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# File0025 - SECTION 7.3 THE NATURAL EXPONENTIAL FUNCTION I...

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Unformatted text preview: SECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION I. Let u = 73w. Then du = —3 dx, so -5 _ 715 . 7 _ 1 Joe ”dab—é 0 e”dU=—%leulolS=-%(e 15—60): %(1-P 15 71. Letu=1+e. Thenduzem dagsofem \/1+e“da:: ffdu:3u3/2+ C: %(1+e\$)3/2+C. 73. f68:1dm2/(1+e_:)dac=m—e‘:—+C 75.Letu:\/;E.Thendu=2 ﬁdx, s0/% dw=2feu du=2e“ +C=2ef+0 77. Areazfo1 (63” e)d:v= [%63‘”— 9’10: (383 —e) —(%——1) 2%83-e+§ @4544 79. V = f017r(e””)2 den :71th1 e21 dm 2 éﬂez‘]; = §(e2 — 1) 81.y—ln(:cl3) e”—x+3 x = 6” — 3. Interchanging a: and y, we get yzemﬁ3,sof“1(w)=ef—3. . We use Theorem 7.1.7. Note that f(0) : 3 + 0 + e0 = 4, so f_1(4) = 0. Also f’(m) = 1 + 6““. Therefore, 1 1 1 1 ( 1(4): 1011(4)) =f’(0):1+e°: . Using the second law of logarithms and Equation 5, we have ln(e“”/ey) = ln ef — 1n ey : 1: — y = 1n(e”” 11). Since In is a one-to—one function, it follows that ew/ey = e” _ y. . (a) Let f(m) = 3”” — 1 — ac. Now f(0) : eo — 1 = 0, and form 2 0, we have f’(m) = em —1 2 0. Now. since f(0)=Oandfisincreasingon[0,00),f(w)ZOformZO => 66—1—2220 : 6321+LL‘. (b) ForO < a: < 1, 3:2 g as, so 812 g e” [since 8” is increasing]. Hence [from (a)] 1 + m2 g e’”2 S e“. 80—: f01(1 1+Im2)dmgfolem2dmgfgewdxze—l<e => ggfolemzdxge. \$2 ‘"‘“ wk .(a) By Exercise 87(a), the result holds for n— — 1. Suppose that e’” > 1 + a: +2 — !-+ -+ % for 1: >0. xk \$k+1 Letf(zt)_e” 1 an 2! ----k! (k+1)!.Thenf'(;c)262—1—w—---—— by assumption. Hence f (:5) is increasing on (O, 00). So 0 g x implies that \$k \$k+l 0=f(0)£f(:n) ex 1 a: k1 h w>1 — (k~1)!’and encee _ +x+ '12—‘— 1+ (k:+1)! 2 mm for a: 2 0. Therefore, for ac 2 0, e“ 2 1 + :c + %- —— + H for every positive integer n, by mathematical induction. (b) Takingn=4andm = 11n(a),wehavee=e1 2 1+ % + % + 513‘ = 2.7083 > 2.7. k k+1 (c)e\$21+m+-+%+<lfl 1 1 1 ' ., + +---+—+ “3 T ea: _ >__ > . 2} 50k — mk—l kl (1+1)! - (19+ 1)! k+1)l wk 9: 6 But \$1320 (k+ 1)! _ 00, so Jingle—k —oo. ...
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