0,): e+l+y = 0 y = -e - Tangent line: y - = (-e - )(x - 0) y : (-e-)x + l l+ln(v ) = e x-y, (,), [ y ] -~-[xy + y] = e x-y - at (, ): [y + ] = t- y y = 0 Tangent line: y - = 0(x - ) y=l f(x) = (3 + x)e -3x f (x) = (3 + x)(-3e -3x) + e -3x = (-7-6x)e -3x f"(x) = (-7-6x)(-3e -3x) - 6e -3" = 3(6x + S)e -3x 74. g(x) = ~x + ex ln x e g (x) = ~ + --x x + ex lnx g"(x) = Xe x -- e x e x ex ln x 4X 3/ X X ex(zx -) + +exlnx 4 x./ x x y = 4e -x y = _4e -x y" = 4e -x y"- y = 4e -x -4e-x = 0 y = e 3x + e -3x y = 3e 3x _ 3e-3X y" = 9e 3x + 9e -3x /,- = 9(e, _-o 77. y = ex(cosxi~x + sin x/~x) 78. y = ex(-.,4~ sin x/~x + ~ cos x/~x)+ ex(cos ~/-~x + sin w/-~x) = ex[( + x/~) cos w/-~x + ( - ~) sin ~x],"= ex[-(~ + )sin ~x+ (~- )cos ~x]+ e ;[( + ~)cos ~x+ (- ~)sin ~x~ = ex[(-, - ~) sin ~x + (- + ~) cos ~x~ -y + 3y= -e ~( + ~)cos ~x+ (- ~)sin ~x~+ 3eX[cos ~x + sin ~x~ = ex[o- )cos (,+ )sin Therefore, -y + 3y = -y" ~ y" - y + 3y = 0. y = ex(3 cos x - 4 sin x) y = ex(-6 sin x - 8cosx) + ex(3 cosx - 4sin x) = e ~(-losin x - 5 cosx) = -SeX( sin x + cos x) y" = -SEX(4 cos x - sin x) - 5e ~( sin x + cos x) = -SEX(5 cos x) = -5e x cos x y"- y = -5e xcos x - (-5eX)( sin x + cos x) = -5eX(3 cos x - 4 sin x) = -Sy Therefore, y" - y = -5y ~ y" - y + 5y = Brooks/Cole, Cengage Learning

7 54 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions ~x _ e-x f (x~ = = 0 when x = 0. f ()-" x" - ex + e-x >0 Relative minimum: (0, ) g(x) = ~l~g -(x-3)/ ",/Z g (x) = - --~z(x- 3)e_(X_3)/ g"(x) = ~(~- - )(x 4)e -(x-3)/ Relative maximum: 3, Points of inflection: t3,0.39 (,-~ze ),l _/~ ) (4, &e_,/~, (,0.4),(4,0.4 )~/;,r ) if(x) ex + e -x >0 e x _ e-x f"(x} = ~ = 0 when x = 0. Point of inflection: (0, 0) 8. g(x) = -~ e-(x-) / g (x) = - --~z(x- )e_(x_)/ g,,(~) = ~;(~- )(x -3)e -( ~-)/ ~ Relative maximum: (~ff), (,0.399) Points of inflection: -/") -/"] ll, -~ffe ), 3,-~ffe ) ~ (,0.4),(3,0.4) 83. o f(x).= xe -x f~ (X) :--X~ -x -t- xe -x = xe-*( -x) = 0when x = 0,. f"(x) = -e-x(x - x ) q- e-x( - x) = e-x(x -4x+)= 0whenx = _+ Relative minimum: (0, 0) Relative maximum: (, 4e -) x = _+~/~ Points of inflection: / ~ (3.44, 0.384),(0.586, 0.9) o 00 Brooks/Cole, Cengage Learning

8 Section 5.4 Exponential Functions." Differentiation and Integration f(x) = xe -~ f (x) = -xe -x + e -~ = e- ~(- x) = 0when x =. s"(4 : + (-e-do- x) = e-"(x-) = 0whenx =. Relative maximum: (, e -l) Point of inflection: (, e -) - (, e -I) / g(t) : + ( + t)e- g (t) : -( + t)e- g"(t) : te -t Relative maximum: (-, + e)~ (-,3.78) Point of inflection: (0, 3) A = (base)(height)= xe -x da dx _4xe_.~ + e_.~ = e-~(- x ) = 0when x - A = x/~e (a) S(c) l Oce -c c e c ce c+x = (C + X)e c ce x = f(c + x) = 0(c + ) -- Ce x --C = X C -- ec+x =CWX X 86. f(x) = - + e3x(4- x) f (x) : e3x(-)+ 3eaX(4 - x) = e 3 ~0-6x ( ) = 0whenx = f"(x) : e3x(-6) + 3e3X(0-6x) = e3x(4-8x) = 0when x = 4 3" Relative maximum: (-}, 96.94) Point of inflection: (-}, ) 00 ( ) -0,5 - ~-~J.5 o 0x x/(-e x) (C) A(X)... e" e x - (.8, 4.59) 9 0 The maximum area is 4.59 for x =.8 and f(x) =.547. x (d) c- e~-i lim c = l~ x--~. 0 + lim c = 0 x---~ oo Answers will vary. Sample answer: As x approaches 0 from the right, the height of the rectangle approaches. As x approaches ~o, the height of the rectangle approaches Brooks/Cole, Cengage Learning

9 56 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 89 f(x) = e x s (x) : ex Let (x, y) = (x, ex)be the point on the graphwhere the tangent line passes through the origin. Equating slopes, ex _ e x - 0 x-o =- x x = 5 y = e, y = e. P int: (~, e / 9. V = 5,000e - 686t, 0 _< t _< 0 (a) (b) d.._~v = _949e_O.686 t dt When t =, -- dv~ dt dv When t = 5,- ~, dt C) 0,000 Tangent line: y- e = e(x-½ y = ex o 90 Let (Xo, Y0) be the desired point on y = e -x e - t cos 4.9t _< 0.5 (3 inches equals one-fourth foot.) Using a graphing utility or Newton s Method, you have t > 7.79 seconds. y = -e -x (Slope of tangent line) = e x (Slope of normal line) y y-e-xo = exo (x- Xo) You want (0, O) to satisfy the equation: _e-xo = -xoexo XO exo -- = 0 = XO exo Using a graphing utility or Newton s Method,the solution is x0 = (0.463, e -0"463 ) ~x h P 0, In P a) o y = h is the regression line for data (h, In P). (b) lnp = ah +b P = e ah+b = ebe ah P = Ce ah, C = e b a = and C = e = 0, So, P = 0,957.7e - ~499h (C) ~,ooo o o

10 Section 5.4 Exponential Functions." Differentiation and Integration 57 dp = (0,957.7)(_0.499)e_0.499 h (d) -~ = e - 499h For h = 5,-- = dh For h = 8, dp -- ~ dh 94. (a) Linear model: V = t Quadratic model: V = 09.5t z t ,000 Quadratic. Linear[, L 0 0 (b) The slope represents the average loss in value per year. (c) Exponential model: V = 30,58.68(0.9074) t = 30,58.68 e t (d) As t -~ 0, V --~ 0 for the exponential model. The value tends to zero (e) When t = 4, V = -07 dollars/year When t = 8, V ~ -366 dollars/year 95. f(x)= e x f(o) = f (x) = e >: f (O) = f"(x) = e,< f"(o) = Pl(x) = +l(x-o)= + x P~(x) = + l(xi O)= )(x- 0) = + x f(x) = e x/:, f(o) =, f (x) : e 4, f (O) :7 f"(x) : e 4, f"(o) = -~ P~(x) = + ½( x-o) : x 7 + Pl(O) :, : 7 = -7 0)+ 0) P~(O) P~(x) : +-~(x- ~(x- : X X = P~ (x) : 4 -x + 7 : -7 The values of f, P~, P~ and their first derivatives agree at x = 0. The values of the second derivatives off and P agree at x = n =.! = = 479,00,600 Stirlings Formula: ~,~~,, ~ 475,687,487! 98 n = 5.5! = =,307,674,368,000 Stirlings Formula: 5t ~,~,,,, ~,300,430,7,00 ~.3004 x 0 l~ The values off, P~, and P and their first derivatives agree at x = Let u = 5x, du = 5 dr. [esx(5"~ dr = e 5x + C )0. Let u = -x 4, d// = i4x3dr. le-x4(_4~t ~x _-e -~4 ~ 0. Let u = x-l, du = dr. l e x- II x-l( dr ) l_x- = -~ e dr = -~e + C 00 Brooks/Cole, Cengage Learning

11 58 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions t). Let u = - 3x, du = -3 dx. Ie,-3.~dx = _½Iel-a,(_3)dx = --~ 03. Let u = x 3,du = 3x dx. = ~x Letu = e x +l, du = e xdx. + C. Let u = e ~ +e -x,du = (e ~ - e-x) dx. (ex + e-x) C e x + e -x iex(e, + )dx = l(ex + )(ex)dx _ (e" + ) --+C ~ Letu = ~,du =~&. 4. fe x + e x + e"g Ie "/~ Let u = -- du = dx. X ~ 7 gl/x dx = e "/; + C J e x = e x +x-e -x + C ~. le-x tan(e-x) dx : -I[tan(e-X)l(-e-X) dx = In cos(e -x ) + C 6. " "~ln{e z~-~) dx =" J{x - l)dx =x-x+c l~7. Let u = + ~-x 7. ~o [~e -x dx = -+ f e-~(-)dx = [--e-~l L Jo e~ - = -~( - e-) - e e 3-x d~ = ~-e3-x~ 4 = -e -~ + = - -- k -3 e 08. Let u = + e ~, du = e ~-x dx. e x f e x ln[ l~ dx = -~al + e -----Tdx =-k ~+ ezx) +c 09. Let u = -e x,du =-e xdx. l~-~,]i- ~x dx =-(-ex) /(-e~)dx = --}( - e x)3/ + C 0. Letu = e" +e -x,du = (e x - e-x) dx. e x _ e-x I~ dx = ln(e x + e -x) + C. Letu = e x-ex,- du = (e x +e-x) dx.! e~ + ~-x dx = lne x-e-x +C e x _ e-x 9. xe-x= dx = --~I~e-~(-x)dx =, L ]o =-+Ee - -~ _ ~ - O/e) _ e - e 0. l~ xe x3/ dx =. Z fo ~x3/(x dx 3 ~-~ \ ~ _ IeX3/~ - 3 ~-~ =--~I-~4 - (e4-3e4 ) 00 Brooks/Cole, Cengage Learning,

12 Section 5.4 Exponential Fidnctions: Differentiation and Integration Let u = --, du = --~ dx. x 3 e3/x 3e3/X _Z~dx 7. Let Id = ax, dld = ax dx. (Assume a : 0.) y = Ixear dx _- --~a f ax ae (ax) dx =a l ~_eax + C _x. Let u = --., du = -x dx. 3. Let u = l+e x,du = e xdx.!o~ dx = I / + = In( + e 6) - ln 4. Let u = 5 - M, du = -e x dx. ~dx = -I 5 - e x~ 5 -- e x = -ln(5 - e) + In 4 In 4 5. Let u = sin ~x, du = ~c cos ~cx dx. l o~r/esin~x cos ~cx dx = -~!~/esin~x(zc cos ~x) dx ~k AO 6. Let u = sec x, du = sec x tan x dx. e~r/ ~. [~/~e ~ ~x sec xtan x dx = J,d~ e... X{sec xtan x)dx ~/~ = l[e~o z~-]~/~ u -~/~ =~[e-l-e -] = "~[~ - -~TJ = e ~ x _ x - _-x - -~e -~ + C ~. f (x) = [.½(e+~ + e-x)ax = ½(e x- e -x) + C, f (O) = C = 0 f(x) = [.½(e~ - e-x)dx = ½(e x + e -x) + C f(o) = + C = ~ C~ = 0 f(x) = ½(e x + e -x) 30. f (x) = sinx+e x dx =-cosx+te + C~ f (0) = - 4- ~ 4- C, = -~ ~ C = _x f (x) = -cos x + e + s(x) = l(- os + + A.~ ) dx _x =-sinx+-~ +x+c f(0) = X + C -- ~" :=~ C = 0 ~x f(x] = x - sin x + =~ 3. (a) Y (b) ffl!is Iit (0,,,I )t s I ~ l Ill! " / i / //--,-,... i i.!// " 5 "- / /,,,/./.,,,,.,....,.,...+~ Sl~l-: -7,... dy = e_xl (0,) dx y= e-xl~ dx = -4e-xl(-~ dx ) = -me -x/ + C (0,): =-4e + C =-4+ C ~ C = 5 y = -4e -x/ + 5 / 6 00 Brooks/Cole, Cengage Learning

13 5 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions ~3. (a) y = Ixe - "~ dx = l~ie- x(-o.4x) dx ~ e- :~ + C = -.5e - x + C 0.4 -~ = -.5e +C =-.5+C ~ C = I y = -.5e - x + ~ /---4 x 37o Jo.,l xe dx, n : Midpoint Rule: Trapezoidal Rule: Simpson s Rule: Graphing utility: o xe-xdx, n = Midpoint Rule:.906 Trapezoidal Rule:.87 Simpson s Rule:.880 Graphing utility: I~] e- 39(t-48)dt Graphing utility: = 47.7% = e 5 - ~ ) t dt = - -e - 3x + = -- e_o.3x : e -a _ e-b = -e -3/ + = x = ln- = -In ln x = ~ =.3 minutes 0.3 4o x(t) = Ae ~ + Be-k ; A, B, k > 0 (a) x (t) = Ake ~ - Bke -~ = 0 Ae kt = Be -kt B A t = ~ By the first Derivative Test, this is a minimum. (b) x"(t)= Ak~e ~ + Bk~e -~ k z is the constant 00 Brooks/Cole, Cengage Learning

14 Section 5.4 Exponential Functions." Differentiation and Integration 5 4. t R lnr (a) In R = t R = e t = 48.78e - 6~55t (b) 4~o - o (C) IR(t)dt =ao[448"78e- 655 dt 637. liters 43. f(x) = ex.domain is (-~o, ~)and range is (0, c~). f is continuous, increasing, one-to-one, and concave upwards on its entire domain. lime x = 0andlime x = o~. X--+--oo 44o Yes. f(x) = Ce x, C a constant. 45. (a) Log Rule: (u = e x + ) 46. (a) (b) Substitution: (u = x ) -4 - (b) When x increases without bound, Ix approaches zero, and e l/x approaches. Therefore, f(x) approaches /( + ) =.So, f(x)has a horizontal asymptote at y =. As x approaches zero from the right, Ix approaches ~o, e ~/x approaches ~o and f(x) approaches zero. As x approaches zero from the left, Ix approaches -~o, e l/x approaches zero, and f(x) approaches. The limit does not exist because the left limit does not equal the right limit. Therefore, x = 0 is a nonremovable discontinuity. 47. l~ e t dt >_ l~ l dt [e ]~ >_ e: - > x ~ e x > + xforx > The graphs of f(x) : In x and g(x) : e x are mirror images across the line y = x. 49. e -x = x ~ f(x) = x-e -x f (x) = + e -~" Xn+l-= Xn xl = X = X --_ X 3 = X --_ X 4 = X 3 y(x,,) _ f (x,) X n f(x,).~ f(x) ~ Z (~) f(x3) ~ f (x3) X n -- e-xn + e -x" Approximate the root off to be x ~ Area = 8_ =!_~e_xd x = -e-x3 ~ = -e- ~ + e~ 3 o a Let z = ca: 8-3Z -- 8z - 3 = 0 8 -z = - + z 3 (3z+l)(z-3) = 0 z = 3 ~ e ~ =3~ a=ln3 e, z = -- ~ = ~ -- impossible 3 3 ] So, a = In Brooks/Cole, Cengage Learning

15 5 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 5. y - a > O,b > O,L > 0 al e-x] b b y tt ~ y" = Oifae -x/b = ~ b L y(b, In a], = ae -(6 ",)/b + a(/a) Therefore, the y-coordinate of the inflection point is L/. 5. f(x) - lnx x (a) f (x) - - In x x = Owhenx = e. On (0, e),f (x) > 0 ~ f is increasing. On (e, oo),f (x) < 0 ~ f is decreasing. (b) For e < A < B, youhave lna lnb -- > A B BInA > AInB In A s > In B a A B > B "4. (c) Because e < ~, from part (b) you have e y Section 5.5 Bases Other than e and Applcations. log ½ = log -3 = -3. 0g7 9 = 0g7 7/3 = o log 7 = 0 4. log,-- = log,,-og,,a = - a 5. (a) (b) 3 =8 log 8 = 3 3-~ = _. 3 log3 ~ = - 00 Brooks/Cole, Cengage Learning