funçoes

Resolu¸˜o ca 3.1
a) f (0) = 0

b) f (1) = 2a − 1

c) f (a) = a2

d) f (−a) = −3a2

e) f (2a) = 0

f) 3f (a) + f (−2a) = −5a2

3.2
a) Df = R

b) Df = {x ∈ R : x − 4 = 0} = R\{4}

c) Df = {x ∈ R : x − 2 ≥ 0} = [2, +∞[

d) Df = {x ∈ R : x − 5 = 0} = R \ {5}

e) Df = [ 4, +∞ [

f) Df = ] 1, +∞ [

3.3

Para f ,

f (1 + 2) = f (3) = 15

e

f (1) + f (2) = 5 + 5 × 2 = 15

e portanto f (1 + 2) = f (1) + f (2).
Para g, g(1 + 2) = g(3) = 2 × 32 = 18

e

g(1) + g(2) = 2 + 2 × 22 = 10

obtemos que g(1 + 2) = g(1) + g(2).
Para h, h(1 + 2) = h(3) =

√

3

e

h(1) + h(2) = 1 +

√

2

e logo h(1 + 2) = h(1) + h(2).
3.4
a)

f (x) g(x) =

2x x2 +1

b) (f ◦ g)(x) = f (g(x)) = f (x2 + 1) = 2(x2 + 1)
c) (g ◦ f )(x) = g(f (x)) = g(2x) = (2x)2 + 1 = 4x2 + 1
3.5
a) O declive ´ a = e 8−3
5−2

= 5 , e a equa¸˜o da recta fica ca 3

5
5
1 y = (x − 2) + 3 = x −
3
3
3
Solu¸˜es dos exerc´ co ıcios

14

b) O declive ´ a = e −5−(−3)
2−(−1)

2
= − 3 , e a equa¸˜o da recta fica ca 2
2
11 y = − (x − 2) + (−5) = − x −
3
3
3
c) O declive ´ a = e y=

3
− 17
−1 − 2
51
5
= 10 = e a equa¸˜o da recta fica ca 1
1
1
5
−6
3 − 2

51
1
3
51
18
(x − ) + = x −
5
2
2
5
5

d) A equa¸˜o da recta fica y = 2(x − 1) + 3 = 2x + 1. ca e) A equa¸˜o da recta fica y = − 1 x. ca 2
f) O declive ´ e b−0 b = − ; e a equa¸˜o da recta fica ca 0−a a b b y = − (x − a) + 0 = − x + b a a
3.6

a) |5x − 3| = 12 ⇔ 5x − 3 = 12 ∨ 5x − 3 = −12 ⇔ x = 3 ∨ x = −

9
5

CS = {− 9 , 3}.
5
2
b) Para 3x + 2 ≥ 0 ⇔ x ≥ − 3 temos,

|3x + 2| = 5 − x ⇔ 3x + 2 = 5 − x ⇔ 4x = 3 ⇔ x = e como

3
4

3
4

≥ − 2 , ´ uma solu¸˜o poss´ ca ıvel.
3 e

Para 3x + 2 < 0 ⇔ x < − 2 temos,
3
|3x + 2| = 5 − x ⇔ 3x + 2 = −(5 − x) ⇔ 2x = −7 ⇔ x = −

7
2

2
− 7 < − 3 , pelo que ´ uma solu¸˜o poss´ e ca ıvel. 2

O conjunto solu¸˜o fica CS = {− 7 , 3 }. ca 2 4
c) |x