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CURSOS LIVRES DE 3 GRAU CLCULO I
ATIVIDADE 01
1) Resolva as seguintes inequaes:
(a) 01
1
< < + < <
> > + > >
= < >
2) Faa o grfico das seguintes funes:
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Obs.: Voc pode usar o Software Graph (acesse o site para download do programa no link:
http://www.padowan.dk/graph) para construir os grficos.
(a) 32 += xy
(b) ( ) 322 += xxxf
(c) ( ) 123 = xxxg
(d) ( ) )1cos( = xxh
Soluo:
a) y = 2x + 3
f(x)=2*x+3
-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-8
-6
-4
-2
2
4
6
8
x
y
b) ( ) 322 += xxxf
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f(x)=-x^ 2+2*x-3
-9 -8 -7 -6 -5 -4 -3 -2 - 1 1 2 3 4 5 6 7 8 9
-8
-6
-4
-2
2
4
6
8
x
y
c) ( ) 123 = xxxg
f(x)=x^ 3-2*x-1
-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-8
-6
-4
-2
2
4
6
8
x
y
d) ( ) )1cos( = xxh
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f(x)=cos(x-1)
-3 -5/2 -2 -3/2 - -/2 /2 3/2 2 5/2 3
-8
-6
-4
-2
2
4
6
8
x
y
3) Para as funes dadas, determine o domnio; o conjunto imagem; analise o sinal; identifique intervalos de
crescimento e decrescimento.
(a) 122 += xxy
(b)
x
y
= 31
Soluo:
a) 122 += xxy
Traando o grfico via o Software Graph:
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f(x)=-x^ 2+2*x-1
-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-8
-6
-4
-2
2
4
6
8
x
y
A partir desse grfico, tiramos algumas concluses:
1) Domnio
D = .
2) Conjunto-Imagem
{ }Im y | y 0=
3) Variao de Sinal
Se x 1 y 0
Se x 1 y 0
= =
4) Intervalos de Crescimento e de Decrescimento da Funo:
A funo decrescente em , +
4) Um estacionamento oferece as seguintes opes de cobrana: (1,0 ponto)
00,5$R por dia;
00,1$R por dia adicionando uma taxa fixa de 00,40$R .
Determine:
a) As funes para o custo Cde estacionamento, porxdias, com os dois tipos de pagamento.
b) Esboce os grficos das duas funes no mesmo sistema de eixo cartesiano.
Soluo:
a) Funo Custo
Tipo 1: ( )C x 5x=
Tipo 2: ( )C x x 40= +
b) Grficos
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(f)15234
592lim
2
2
5 +
xx
xx
x
Soluo:
(a)2
2x 2
x 5x 6lim
x x 2+ +
=+
( ) ( )
( ) ( )
22
2 2x 2
2
2x 2
2 5 2 6x 5x 6lim
x x 2 2 2 2
x 5x 6 4 10 6 10 10 0lim
x x 2 4 2 2 4 4 0
+ ++ +=
+ +
+ + + = = =
+
Levantando a indeterminao00
;
( ) ( )( ) ( )
2
2x 2 x 2x 2 N xx 5x 6lim lim
x x 2 x 2 D x + + =
+
Aplicando o Dispositivo de Briot-Ruffini:
1 5 6 -2
1 3 0
Assim: N(x) = x + 3
1 1 -2 -2
1 -1 0
Assim: D(x) = x 1
Substituindo:
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( ) ( )
( ) ( )
( )
2
2x 2 x 2
2
2x 2 x 2
x 2 N xx 5x 6lim lim
x x 2 x 2 D x
x 2x 5x 6lim lim
x x 2
+ +=
+
+ +=
+
( )
( )
x 3
x 2
+
( )
( )
( )
2
2x 2 x 2
2
2x 2
x 1
x 3x 5x 6 2 3 1lim lim
x x 2 x 1 2 1 3
x 5x 6 1lim
x x 2 3
++ + += = =
+
+ +=
+
(b)xx
xx
x 6
3lim
2
2
++
2
2
2x x
3x
x3x xlim lim
x 6x
+
=+
0
2
1
6x 1
x
+
+
( )
0
2 2
2 2x x x
2
2x
3x x xlim lim lim 1 1x 6x x3x x
lim 1x 6x
+ = = = + +
= +
(c)36
6lim
36 +
x
x
x
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( )( )
( )( )
( )( ) ( )
( ) ( )
( )
x 36
22
x 36 x 36
x 36 x 36 x 36
x 6 36 6 0lim
x 36 36 36 0
Assim:
x 6 x 6 x 6lim lim
x 36 x 6 x 36 x 6
x 36x 6 x 36lim lim lim
x 36 x 36 x 6
= =
+ +
+ =
+ + + +
= =
+ + + ( )x 36 ( )
x 36 x 36
x 36
x 6
x 6 1 1 1 1lim limx 36 6 6 12x 6 36 6
x 6 1lim
x 36 12
+
= = = = + ++ +
=
+
(d)x 2
xlim
2 x+= +
x 2+ ( )
xf x
2 x=
-1,7 5,6667
-1,8 9,0000
-1,9 19,000
-1,99 190,00
-1,999 1.999
-1,9999 19.999
-1,99999 199.999
-1,999999 1.999.999
etc etc
( )x 2x 2x x 2 2 2
lim lim 12 x 2 x 2 2
2porque 0
+ +
+ + = = = = = +
+
+
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(e)0
1 coslimx
x
x
( )
( ) ( )
( ) ( )
( )
x 0
2
x 0x 0
2
x 0 x 0x 0
x 0x 0
1 01 x 1 1 0
x 0 0 0
Assim
1 x 1 x 1 x
x 1 x x 1 x
1 x sen x senx senx
x x 1 x x 1 x
1 x senx
x x
= = =
+ =
+ +
= = + +
=
coscos
:
cos cos coslim
cos cos
coslim limcos cos
coslim
lim
lim
lim
lim( )
( )
1
x 0
x 0
sen 0senx 0 00
1 x 1 0 1 1 2
1 x0
x
= = = =+ + +
=
limcos cos
coslim
(f) 15234
592lim 2
2
5 +
xx
xx
x
( ) ( )
( ) ( )
22
22x 5
2 5 9 5 52x 9x 5 50 50 0
4x 23x 15 115 115 04 5 23 5 15
= = =
+ +lim
Levantando a indeterminao00
:
( ) ( )( ) ( )
2
2x 5 x 2
x 5 N x2x 9x 5lim lim4x 23x 15 x 5 D x = +
Aplicando o Dispositivo de Briot-Ruffini:
2 -9 -5 5
2 1 0
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Assim: N(x) = 2x +1
4 -23 15 5
4 -3 0
Assim: D(x) = 4x 3
Substituindo:
( ) ( )
( ) ( )
( )
2
2x 5 x 5
2
2x 5 x 5
x 5 N x2x 9x 5lim lim
4x 23x 15 x 5 D x
x 52x 9x 5lim lim
4x 23x 15
=
+
=
+
( )
( )
2x 1
x 5
+
( )
( )
( )
2
2x 5 x 5
2
2x 5
4x 3
2x 12x 9x 5 2 5 1 10 1 11lim lim
4x 23x 15 4x 3 4 5 3 20 3 17
2x 9x 5 11lim
4x 23x 15 17
+ + += = = =
+
=
+
6) Seja a funo, representada graficamente da seguinte forma:
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(a) Determine ( )1f ( )4f ( )3f
(b) Analise o sinal da funo ( )xf
(c) )(lim3
xfx
+
(d) )(lim3 xfx Soluo:
(a) Determine ( )1f ( )4f ( )3f
Observando o grfico, temos que:
f(-1) = - 1 f(4) = 3 f(3) = 1
(b) Analise o sinal da funo ( )xf
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( )f x 0 x 3
f x 0 x 3
< ( )
( )
( )
f x 0 x 3
f x 1 x 3
f x 0 x 3
< > ( )
(c) )(lim3
xfx
+
x 3f x 3
+=( )lim
(d) )(lim3
xfx
x 3 f x 1 = ( )lim
7) Considere a funo:
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