7.4_23 - secnou 14 GENERAL LOGAHITHMIC AND EXPONENTIAL...

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Unformatted text preview: secnou 14* GENERAL LOGAHITHMIC AND EXPONENTIAL FUNCTIONS 23. h(t) = t3 — 3t => h’(t) = 312 — 3t 1113 Z73 25. Using Formula 4 and the Chain Rule, y 2 5—1/1 => 3/ = 5—1/z(1n5)[—1-(—x—2)] = 5-1/$(1ns)/gc2 27. f(u) = (2" +2”)10 : f’(u) = 10(2“ + 2‘“)9 Edd (2” + 2‘") = 10(2" + 2‘“)9 [2“1112 + 2*“ 1112 - (4)] = 101n2(2" + 2—u)9(2“ — 2‘“) 2x . =1 2— ’ : 2 = 29 W”) °g3($ 4) 3 f (w) ($2—4)ln3( 3” (912—4) 1113 31.312231 => lnyzmlnm => y’/y=ln$+x(1/m) => y’zwmflnar—l-l) , . . 33.y=m5i” => lny=sinmlnsc :> %=cosxlnm+31:x © yl=cc5in$(cosmln$+smm> an 35 —(l )3: ¢ In —mlnlnw :> y—I—lnlnm—f—m-i l : '—(ln )9” lnl + 1 .y— 11w 3]— y- Inn: on y— as my lnm z y/ 62 r 1 1 37.y=me :> lny:e””ln:c => Ezemlnx+g : yl=xe font—FE) 39. = 10”” => 3/ 2 10m In 10, so at 1, 10 , the $10 e ofthe tangent line is 101 y P is y — 10 2 mm 10(110 — 1), org 2 (101n10)$ + 10(1 — In 10). 2 t 2 2 1 _ 41./ 10tdt=[10 ] _ 10 10 100 10_ 90 1 111101—11110—11110: 11110 ”11110 43. m dw = / W dcc = L ln—w dm. Now put 11 : ln ac, so du = 1 dm, and the expression a: m In 10 a: m 1 In 10 = mm 10, and its equation 1 _ 1 1 2 _ 1 2 becomes lnlO/udu— 11110 (211 +C1) _ 21nwflnac) +0. Or: The substitution u = log10 .1; gives du = $12130 and we get/ 1020 x dm : % ln 10(log10 :02 + C. 45. Letu = sin0. Then du = cos6d0 and/ 35m9 cos0d9 = /3”du 2 1131—3 +C’ = $35k” + C. 0 1 47. A 2/ (2m — 53) d2: +/ (5”E — 21) d9; —1 0 _ 2:1: 5a: 0 52: 2x 1 *im—mifiim‘ml. _ 1 1 1/2 1/5 5 2 1 1 (__)(__)<__>(__> _ 16 1 ‘51n5_2_1fi 49. We see that the graphs ofy = 2m and y = 1 + 3"” m m 0.6. We let flan) : 2m — 1 — 3—ac intersect at 4-1 and calculate f’(m) = 2“” ln 2 + 3” ln 3, and using the formula acn+1 2 son — flat”) /f’($n) (Newton’s Method), we get $1 = 0.6, :52 w 303 % 0.600967. So, correct to six decimal places, the root ' _1 4m 2 ( occurs at a: = 0.600967. ...
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