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28CAPÍTULO 2
SEÇÃO 2.10 – página 20
1. S e f ( x ) =
a) f (0 ) =
x2 − 4
, achar: x −1
02 − 4 − 4
=
= 4.
0 −1
−1
b) f (− 2 ) =
(− 2)2 − 4 = 4 − 4
(− 2) − 1 − 3
= 0.
2
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
x −3
2
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
(t ) − 4 = t f (t ) = t −1 t 22
f)
4
2
2
2
2. Se f ( x ) =
a)
−4
.
−1
3x − 1
, determine: x−7 5 f (− 1) − 2 f (0 ) + 3 f (5)
7
f (− 1) =
f (0 ) =
3(− 1) − 1 − 3 − 1 − 4 1
=
=
=
−1− 7
−8
−8 2
3× 0 −1 −1 1
=
=
0−7
−7 7
29
f (5) =
3(5) − 1 15 − 1 14
=
=
= −7
5−7
−2
−2
Portanto,
5 f (− 1) − 2 f (0 ) + 3 f (5)
=
7
1
1
− 2 + 3 (− 7 )
2
7
7
5
2
− − 21
7
2
7
35 − 4 − 294 1
⋅
14
7
− 263 1 − 263
⋅=
14 7
98
5
=
=
=
=
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
c) f (3 x − 2 ) =
3(3 x − 2 ) − 1 9 x − 7
=
.
3x − 2 − 7
3x − 9
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
30
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
f [ f (5)] = f (− 7 ) =
3(− 7 ) − 1 − 21 − 1 − 22 11
=
=
=.
(− 7 ) − 7
− 14
− 14 7
3. Dada a