Odpowiedź:
zad 2
a)
f(x) = (3x-2)/5
3x - 2 = 0
3x = 2
x₀ - miejsce zerowe = 2/3
Df: x ∈ R
b)
f(x) = x(x - 5)(2x- 1)
x = 0 ∨ x - 5 = 0 ∨ 2x - 1 = 0
x = 0 ∨ x = 5 ∨ x = 1/2
x₀ = { 0 , 1/2 , 5 }
c)
f(x) = √(0,5x - 4 )
0,5x - 4 ≥ 0
0,5x ≥ 4
x ≥ 4 : 0,5
x ≥ 8
Df: x ∈ < 8 , + ∞ )
0,5x - 4 = 0
0,5x = 4
x₀ = 4 : 0,5 = 8
d)
f(x) = [(8x-18)(x + 1)]/(x² - 9)
założenie:
x² - 9 ≠ 0
(x - 3)(x + 3) ≠ 0
x - 3 ≠ 0 ∧ x + 3 ≠ 0
x ≠ 0 ∧ x ≠ - 3
Df: x ∈ R \ {- 3 , 3 }
8x - 18 = 0 ∨ x + 1 = 0
8x = 18 ∨ x = - 1
x = 18/8 ∨ x = - 1
x = 2 2/8 ∨ x = - 1
x = 2 1/4 ∨ x = - 1
x₀ = { - 1 , 2 1/4 }
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Verified answer
Odpowiedź:
zad 2
a)
f(x) = (3x-2)/5
3x - 2 = 0
3x = 2
x₀ - miejsce zerowe = 2/3
Df: x ∈ R
b)
f(x) = x(x - 5)(2x- 1)
x = 0 ∨ x - 5 = 0 ∨ 2x - 1 = 0
x = 0 ∨ x = 5 ∨ x = 1/2
x₀ = { 0 , 1/2 , 5 }
Df: x ∈ R
c)
f(x) = √(0,5x - 4 )
0,5x - 4 ≥ 0
0,5x ≥ 4
x ≥ 4 : 0,5
x ≥ 8
Df: x ∈ < 8 , + ∞ )
0,5x - 4 = 0
0,5x = 4
x₀ = 4 : 0,5 = 8
d)
f(x) = [(8x-18)(x + 1)]/(x² - 9)
założenie:
x² - 9 ≠ 0
(x - 3)(x + 3) ≠ 0
x - 3 ≠ 0 ∧ x + 3 ≠ 0
x ≠ 0 ∧ x ≠ - 3
Df: x ∈ R \ {- 3 , 3 }
8x - 18 = 0 ∨ x + 1 = 0
8x = 18 ∨ x = - 1
x = 18/8 ∨ x = - 1
x = 2 2/8 ∨ x = - 1
x = 2 1/4 ∨ x = - 1
x₀ = { - 1 , 2 1/4 }