Teste5a Resol
5
8/20/2019 Teste5a Resol http://slidepdf.com/reader/full/teste5a-resol 1/5 f g R f × g 2sin 2 x + sin x = 0 − π 2 , π 2 − π 2 π 2 − π 3 0 0 π 3 − π 6 0 h (x) = 2x+1 3−x + 4 lim x→+∞ h(x) 4 0 2 −2
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Transcript of Teste5a Resol
8/20/2019 Teste5a Resol
http://slidepdf.com/reader/full/teste5a-resol 1/5
f
g
R
f × g
2sin2 x + sin x = 0
−π2
, π2
−π2
π2
−π3
0
0
π3
−π6
0
h (x) = 2x+13−x
+ 4
limx→+∞ h(x)
4 0 2
−2
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f
x = −1
y = 2x − 3
f ◦ g
g (x) = x + 1
x = 0
y = 2x − 1
x = −2
y = 2x − 2
x = −2
y = 2x − 1
x = 0
y = 2x − 2
f
g
(0, b)
f
g
a1
a2
m
f
f
g
x = a1
y = −m2
x = a1
y = − 1m2
x = a2
y = −m2
x = a2
y = − 1m2
t
R (t) = 2 + 6t
t + 1 , t ≥ 0
R (0)
R (0) = 2
t = 0
2
t 4, 5
R (t) ≤ 4, 5
⇔ 2+6tt+1
≤ 4, 5
⇔ 2+6t−4,5(t+1)t+1
≤ 0
⇔ 1,5t−2,5t+1
≤ 0
t −∞ −1 53
+∞1, 5t − 2, 5
−4
0
t + 1
0
83
1,5t−2,5t+1
0
−1, 53
0, 5
3
R
R
y = 6
6
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r (t) = R (t) × 1 − t2
t ∈ ]0, 1[
r (t)
t
r (t) = R (t) × 1 − t2
= 2 + 6t
t + 1 ×
1 − t2
= 2 + 6tt + 1
× (1 − t) (1 + t)
= (2 + 6t) (1 − t) ∧ t = −1
= −6t2 + 4t + 2 ∧ t = −1
y = −6t2 + 4t + 2
V − b
2a; − ∆
4a
V 13
; 83
t = 13
f (x) = 3cos (π − x) + sinπ2 + x
g (x) = πx−1
f (x) = −2cos x
f (x) = 3 cos (π − x) + sinπ
2 + x
= −3cos x + cos x
= −2cos x
f
Df = [−2;2]
g
x = 1
y = 0
g
[0, 2]
tvm [0, 2] = g (2) − g (0)
2 − 0
=
π
2−1 − π
0−12
= π + π
2
= 2π
2= π
f ◦ g (2)
g ◦ f (0)
f ◦ g (2) = f (g (2)) = f
π2−1
= f (π) = −2cos π = −2 × (−1) = 2
g ◦ f (0) = g (f (0)) = g (−2 cos 0) = g(−2) = π−2−1 = −π3
g ◦ f
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g ◦ f (x) = g (f (x))
= g (−2cos x)
= π
−2cos x − 1
Df = R
Dg =
{x
∈R : x
−1
= 0
}
Dg
◦f =
{x
∈R :
−2cos x
−1
= 0
}
−2cos x − 1 = 0
⇔ cos x = −1
2
⇔ x = ±2π
3 + 2kπ, k ∈ Z
Dg◦f = R\x ∈ R : x = ±2π3
+ 2kπ, k ∈ Z
V (4, −5, 3)
C (0, 1, −1)
x − 4
3 =
y + 5
−8 =
z − 3
2
−−→CV = (4, −6, 4)
C
−−→CV
x
4 =
y − 1
−6 =
z + 1
4
C
−−→CV
4 (x − 0) − 6 (y − 1) + 4 (z + 1) = 0
⇔ 4x − 6y + 4z + 10 = 0
⇔ 2x − 3y + 2z + 5 = 0
(x, y, z) = (4, −5, 3) + k (3, −8, 2) , k ∈ R
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x = 4 + 3k
y = −5 − 8k
z = 3 + 2k
2x − 3y + 2z = −5
⇔
−−−2 (4 + 3k) − 3 (−5 − 8k) + 2 (3 + 2k) = −5
⇔
−−−34k = −34
⇔
x = 1
y = 3
z = 1
k = −1
P (1, 3, 1)
−−→V C
=
42 + (−6)2 + 42 =
√ 68 = 2
√ 17
−−→P C
=
(−1)2 + (−2)2 + (−2)2 =
√ 9 = 3
V = 32π×2√ 17
3 = 6
√ 17π
h R
t.v.m [−a, a] = 0
a
t.v.m [−a, a] = 0
⇔ f (a)−f (−a)a−(−a)
= 0
⇔ f (a) − f (−a) = 0 ∧ 2a = 0
⇔ f (−a) = f (a) ∧ a = 0
a = 0
f (−a) = f (a)
a =0
f (−0) = f (0)
