Ответ:
1) f(x) = 2x³ - 3x, x₀ = 1
f(1) = 2(1)³ - 3(1) = 2 - 3 = -1
f'(x) = 6x² - 3
f'(1) = 6(1)² - 3 = 6 - 3 = 3
y = f'(x₀)(x - x₀) + f(x₀)
y = 3(x - 1) - 1
y = 3x - 3 - 1
y = 3x - 4
2) f(x) = 0,5x² - 2x + 2, x₀ = 0
f(0) = 0,5(0)² - 2(0) + 2 = 0 - 0 + 2 = 2
f'(x) = x - 2
f'(0) = 0 - 2 = -2
y = f'(x₀)(x - x₀) + f(x₀)
y = -2(x - 0) + 2
y = -2x + 2
3) f(x) = cos x, x₀ = π/2
f(π/2) = cos(π/2) = 0
f'(x) = -sin x
f'(π/2) = -sin(π/2) = -1
y = f'(x₀)(x - x₀) + f(x₀)
y = -1(x - π/2) + 0
y = -x + π/2
4) f(x) = (x² - 4x) / (x - 2), x₀ = 3
f(3) = (3² - 4(3)) / (3 - 2) = (9 - 12) / 1 = -3
f'(x) = ((2x - 4)(x - 2) - (x² - 4x)(1)) / (x - 2)²
f'(x) = (2x² - 4x - 4x + 8 - x² + 4x) / (x - 2)²
f'(x) = (x² - 4x + 8) / (x - 2)²
f'(3) = (3² - 4(3) + 8) / (3 - 2)² = (9 - 12 + 8) / 1 = 5
y = f'(x₀)(x - x₀) + f(x₀)
y = 5(x - 3) - 3
y = 5x - 15 - 3
y = 5x - 18
Ответ: