Solução_Calculo_Stewart_6e

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F. TX.10 x +1 (d) (g ◦ g)(x) =g(g(x)) = g = x +2 x +1 x +2 +1 = x +1 x +2 +2 SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 21 x +1+1(x +2) x +2 x +1+2(x +2) x +2 Since g(x) is not defined for x = −2 and g(g(x)) is not defined for x = − 5 3 , the domain of (g ◦ g)(x) is D = x | x 6=−2, − 5 3 . 37. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x − 1)) = f(2(x − 1)) = 2(x − 1) + 1 = 2x − 1 39. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 3 +2))=f[(x 3 +2) 2 ] = f(x 6 +4x 3 +4)= (x 6 +4x 3 +4)− 3= √ x 6 +4x 3 +1 = x +1+x +2 2x +3 = x +1+2x +4 3x +5 41. Let g(x) =x 2 +1and f(x) =x 10 .Then(f ◦ g)(x) =f(g(x)) = f(x 2 +1)=(x 2 +1) 10 = F (x). 43. Let g(x) = 3√ x and f(x) = x 1+x .Then(f ◦ g)(x) =f(g(x)) = f( 3√ x )= 3√ x 1+ 3√ x = F (x). 45. Let g(t) =cost and f(t) = √ t.Then(f ◦ g)(t) =f(g(t)) = f(cos t) = √ cos t = u(t). 47. Let h(x) =x 2 , g(x) =3 x ,andf(x) =1− x. Then (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 2 )) = f 3 x2 =1− 3 x2 = H(x). 49. Let h(x) = √ x, g(x) =secx, andf(x) =x 4 .Then (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g( √ x )) = f(sec √ x )=(sec √ x ) 4 =sec 4 ( √ x )=H(x). 51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus,f(g(2)) = f(5) = 4, because the point (5, 4) is on the graph of f. (b) g(f(0)) = g(0) = 3 (c) (f ◦ g)(0) = f(g(0)) = f(3) = 0 (d) (g ◦ f)(6) = g(f(6)) = g(6). This value is not defined, because there is no point on the graph of g that has x-coordinate 6. (e) (g ◦ g)(−2) = g(g(−2)) = g(1) = 4 (f ) (f ◦ f)(4) = f(f(4)) = f(2) = −2 53. (a) Using the relationship distance = rate · time with the radius r as the distance, we have r(t) =60t. (b) A = πr 2 ⇒ (A ◦ r)(t) =A(r(t)) = π(60t) 2 = 3600πt 2 . This formula gives us the extent of the rippled area (in cm 2 )atanytimet. 55. (a) From the figure, we have a right triangle with legs 6 and d, and hypotenuse s. By the Pythagorean Theorem, d 2 +6 2 = s 2 ⇒ s = f(d) = √ d 2 +36. (b) Using d = rt,wegetd =(30km/hr)(t hr) =30t (in km). Thus, d = g(t) =30t. (c) (f ◦ g)(t) =f(g(t)) = f(30t) = (30t) 2 +36= √ 900t 2 +36. This function represents the distance between the lighthouse and the ship as a function of the time elapsed since noon.
F. TX.10 22 ¤ CHAPTER 1 FUNCTIONS AND MODELS 57. (a) (b) H(t) = 0 if t
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