01_Exercícios do Estudo de Funções_Resolvidos
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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. 
 
 3º) Assíntota inclinada: 
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. 
 
 3º) Assíntota inclinada: 
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. 
 
 3º) Assíntota inclinada: 
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. 
 
 3º) Assíntota inclinada: 
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. 
 
 3º) Assíntota inclinada: 
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
 é assíntota inclinada. 
 
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 ,