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# 01_Exercícios do Estudo de Funções_Resolvidos

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```x ax
\uf0ae\uf0b1\uf0a5
\uf03d \uf02d
ambos finitos)
Temos 11lim lim 0
1x x
x
xa
x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02d\uf03d \uf03d \uf03d
\uf02d
e
0
1
1lim
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf02d
\uf0e9 \uf0f9\uf03d \uf03d
\uf0ea \uf0fa\uf0eb \uf0fb
(finitos).
Logo, r: y = 0 x + 1 é assíntota inclinada (coincide com a horizontal).

Observação: Utilizamos
x\uf0ae\uf0b1\uf0a5 apenas por comodidade, visto que os
limites têm o mesmo valor.

b) 2
( ) , 1
1
x
f x x
x
\uf03d \uf0b9
\uf02d

1º) Assíntota horizontal:
Temos que 2 2
lim lim lim
1x x x
x x
x
x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf03d \uf03d \uf03d \uf0b1\uf0a5
\uf02d
. Logo, não há assíntota
horizontal.
2º) Assíntota vertical

Vemos que x = 1 não pertence ao domínio de f, mas é ponto de acumula-
ção do domínio de f e, também, que
2
1 1
1
lim ( ) lim
1 0x x
x
f x
x\uf0ae \uf02d \uf0ae \uf02d \uf02d
\uf0e9 \uf0f9
\uf03d \uf03d \uf03d \uf02d\uf0a5\uf0ea \uf0fa
\uf02d \uf0eb \uf0fb
e 2
1 1
1
lim ( ) lim
1 0x x
x
f x
x\uf0ae \uf02b \uf0ae \uf02b \uf02b
\uf0e9 \uf0f9
\uf03d \uf03d \uf03d \uf02b\uf0a5\uf0ea \uf0fa
\uf02d \uf0eb \uf0fb
.
O fato de haver limite tendendo ao infinito teremos r:
1x \uf03d como
assíntota vertical.

y a x b\uf03d \uf02b

(Deverá ocorrer
( )
lim
x
f x
a
x\uf0ae\uf0b1\uf0a5
\uf03d
e
\uf05b \uf05dlim ( )
x
b f x ax
\uf0ae\uf0b1\uf0a5
\uf03d \uf02d
ambos finitos)
Temos
2
1lim lim lim 1
1x x x
x
x xxa
x x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02d\uf03d \uf03d \uf03d \uf03d
\uf02d
e 2
1.
1
lim
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf02d
\uf0e9 \uf0f9
\uf03d \uf03d\uf0ea \uf0fa
\uf0eb \uf0fb

2
1 1
( 1)
lim lim 1
x x
x x
x x
x x
\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5\uf02d \uf02d
\uf02d \uf02d\uf0e9 \uf0f9 \uf0e9 \uf0f9\uf03d \uf03d\uf03d \uf03d\uf0ea \uf0fa \uf0ea \uf0fa\uf0eb \uf0fb\uf0eb \uf0fb
(ambos finitos).
Logo, r: y = 1 x + 1 é assíntota inclinada.

c) 3
2
8
( ) , 0
x
f x x
x
\uf02b
\uf03d \uf0b9

1º) Assíntota horizontal:
Temos que 3 3
2 2
8
lim lim lim
x x x
x x
x
x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02b
\uf03d \uf03d \uf03d \uf0b1\uf0a5
. Logo, não há assíntota
horizontal.

12

2º) Assíntota vertical

Vemos que x = 0 não pertence ao domínio de f, mas é ponto de acumula-
ção do domínio de f e, também, que
3
20 0
8 8
lim ( ) lim
0x x
x
f x
x\uf0ae \uf02d \uf0ae \uf02d
\uf02b \uf0e9 \uf0f9
\uf03d \uf03d \uf03d \uf02b\uf0a5\uf0ea \uf0fa
\uf0eb \uf0fb
e 3
20 0
8 8
lim ( ) lim
0x x
x
f x
x\uf0ae \uf02b \uf0ae \uf02b \uf02b
\uf0e9 \uf0f9\uf02b
\uf03d \uf03d \uf03d \uf02b\uf0a5\uf0ea \uf0fa
\uf0eb \uf0fb
.
O fato de haver limite tendendo ao infinito teremos r:
0x \uf03d como
assíntota vertical.

y a x b\uf03d \uf02b

(Deverá ocorrer
( )
lim
x
f x
a
x\uf0ae\uf0b1\uf0a5
\uf03d
e
\uf05b \uf05dlim ( )
x
b f x ax
\uf0ae\uf0b1\uf0a5
\uf03d \uf02d
ambos finitos)
Temos
3
3 32
3 3
8
8
lim lim lim 1
x x x
x
x xxa
x x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02d
\uf02d
\uf03d \uf03d \uf03d \uf03d
e 3
2
1.
8
lim
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf02d\uf0e9 \uf0f9
\uf03d \uf0ea \uf0fa
\uf0eb \uf0fb

3 2
2 2
88 ( )
lim lim 0
x x
x
x x
x x
\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02d\uf02d \uf02d\uf0e9 \uf0f9 \uf0e9 \uf0f9\uf03d \uf03d \uf03d\uf0ea \uf0fa \uf0ea \uf0fa\uf0eb \uf0fb\uf0eb \uf0fb
(ambos finitos).
Logo, r: y = 1 x + 0 é assíntota inclinada.

d)
sen
( ) , 0
x
f x x
x
\uf03d \uf0b9

1º) Assíntota horizontal:
Temos que
finitosen [ ]
lim lim 0
x x
x
x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf03d \uf03d
(finito). Logo,
: 0r y \uf03d é assíntota
horizontal.

2º) Assíntota vertical
Vemos que x = 0 não pertence ao domínio de f, mas é ponto de acumula-
ção do domínio de f e, também, que

0 0
sen
lim ( ) lim 1
x x
x
f x
x\uf0ae \uf0ae
\uf03d \uf03d
(finito), limite fundamental.
O fato de não haver limite tendendo ao infinito implica que não existe
assíntota vertical.

y a x b\uf03d \uf02b

(Deverá ocorrer
( )
lim
x
f x
a
x\uf0ae\uf0b1\uf0a5
\uf03d
e
\uf05b \uf05dlim ( )
x
b f x ax
\uf0ae\uf0b1\uf0a5
\uf03d \uf02d
ambos finitos)
Temos
2 2
finito
sen
sen [ ]
lim lim lim 0
x x x
x
xxa
x x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf03d \uf03d \uf03d \uf03d
e
sen
0.lim
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf0e9 \uf0f9\uf03d
\uf0ea \uf0fa\uf0eb \uf0fb
sen
lim 1
x
x
x\uf0ae\uf0b1\uf0a5
\uf03d \uf03d
(ambos finitos).
Logo, r: y = 0 x + 1 é assíntota inclinada (coincide com a horizontal).

13

e)
2( ) , 1( 1)
x
f x x
x
\uf03d \uf0b9
\uf02d

1º) Assíntota horizontal:
Temos que
2 2
finito
1
lim lim lim 0 ( )
( 1)x x x
x x
x x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf03d \uf03d \uf03d
\uf02d
. Logo, r:
0y \uf03d
é
assíntota horizontal.

2º) Assíntota vertical
Vemos que x = 1 não pertence ao domínio de f, mas é ponto de acumula-
ção do domínio de f e, também, que

2
1 1
1
lim ( ) lim
( 1) 0x x
x
f x
x\uf0ae \uf02d \uf0ae \uf02d
\uf02b
\uf03d \uf03d \uf03d \uf02b\uf0a5
\uf02d
\uf0e9 \uf0f9
\uf0ea \uf0fa
\uf0eb \uf0fb
e
2
1 1
1
0
( )
( 1)
lim lim
x x
x
f x
x
\uf02b
\uf0ae \uf02b \uf0ae \uf02b \uf02d
\uf03d \uf03d \uf03d \uf02b\uf0a5
\uf0e9 \uf0f9
\uf0ea \uf0fa
\uf0eb \uf0fb

O fato de haver limite tendendo ao infinito teremos r:
1x \uf03d como
assíntota vertical.

y a x b\uf03d \uf02b

Temos 2
2
1( 1)
lim lim 0
( 1)x x
x
x
a
x x\uf0ae \uf0ae\uf0b1\uf0a5 \uf0b1\uf0a5
\uf02d
\uf03d \uf03d \uf03d
\uf02d
e
2
0
( 1)
lim 0
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf02d
\uf03d \uf03d
\uf0e9 \uf0f9
\uf0ea \uf0fa\uf0eb \uf0fb
(finitos).
Logo, r: y = 0 x + 0 é assíntota inclinada (coincide com a horizontal).

f) 2
( ) , 2
2
x
f x x
x
\uf03d \uf0b9 \uf02d
\uf02b

1º) Assíntota horizontal:
Temos que 2 2
lim lim lim
2x x x
x x
x
x x\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf03d \uf03d \uf03d \uf0b1\uf0a5
\uf02b
. Logo, não há assíntota
horizontal.

2º) Assíntota vertical
Vemos que x =
\uf02d
2 não pertence ao domínio de f, mas é ponto de acumula-
ção do domínio de f e, também, que
2
( 2) ( 2)
4
0
lim ( ) lim
2x x
x
f x
x
\uf02d
\uf0ae \uf02d \uf02d \uf0ae \uf02d \uf02d
\uf0e9 \uf0f9
\uf03d \uf03d \uf03d \uf02d\uf0a5\uf0ea \uf0fa\uf02b \uf0eb \uf0fb
e 2
( 2) ( 2)
4
0
lim ( ) lim
2x x
x
f x
x
\uf02b
\uf0ae \uf02d \uf02b \uf0ae \uf02d \uf02b
\uf0e9 \uf0f9
\uf03d \uf03d \uf03d \uf02b\uf0a5\uf0ea \uf0fa\uf02b \uf0eb \uf0fb

O fato de haver limite tendendo ao infinito teremos r:
2x \uf03d \uf02d como
assíntota vertical.

y a x b\uf03d \uf02b

Temos
2
2
2
lim lim lim 1
x x x
x
x xx
x x x
a
\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02b
\uf02b
\uf03d \uf03d \uf03d \uf03d
e 2
1.
2
lim
x
x
x
x
b
\uf0ae\uf0b1\uf0a5
\uf02d
\uf02b
\uf0e9 \uf0f9
\uf03d \uf03d\uf0ea \uf0fa
\uf0eb \uf0fb

2
2
2 2
( 2)
lim lim 2
x x
x x
x x
x x
\uf0ae\uf0b1\uf0a5 \uf0ae\uf0b1\uf0a5
\uf02d
\uf02b \uf02b
\uf02d \uf02b\uf0e9 \uf0f9 \uf0e9 \uf0f9\uf03d \uf03d \uf03d \uf02d\uf0ea \uf0fa \uf0ea \uf0fa\uf0eb \uf0fb\uf0eb \uf0fb
(ambos finitos).
Logo, r:
1. 2y x\uf03d \uf02d

14

7) Esboçar o gráfico das funções dadas abaixo:
a) b)

c) d)

e) f)

g) h)

f(x) = x - x
3 2
2 0
3
x

f(x) = x - x
4 2
1
0
2
y
x-1

f(x) = x . x
1
0
y
x
ln

x
y
f(x) = x - xarctg

x
y
f(x) = x e
x
0

x
y
f(x) =
20
x -
x
2
1

x
y
f(x) = e
0
/x1

x
y
f(x) =
0
x -
2
1
x
( )
1

15

i)

8) A função f, real de variável real, tem seu gráfico cartesiano descrito abaixo.
Sabendo- se que possui derivadas até terceira ordem, pede os esboços gráficos de
' e ''.f f

a) I1(0, 3) e I2(?, 2) são pontos de inflexão de f.
m1(-1, 2) e m2(3, 0) são pontos mínimos de f.
M(1, 4) é ponto máximo de f.

b)
'( ) 0f x \uf03c em , 1 1, 3(] [ ] [)x \uf02d\uf0a5 \uf02d\uf0ce \uf0c8, visto que f é decrescente nestes intervalos.
'( ) 0f x \uf03e em 1,1 ,(] [ ]3 [)x \uf02d \uf02b\uf0a5\uf0ce \uf0c8 ,```