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8/19/2019 M1_1_zavrsni_sve_1415_130_140
1/2
(1 − i) · z 4 + (1 + i) · z = 0, z ∈ C,
|z − 1 + i| ≤ 3
2.
2x1 + 3x2 − x3 + x4 = 28x1 − 2x2 + x3 − x4 = 3
−x1 + 6x2
− x3 + 2x4 = 7
5x1 − x2 + 5x4 = −1.
x2
−→a = (4, 2, −4)
−→b = (8, 0, 6)
λ ∈R
−→c = (2 − λ, 1, 3)
(−→a , −→c ) = −→
b , −→c
.
f (x) = sin√ x + π2 g (x) = √ 1 + ln (x2 + 1)
(f ◦ g)′ (x) .
f (x) = x√
x2 − 1 .
x0 = 1 f (x) = l n(5x)
Ax = b
n
A
Ax = b
R3
T (0, −2, 1)
f (x)= 2x3.
8/19/2019 M1_1_zavrsni_sve_1415_130_140
2/2
z 1 = 0 z 2 =
√ 32 − i
2
x2 = 2
λ = 28
(f ◦ g)′ (x) = cos
√ 1 + ln (x2 + 1) + π2 · 1
2√ √
1+ln(x2+1)+π2· x(x2+1)·
√ 1+ln(x2+1)
Df = ⟨−∞, −1⟩ ∪ ⟨1, +∞⟩
x = −1 x = 1 y = −1 y = 1
x ∈ ⟨1, +∞⟩ x ∈ ⟨−∞, −1⟩ f (x) = ln 5 + ∞
∑n=1
(−1)n−1
n (x − 1)n ⟨0, 2]