Решение
1. y = (2x-3)*x² в x0=1
y = [(2x-3)*x²]` = 2*x² + 2x*(2x - 3) = 2x² + 4x² - 6x = 6x² - 6x
y(1) = 6*1² - 6*1 = 0
2. y = (2+3x)/5x² в x0=2
y` = (2+3x/5x2)` = [3*5x² - 10x*(2 + 3x)]/(5x²)² =
= (15x² - 20x - 30x²)/(25x⁴) = (- 30x² - 20x)/(25x⁴) =
= [(- 5x)*(6x + 4)] / (25x⁴) = - (6x + 4)/(5x³)
y(2) = - (6*2 + 4)/(5*2³) = - (16/40) = - 2/5 = - 0,4
3. y = 3*ln(2x³ - 3x) в x0=2
y` = 3/(2x³ - 3x) * (6x² - 3) = (18x² - 9)/<span>(2x³ - 3x)
y(2) = </span>(18*2² - 9)/<span>(2*2³ - 3*2) = 63/10 = 6,3</span>
Y=8x-3
1)x=3, y=8.3-3=24-3=21
2)y=2, 2=8x-3, 8x=5, x=5/8
3)K(-2,-13), 8.(-2)-3=-16-3=-19, Točka K ne prinadležit grafiky funkcii.
1 - bx - x + b = 1 + b - (bx + x) = (1 + b) - x(b + 1) = (b + 1) - x(b + 1) = (b + 1)(1 - x).
Решение.....................