Solução_Calculo_Stewart_6e

F.
TX.10
20 ¤ CHAPTER 1 FUNCTIONS AND MODELS
29. f(x) =x 3 +2x 2 ; g(x) =3x 2 − 1. D = R for both f and g.
(f + g)(x) =(x 3 +2x 2 )+(3x 2 − 1) = x 3 +5x 2 − 1, D = R.
(f − g)(x) =(x 3 +2x 2 ) − (3x 2 − 1) = x 3 − x 2 +1, D = R.
(fg)(x) =(x 3 +2x 2 )(3x 2 − 1) = 3x 5 +6x 4 − x 3 − 2x 2 , D = R.
f
(x) = x3 +2x 2
g 3x 2 − 1 , D = x | x 6=± √ 1
since 3x 2 − 1 6= 0.
3
31. f(x) =x 2 − 1, D = R; g(x) =2x +1, D = R.
(a) (f ◦ g)(x) =f(g(x)) = f(2x +1)=(2x +1) 2 − 1=(4x 2 +4x +1)− 1=4x 2 +4x, D = R.
(b) (g ◦ f)(x) =g(f(x)) = g(x 2 − 1) = 2(x 2 − 1) + 1 = (2x 2 − 2) + 1 = 2x 2 − 1, D = R.
(c) (f ◦ f)(x) =f(f(x)) = f(x 2 − 1) = (x 2 − 1) 2 − 1=(x 4 − 2x 2 +1)− 1=x 4 − 2x 2 , D = R.
(d) (g ◦ g)(x) =g(g(x)) = g(2x +1)=2(2x +1)+1=(4x +2)+1=4x +3, D = R.
33. f(x) =1− 3x; g(x) =cosx. D = R for both f and g, and hence for their composites.
(a) (f ◦ g)(x) =f(g(x)) = f(cos x) =1− 3cosx.
(b) (g ◦ f)(x) =g(f(x)) = g(1 − 3x) =cos(1− 3x).
(c) (f ◦ f)(x) =f(f(x)) = f(1 − 3x) =1− 3(1 − 3x) =1− 3+9x =9x − 2.
(d) (g ◦ g)(x) =g(g(x)) = g(cos x) =cos(cosx)
[Note that this is not cos x · cos x.]
35. f(x) =x + 1 +1
, D = {x | x 6= 0}; g(x) =x , D = {x | x 6=−2}
x x +2
x +1
(a) (f ◦ g)(x) =f(g(x)) = f = x +1
x +2 x +2 + 1 = x +1
x +1 x +2 + x +2
x +1
x +2
=
(x +1)(x +1)+(x +2)(x +2)
(x +2)(x +1)
=
x 2 +2x +1 + x 2 +4x +4
(x +2)(x +1)
Since g(x) is not defined for x = −2 and f(g(x)) is not defined for x = −2 and x = −1,
the domain of (f ◦ g)(x) is D = {x | x 6=−2, −1}.
(b) (g ◦ f)(x) =g(f(x)) = g x + 1
=
x
x + 1
+1
x
x + 1 =
+2
x
x 2 +1+x
x
x 2 +1+2x
x
Since f(x) is not defined for x =0and g(f(x)) is not defined for x = −1,
the domain of (g ◦ f)(x) is D = {x | x 6=−1, 0}.
(c) (f ◦ f)(x)=f(f(x)) = f x + 1
= x + 1
+ 1
x x x + 1 x
= x(x) x 2 +1 +1 x 2 +1 + x(x)
x(x 2 +1)
= x4 +3x 2 +1
, D = {x | x 6= 0}
x(x 2 +1)
= x + 1 x + 1
x 2 +1
x
= x4 + x 2 + x 2 +1+x 2
x(x 2 +1)
= x2 + x +1
x 2 +2x +1 = x2 + x +1
(x +1) 2
= x + 1 x + x
x 2 +1
= 2x2 +6x +5
(x +2)(x +1)