AD1 Matematica para Computacao 2014 2 Gabarito

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Curso Superior de Tecnologia em Sistemas de Computac¸a˜o
Disciplina: Matema´tica para Computac¸a˜o
AD1 - 2o semestre de 2014 — Gabarito
Questo˜es
1. (1,0 ponto)
Diga qual o domı´nio e a imagem das seguintes func¸o˜es:
(a) f(x) =
{
x2 se 2 ≤ x ≤ 4
x+ 1 se 1 ≤ x < 2
(b) f(x) = x2 + 4
(c) f(x) =
√
x2 + 4
(d) f(x) =
√
x2 − 4
(e) f(x) =
x
x+ 3
(f) f(x) =
2x
(x− 2)(x + 1)
(g) f(x) =
1√
9− x2
(h) f(x) =
x2 − 1
x2 + 1
(i) f(x) =
√
x
2− x
Soluc¸a˜o:
(a) f(x) =
{
x2 se 2 ≤ x ≤ 4
x+ 1 se 1 ≤ x < 2
Dom f = {x ∈ IR tais que 1 ≤ x ≤ 4}
Im f = {x ∈ IR tais que 2 ≤ x < 3 e 4 ≤ x ≤ 16}
(b) f(x) = x2 + 4
Dom f = {x ∈ IR}
Im f = {x ∈ IR}
(c) f(x) =
√
x2 + 4
Dom f = {x ∈ IR}
Im f =
{
x ∈ IR tais que
√
4 ≤ x <∞
}
(d) f(x) =
√
x2 − 4
Para a func¸a˜o ser definida x2 − 4 ≥ 0 ou x2 ≥ 4, isto e´ x ≥ 4 ou x ≤ −4
logo
Dom f = {x ∈ IR tais que x ≤ −4 ou 4 ≤ x}
Im f = {x ∈ IR tais que 0 ≤ x <∞}
(e) f(x) =
x
x+ 3
Dom f = {x ∈ IR tais que x 6= −3}
Im f = {x ∈ IR}
(f) f(x) =
2x
(x− 2)(x + 1)
Dom f = {x ∈ IR tais que x 6= −1 e x 6= 2}
Im f = {x ∈ IR}
(g) f(x) =
1√
9− x2
Para a func¸a˜o ser definida 9− x2 > 0 ou 9 > x2, isto e´ − 3 < x < 3
logo
Dom f = {x ∈ IR tais que − 3 < x < 3}
Im f =
{
x ∈ IR tais que 1
9
≤ x <∞
}
(h) f(x) =
x2 − 1
x2 + 1
Dom f = {x ∈ IR}
Im f = {x ∈ IR tais que −∞ < x < 1}
(i) f(x) =
√
x
2− x
Dom f = {x ∈ IR tais que x 6= 2}
Im f = {x ∈ IR tais que 0 ≤ x <∞}
2. (1,0 ponto) ————————————————————————————
Encontre os seguintes limites:
(a) lim
x→0
1
x2
(b) lim
x→1
−1
(x− 1)2
(c) lim
x→+∞
1
x
(d) lim
x→+∞
(
2 +
1
x2
)
Soluc¸a˜o:
(a) lim
x→0
1
x2
= +∞
(b) lim
x→1
−1
(x− 1)2 = −∞
(c) lim
x→+∞
1
x
= 0
(d) lim
x→+∞
(
2 +
1
x2
)
=
(
lim
x→+∞
2 + lim
x→+∞
1
x2
)
= (2 + 0) = 2
3. (1,0 ponto) ————————————————————————————
Encontre os seguintes limites:
(a) lim
x→4
√
25− x2
(b) lim
x→−5
x2 − 25
x+ 5
(c) lim
x→4
x− 4
x2 − x− 12
(d) lim
x→2
x2 + x− 2
(x− 1)2
Soluc¸a˜o:
(a) lim
x→4
√
25− x2 =
√
lim
x→4
(25− x2)
=
√ (
lim
x→4
25− lim
x→4
x2
)
=
√
(25− 16)
=
√
9 = 3
(b) lim
x→−5
x2 − 25
x+ 5
= lim
x→−5
(x+ 5)(x− 5)
x+ 5
= lim
x→−5
(x− 5)
= ( lim
x→−5
x− lim
x→−5
5)
= (−5− 5) = −10
(c) lim
x→4
x− 4
x2 − x− 12 = limx→4
x− 4
(x+ 3)(x− 4)
= lim
x→4
1
(x+ 3)
=
1
7
(d) lim
x→2
x2 + x− 2
(x− 1)2 = limx→2
(x− 1)(x + 2)
(x− 1)2
= lim
x→2
(x+ 2)
(x− 1)
=
(2 + 2)
(2− 1) =
4
1
= 4
4. (1,0 ponto) ————————————————————————————
Nos itens a seguir lim
x→±∞
pode ser interpretado como lim
x→+∞
ou lim
x→−∞
. Calcule enta˜o os
limites:
(a) lim
x→±∞
x2 + x− 2
4x3 − 1
(b) lim
x→±∞
2x3
x2 + 1
(c) lim
x→±∞
(x5 − 7x4 − 2x+ 5)
Soluc¸a˜o:
(a) lim
x→±∞
x2 + x− 2
4x3 − 1 = limx→±∞
x2/x3 + x/x3 − 2/x3
4x3/x3 − 1/x3
= lim
x→±∞
1/x + 1/x2 − 2/x3
4− 1/x3
=
lim
x→±∞
1/x + lim
x→±∞
1/x2 − lim
x→±∞
2/x3
lim
x→±∞
4− lim
x→±∞
1/x3
=
0 + 0− 2 · 0
4− 0 =
0
4
= 0
(b) lim
x→±∞
2x3
x2 + 1
= lim
x→±∞
2x3/x2
x2/x2 + 1/x2
= lim
x→±∞
2x
1 + 1/x2
=
lim
x→±∞
2x
lim
x→±∞
1 + lim
x→±∞
1/x2
=
±∞
1 + 0
= ±∞
(c) lim
x→±∞
(x5 − 7x4 − 2x+ 5) = lim
x→±∞
x5(1− 7x
4
x5
− 2x
x5
+
5
x5
)
= lim
x→±∞
x5(1− 7
x
− 2
x4
+
5
x5
)
= lim
x→±∞
x5 · lim
x→±∞
(1− 7
x
− 2
x4
+
5
x5
)
= lim
x→±∞
x5 · ( lim
x→±∞
1− lim
x→±∞
7
x
− lim
x→±∞
2
x4
+ lim
x→±∞
5
x5
)
= lim
x→±∞
x5 · (1− 0− 0 + 0)
= lim
x→±∞
x5 · (1)
= lim
x→±∞
x5 = ±∞
5. (1,0 ponto) ————————————————————————————
Estude a continuidade das seguintes func¸o˜es:
(a) f(x) =
{
x2 se x 6= 2
0 se x = 2
(b) f(x) =
| x |
x
Soluc¸a˜o:
(a) f(x) =
{
x2 se x 6= 2
0 se x = 2
No ponto x = 2
lim
x→2
f(x) = 4
ja´ que
lim
x→2−
f(x) = lim
x→2−
x2 = 4
e
lim
x→2+
f(x) = lim
x→2+
x2 = 4
mas
f(2) = 0
sendo enta˜o o valor do limite diferente do valor da func¸a˜o no ponto. Da´ı f e´ des-
cont´ınua no ponto x = 2.
(b) f(x) =
| x |
x
A func¸a˜o na˜o e´ definida no ponto x = 0. Ale´m disso, como | x | e´ definida por
| x | =
{ −x se x < 0
x se x ≥ 0
teremos
lim
x→0−
| x |
x
= lim
x→0−
− x
x
= lim
x→0−
−1 = −1
e
lim
x→0+
| x |
x
= lim
x→0−
x
x
= lim
x→0−
1 = 1
e
lim
x→0−
| x |
x
6= lim
x→0+
| x |
x
=⇒ lim
x→0
| x |
x
na˜o existe
Portanto, f e´ descont´ınua em x = 0.
6. (1,0 ponto) ————————————————————————————
Calcule as derivadas das func¸o˜es abaixo, usando sua definic¸a˜o por limite, isto e´:
f ′(x) = lim
h→0
f(x+ h)− f(x)
h
(a) f(x) =
1
x− 2
(b) f(x) =
2x− 3
3x+ 4
Soluc¸a˜o:
(a) f(x) =
1
x− 2
f ′(x) = lim
h→0
f(x+ h)− f(x)
h
= lim
h→0
1
(x+ h)− 2 −
1
x− 2
h
= lim
h→0
(x− 2)− (x+ h− 2)
(x+ h − 2)(x − 2)
h
= lim
h→0
(x− 2)− (x+ h− 2)
h(x+ h − 2)(x− 2)
= lim
h→0
(x− 2)− (x+ h− 2)
h(x+ h − 2)(x− 2)
= lim
h→0
− 1
(x+ h− 2)(x− 2)
= − 1
(x− 2)(x− 2)
= − 1
(x− 2)2
(b) f(x) =
2x− 3
3x+ 4
f ′(x) = lim
h→0
f(x+ h)− f(x)
h
= lim
h→0
2(x+ h)− 3
3(x+ h) + 4
− 2x− 33x+ 4
h
= lim
h→0
(2(x+ h) − 3)(3x + 4)− (2x− 3)(3(x+ h) + 4)
h(3(x+ h) + 4)(3x + 4)
= lim
h→0
(2x+ 2h − 3)(3x+ 4) − (2x− 3)(3x + 3h + 4)
h(3(x + h) + 4)(3x+ 4)
= lim
h→0
(6x2 + 8x+ 6hx + 8h− 9x− 12) − (6x2 + 6hx + 8x− 9x− 9h− 12)
h(3(x+ h) + 4)(3x + 4)
= lim
h→0
6x2 + 8x+ 6hx + 8h− 9x− 12− 6x2 − 6hx − 8x + 9x+ 9h+ 12
h(3(x+ h) + 4)(3x+ 4)
= lim
h→0
8h+ 9h
h(3x + 3h+ 4)(3x + 4)
= lim
h→0
17h
h(3x + 3h+ 4)(3x + 4)
= lim
h→0
17
(3x+ 3h + 4)(3x+ 4)
=
17
(3x+ 4)(3x+ 4)
=
17
(3x+ 4)2
7. (1,0 ponto) ————————————————————————————
Ache a derivada de f(x) = | x |.
Soluc¸a˜o:
Sabemos que
| x | =


−x se x < 0
0 se x = 0
x se x > 0
Se x < 0
f ′(x) = (−x)′ = −1
e se x > 0
f ′(x) = (x)′ = 1
ou
se x < 0
f ′
−
(0) = lim
h→0−
f(x+ h)− f(x)
h
= lim
h→0−
− (x+ h)− (−x)
h
= lim
h→0−
− h
h
= −1
e se x > 0
f ′+(0) = lim
h→0+
f(x+ h)− f(x)
h
= lim
h→0+
(x+ h)− (x)
h
= lim
h→0+
h
h
= 1
Portanto a derivada na˜o existe em x = 0. Resumindo
x < 0 =⇒ f ′(x) = −1
x = 0 =⇒ na˜o existe
x > 0 =⇒ f ′(x) = 1
8. (1,0 ponto) ————————————————————————————
Diferencie as func¸o˜es:
(a) f(x) = 2x1/2 + 6x1/3 − 2x3/2
(b) f(x) =
2
x1/2
+
6
x1/3
− 2
x3/2
− 4
x3/4
(c) s(t) = (t2 − 3)4
(d) w(y) = (y2 + 4)2(2y3 − 1)3
Soluc¸a˜o:
(a) f(x) = 2x1/2 + 6x1/3 − 2x3/2
f ′(x) =
[
2x1/2 + 6x1/3 − 2x3/2
]′
=
[
2
(
1
2
)
x1/2−1 + 6
(
1
3
)
x1/3−1 − 2
(
3
2
)
x3/2−1
]
=
(
2
2
)
x−1/2 +
(
6
3
)
x−2/3 −
(
6
2
)
x1/2
=
1
x1/2
+
2
x2/3
− 3x1/2
=
1√
x
+
2
3
√
x2
− 3√x
(b) f(x) =
2
x1/2
+
6
x1/3
− 2
x3/2
− 4
x3/4
f ′(x) =
[
2x−1/2 + 6x−1/3 − 2x−3/2 − 4x−3/4
]
′
=
[
2
(
− 1
2
)
x−1/2−1 + 6
(
− 1
3
)
x−1/3−1 − 2
(
− 3
2
)
x−3/2−1 − 4
(
− 3
4
)
x−3/4−1
]
= − 2
2
x−3/2 − 6
3
x−4/3 +
6
2
x−5/2 +
12
4
x−7/4
= − 1
2
√
x3
− 2
3
√
x4
+
3
2
√
x5
+
3
4
√
x7
(c) s(t) = (t2 − 3)4
s′(t) = 4(t2 − 3)4−1(2t)
= 8t(t2 − 3)3
(d) w(y) = (y2 + 4)2(2y3 − 1)3
w′(y) =
[
(y2 + 4)2(2y3 − 1)3
]
′
=
[
(y2 + 4)2
]
′
(2y3 − 1)3+ (y2 + 4)2
[
(2y3 − 1)3
]
′
= 2
[
(y2 + 4)(2−1)
]
(2y)(2y3 − 1)3 + (y2 + 4)2(3)
[
(2y3 − 1)(3−1)
]
(6y2)
= 2(y2 + 4)(2y)(2y3 − 1)3 + (y2 + 4)2(3)(2y3 − 1)2(6y2)
= 4y(y2 + 4)(2y3 − 1)3 + 18y2(y2 + 4)2(2y3 − 1)2
= 2y(y2 + 4)(2y3 − 1)2
[
2(2y3 − 1) + 9y(y2 + 4)
]
= 2y(y2 + 4)(2y3 − 1)2
[
4y3 − 2 + 9y3 + 36y
]
= 2y(y2 + 4)(2y3 − 1)2(13y3 + 36y − 2)
9. (1,0 ponto) ————————————————————————————
Se y = x2 − 4x e x = √2t2 + 1, ache dy/dt quando t = √2.
Soluc¸a˜o:
y = x2 − 4x −→ y′ = dy
dx
= 2x− 4
e
x =
√
2t2 + 1 −→ x′ = dx
dt
=
1
2
· 1√
2t2 + 1
· (4t) = 2t√
2t2 + 1
Pela regra da cadeia
dy
dt
=
dy
dx
· dx
dt
= (2x− 4) 2t√
2t2 + 1
= (2(
√
2t2 + 1) − 4) 2t√
2t2 + 1
=
(4t
√
2t2 + 1− 8t)√
2t2 + 1
avaliando em t =
√
2
dy
dt
(√
2
)
=
(
4
√
2
√
2(
√
2)2 + 1− 8√2
)
√
2(
√
2)2 + 1
=
(
4
√
2
√
5− 8√2
)
√
5
=
4
√
2
(√
5− 2
)
√
5
10. (1,0 ponto) ————————————————————————————
Calcule as primeiras e segundas derivadas das seguintes func¸o˜es:
(a) f(x) = 3x1/2 − x3/2 + 2x−1/2
(b) f(x) = 2x2
√
2− x
(c) f(x) =
(
x2 − 1
2x3 + 1
)4
Soluc¸a˜o:
(a) f(x) = 3x1/2 − x3/2 + 2x−1/2
f ′(x) = 3 · 1
2
· x1/2−1− 3
2
· x3/2−1 + 2 · − 1
2
· x−1/2−1
=
3
2
· x−1/2 − 3
2
· x1/2 − 2
2
· x−3/2
=
3
2x1/2
− 3x
1/2
2
− 1
x3/2
f ′′(x) =
3
2
· − 1
2
· x−3/2 − 3
2
· 1
2
· x−1/2 − − 3
2
· x−5/2
= − 3
4
· x−3/2 − 3
4
· x−1/2 + 3
2
· x−5/2
= − 3
4x3/2
− 3
4x1/2
+
3
2x5/2
= − 3
4
2
√
x3
− 3
4
√
x
+
3
2
2
√
x5
(b) f(x) = 2x2
√
2− x
f ′(x) =
[
2x2
√
2− x
]′
= 2
[(
x2
)
′√
2− x+ x2
(√
2− x
)
′
]
= 2
[
2x
√
2− x+ x2
(
1
2
)(
1√
2− x
)
(−1)
]
= 2
[
2x
√
2− x− x
2
2
√
2− x
]
= 2
[
4x(2− x)− x2
2
√
2− x
]
=
8x − 5x2√
2− x
f ′′(x) =
[
8x− 5x2√
2− x
]′
=

 [8x− 5x
2]
′ ·
[√
2− x
]
− [8x− 5x2] ·
[√
2− x
]
′
[√
2− x
]2


=
[8− 10x] ·
[√
2− x
]
− [8x − 5x2] ·
[
1
2
1√
2− x (−1)
]
[√
2− x
]2
=
[8− 10x] ·
[√
2− x
]
+ [8x− 5x2] ·
[
1
2
√
2− x
]
[√
2− x
]2
=
2
√
2− x [8− 10x] ·
[√
2− x
]
+ [8x− 5x2]
2
√
2− x
[√
2− x
]2
=
2(2− x) [8− 10x] + [8x− 5x2]
2
√
2− x
[√
2− x
]2
=
2(16 − 20x − 8x+ 10x2) + 8x− 5x2
2(2− x)√2− x
=
32− 40x − 16x+ 20x2 + 8x − 5x2
2(2− x)√2− x
=
32− 48x + 15x2
2(2− x)√2− x
(c) f(x) =
(
x2 − 1
2x3 + 1
)4
f ′(x) = 4
(
x2 − 1
2x3 + 1
)3 [
x2 − 1
2x3 + 1
]
′
= 4
(
x2 − 1
2x3 + 1
)3 [
(x2 − 1)′ · (2x3 + 1)− (x2 − 1) · (2x3 + 1)′
(2x3 + 1)2
]
= 4
(
x2 − 1
2x3 + 1
)3 [
(2x) · (2x3 + 1) − (x2 − 1) · (6x2)
(2x3 + 1)2
]
= 4
(
x2 − 1
2x3 + 1
)3 [
(4x4 + 2x)− (6x4 − 6x2)
(2x3 + 1)2
]
= 4
(
x2 − 1
2x3 + 1
)3 [
2x− 2x4 + 6x2
(2x3 + 1)2
]
=
8(x2 − 1)3(x− x4 + 3x2)
(2x3 + 1)5
f ′′(x) = Item dispensado. Na˜o e´ necessa´rio fazer.