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28

CAPÍTULO 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 −32

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 afunção 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
11
1− 2 −1
= −1=
=
.
22
2
2
− 2 2 4 6
f (− 2 3) = − 2 3 − 2
 = + = = 2.
3 3 3 3
f (1 2 ) = 1 2 − 2

f (a ) = a − 2 a
= a −2 a
= −a

4. S e f ( x ) =

ax + b
e d = − a , mostre que f ( f ( x )) = x
cx + d

31

f (x ) =

ax + b
cx −a

 ax + b 
f ( f ( x )) = f 

 cx − a 
 ax + b 
a⋅
+b
 cx − a 
=
 ax + b 
c⋅
 + (− a )
 cx − a 
a (ax + b )
+b
= cx − a
c (ax + b )
−a
cx − a
a (ax + b ) + b(cx − a )
cx − a
=

cx − a
c(ax + b ) − a(cx − a )
a 2 x + ab + bcx − ab
cax + cb − acx + a 2
a 2 x + bcx x a 2 + bc
=x
=
=2
cb + a 2
a + bc
=

(

)

5. Se f ( x ) = x 2 + 2 x ,achar

f (a + h ) − f (a )
, h ≠ 0 e interpretar o resultado
h

geometricamente.
2

(

)

f (a + h ) − f (a ) (a + h ) + 2(a + h ) − a 2 + 2a
=
h
h
2
2
a + 2ah + h + 2a + 2h − a 2 − 2a h(2a + h + 2 )
=
=
h
h
= 2a + 2 + h
A Figura que segue mostra a interpretação geométrica. Nesta figura, α é o ângulo
formado pela reta que passa pelos pontos (a, f (a )) e (a + h, f (a + h )) eo eixo
positivo dos x . O quociente obtido representa a tangente do ângulo α .

32

6. Dada Φ( x ) =

x −1
. Forme as expressões Φ(1 x ) e 1 Φ( x ) .
2x + 7

1− x
1 x −1
1− x
x
1− x
Φ(1 x ) =
=x=
.

=
1
2 + 7x
x 2 + 7x 2 + 7x
2⋅ + 7
x
x

1
1
2x + 7
=
=
.
x −1
Φ( x)
x −1
2x + 7

7. Dada a função f ( x ) = x 2 + 1 , mostrar que para a ≠ 0 f (1 a ) = f (a ) a2 .
2

1
1+ a2
f (a )
1
f (1 a ) =   + 1 = 2 + 1 =
= 2 , para a ≠ 0 .
2
a
a
a
a

8. Dada a função f ( x ) =

1
h
, mostrar que f (1 + h ) − f (1) = −
. Calcular.
x
1+ h

33

f (a + h ) − f (a ) .
f (1 + h ) − f (1) =

1
1 1−1− h − h
−=
=
1+ h 1
1+ h
1+ h

f (a + h ) − f (a ) =

1
1 a−a−h
−h
−=
=
a + h a a (a + h ) a (a + h )

9. Seja f (n )...
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