# 7.3_69 - 69 71 73 75 77 79 81 85 87 89 SECTION 7.3 THE...

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Unformatted text preview: 69. 71. 73. 75. 77. 79. 81. 85. 87. 89. SECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION Let u— — —3\$. Then du— — —3dm so I? d._ — —% 0 15 ewu— — elem” = aw —e°> = 1(1 _.—15). Letu=1+em. Thenduzem dac, sofe 1/1+ewdm:f\/_du=—u 3”+0:§(1+em)3/2+C. /ew +1dm=/(1+e_z)dm=\$—e‘m+0 em Letu:ﬁ.Thendu=~1—dm,so/&dm=2feudu=2eu+C=2eﬁ+Cﬂ 2x5 W? Area: fol (63w —ez) da: 2 [éem —em];= (163 —e)— (g -1) = §e3 —e+§ m4.644 V = fol 7r(ea”)2 dm 2 7rf0162z dac = é’rt[e2””]; = §(e2 — 1) y_ln(ac : 3) -> ey—m+3 as = e y — 3. Interchanging a: and y, we get yzez—3,sof—1(m)=em—3. . We use Theorem 7.1.7. Note that f(0) = 3 + 0 + e0 = 4, so f‘v1(4) = 0. Also f’(ac) = 1 + 6“”. Therefore, _1, 1 _ 1 _ 1 _1 (f )(4)_f’(f‘1(4)) f’(0) 1+e° 2' Using the second law of logarithms and Equation 5, we have ln(e“‘/ey) = ln 6”” — ln ey = w — y = ln(e“” ’ 9). Since ln is a one-to—one function, it follows that ew/ey 2 ex _ y. (a) Let ﬂat) 2 eat — 1 — ac. Now f(0) = 60 — 1 = 0, and form 2 0, we have f’(m) = e” — 1 Z 0. Now, since f(0) = Oandfisincreasingon [0,00), f(:c) 2 Oform Z 0 => em — 1 — ac 2 0 => em 2 1 +1.: (b) For 0 < a: < 1 :52 < at, so e902 < 6“ [since e“E is increasing]. Hence [from (a)] 1 + m2 S 622 S e“. So—=f01-2(1+m)dm<folem2dm£folezdm=e—l<e => ggfolemzdwge. (a) By Exercise 87(a) the result holds for n— — 1. Suppose that e” > 1 + :2 +372 — !-+ -+ % for :1: > 0. 2 k: k+1 k: x x a: w :w_ _ _________ .Th ’ 2w- _ _...__> Let f(\$) e 1 w 2! k! (k+ 1)! enf (at) e 1 m k! _ 0 by assumption. Hence f(ac) is increasing on (0, 00). S0 0 g 3: implies that mic \$k+1mlc \$k+1 =f(0)5f(z)—e°‘ 1 ac k! (k__1)!,,andhencee“ >1+m+- -+'—+(k:+1)! \$2 71 for m > 0. Therefore, for m > 0,61 > 1 + m + % —— ' ~ + % for every positive integer n, by mathematical induction. . (b) Takingn=4andm=1in(a),wehavee=el 2 1+g+ é +51; =2.708§> 2.7. wk k+1 m m e 1 1 a: a; m > 1 — — > > . (0)6 - +9” "I“ k!!+(k+1) T mk‘EB—k+wk—1+ 'l+m1+(k+1) (k+1)! a: 6“” 311211.120 (k+1)l — —00, so \$121010; 2 oo. '271 ...
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