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Chapter 4 6 - 268 43 45 47 49 50 51 52 53 54 CHAPTER 4...

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Unformatted text preview: /, 268 43. 45. 47. 49. 50. 51. 52. 53. 54. CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f(0) = 2cos€+sin26 => f'(0) 2 22sin6+2sin90080. yf’(6) 2 0 2 25i119(cos0 2 1) 2 0 => sinfi 2 0 or c050 2 1 => 9 2 mr (ii an integer) or 0 2 27m. The solutions 6 2 mt include the solutions 0 2 27m. so the critical numbers are 0 2 mr. .g(6)240—tan9 => g'(9)242sec20. g’(0)20 => sec2624 => 5e002i2 2 c030 2 i—% 2 6 2 g + 27m. 5?" + 21m. 2?" + 27177. and 47" + 27m are critical numbers. Note: The values of 6 that make g’(0) undefined are not in the domain of g. f($)2arlnx 2> f'(x)2x(1/m)+(lnac)-12lnm+1. f'(av)20 42> lnx2—1 <:> as 2 e‘1 2 1/6. Therefore. the only critical number is a: 2 1/6. f(m) 2 3062“” 2> f’(:v) 2 m(262$) + e21 2 e2m(2x —I— 1). Since 621 is never 0. we have f’(:z:) 2 0 only when 2:5 + 1 2 0 <:> a: 2 2%. So 2% is the only Critical number. fins) 2 331:2 2 12:16 + 5. [0.3]. f’(ac) 2 6x — 12 2 0 2) a: 2 2. Applying the Closed Interval Method. we find that f(0) 2 5. f(2) 2 27. and f(3) 2 24. So f(0) 2 5 is the absolute maximum value and f(2) 2 27 is the absolute minimum value. f(a:) 2 $3 2 33: + 1, [0,3]. f’(a:) 2 3:102 2 3 2 0 42) m 2 :1, but 21 is not in [0.3]. f(0) 2 1. f(1) 2 21. and f(3) 2 19. So f(3) 2 19 is the absolute maximum value and f(1) 2 21 is the absolute minimum value. f(;v) : 2353 - 318 — 12x + 1. [22,3]. f’(a:) = 6x2 _ 61: — 12 = 60:2 — x — 2) : 6(x — 2)(cc + 1) = 0 42> x 2 2. 21. f(—2) 2 23. f(—1) 2 8. f(2) 2 219. and f(3) 2 28. So f(21) 2 8 is the absolute maximum value and f (2) 2 219 is the absolute minimum value. f(:l3) : m3 — 63:2 + 9m + 2. [—1.4]. f’(m) : 39:2 — 12$ + 9 : 3(m2 — 43: + 3) : 3(z 21x9: — 3) : 0 4:» :c 2 1.3. f(21) 2 214. f(1) 2 6. f(3) 2 2, and f(4) 2 6. So f(1) 2 f(4) 2 6 is the absolute maximum value and f(21) 2 214 is the absolute minimum value. f(x) 2 $4 2 2362 +3. [22.3]. f'(x) 2 4m3 24$ 2 4m($2 2 1) 2 4$(a:+ 1)(a: 2 1) 2 0 <=> w 2 21,0,1. f(—2) — 11. f( 1) 2. f(0) 3. f(1) 2. f(3) 66. So f(3) 2 66 is the absolute maximum value and f(::1) 2 2 is the absolute minimum value. f(:n) 2 (:52 — 1)3. [21,2]. f'(:r) 2 3(m2 2 1).2(2m) 2 6M1: + 1)2 (m 2 1)2 2 0 (2) m 2 21,0,1. f(::1) 2 O. f(0) 2 21. and f(2) 2 27. So f(2) 2 27 is the absolute maximum value and f(0) 2 21 is the absolute minimum value. 2 g _ 2 (1? +1) I(Zm)2 1 m —0 <—‘> x2:I:1.but21isnotin[0,2]. W) : (m2 + 1)2 _ ($2 + 1)2 f(0) 2 0. f(1) 2 l f(2) 2 %. So f(1) 2 % is the absolute maximum value and f(0) 2 0 is the absolute 2. (I: x2+1’ [0121- f'($) 2 minimum value. $2 — 4 (.732 + 4)(2a:) — ($2 2 4)(2m) _ 16:1: ft”) : (1)2 +4~ [—4941' f,(33) * ($2+4)2 ($2+4)2 " 0 * 1: — 0. f(j:4) 2 :—§ 2 g and f(0) 2 21. So f(j:4) 2 g is the absolute maximum value and f(0) 2 21 is the absolute minimum value. ...
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