Exercicios resolvidos

28

CAPÍTULO 2
SEÇÃO 2.10 – página 20
1. Se f ( x ) = x2 − 4 , achar: x −1

a) f (0 ) =

02 − 4 − 4 = = 4. 0 −1 −1

b) f (− 2 ) =

(− 2)2 − 4 = 4 − 4 (− 2) − 1 − 3
2

= 0.

1 1   −4 −4 1 − 4t 2 t 1 − 4t 2 t  t2 = = ⋅ = . c) f (1 t ) = 1 1− t 1− t t2 t −t2 −1 t t d)

( x − 2 )2 − 4 = x 2 − 4 x + 4 − 4 = f (x − 2) = x − 2 −1
2

x −3

x 2 − 4x . x−3

1 1   −4 −4 1 − 16 2 − 15 15 2 4 e) f (1 2) = = = ⋅ = = . 1 1 4 1− 2 − 2 2 −1 −1 2 2
f)

(t ) − 4 = t f (t ) = t −1 t
2 2 2 2

4 2

−4 . −1

2. Se f ( x ) =

3x − 1 , determine: x−7

a)

5 f (− 1) − 2 f (0 ) + 3 f (5) 7

f (− 1) =

3(− 1) − 1 − 3 − 1 − 4 1 = = = −1− 7 −8 −8 2

f (0 ) =

3× 0 −1 −1 1 = = 0−7 −7 7

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f (5) =

3(5) − 1 15 − 1 14 = = = −7 5−7 −2 −2

Portanto, 5 f (− 1) − 2 f (0 ) + 3 f (5) = 7

5 =

= = =

1 1 − 2 + 3 (− 7 ) 2 7 7 5 2 − − 21 2 7 7 35 − 4 − 294 1 ⋅ 14 7 − 263 1 − 263 ⋅ = 14 7 98

b)

[ f (− 1 2)]2

 −1  −1 3  = 2  − 1 −7     2 
2

2

 −3−2  2    2  = −5 ⋅ 2  = 1   9  − 1 − 14   2 − 15     2  3(3 x − 2 ) − 1 9 x − 7 = . 3x − 2 − 7 3x − 9

c) f (3 x − 2 ) =

d) f (t ) + f (4 t ) =

4 3  − 1 3t − 1  t  3t − 1 12 − t t = + = + ⋅ 4 t −7 t −7 t 4 − 7t −7 t (3t − 1) (4 − 7t ) + (12 − t ) (t − 7 ) = 12t − 21t 2 − 4 + 7t + 12t − 84 − t 2 + 7t = 4t − 7t 2 − 28 + 49 − 7t 2 + 53t − 28 − 22t 2 + 38t − 88 = . − 7t 2 + 53t − 28

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e)

f (h ) − f (0 ) = h
 3h − 1 3 ⋅ 0 − 1  1 = − ⋅ 0−7  h  h−7  3h − 1 1  1 = − ⋅  h−7 7 h 21h − 7 − 1(h − 7 ) 1 = ⋅ 7(h − 7 ) h 20h 1 = ⋅ 7(h − 7 ) h 20 = 7(h − 7 )

f) f [ f (5)]

f (5) =

3 ⋅ 5 − 1 14 = −7 = 5−7 −2
3(− 7 ) − 1 − 21 − 1 − 22 11 = = = . (− 7 ) − 7 − 14 − 14 7

f [ f (5)] = f (− 7 ) =

3. Dada a função f ( x ) = x − 2 x , calcular. f (− 1) , f (1 2 ) e f (− 2 3) . Mostrar que

f (a ) = − a .

f (− 1) = − 1 − 2(− 1) = 1 + 2 = 3
1 1 1− 2 −1 = −1= = . 2 2 2 2 − 2 2 4 6 f (− 2 3) = − 2 3 − 2  = +