Question

Решите ур.касательной

Answer 1

$f(x)=2x^3-2x^2-10x+10 \\ y=f(x_0)+f'(x_0) \cdot (x-x_0) \\ f(2) = 2 \cdot 2^3 - 2 \cdot 2^2 - 10 \cdot 2 + 10 = 16 - 8 - 20 + 10 = -2 \\ f'(x)=(2x^3-2x^2-10x+10)'=6x^2-4x-10 \\ f'(2)=6 \cdot 2^2 - 4 \cdot 2 - 10 = 24 - 8 - 10 = 6 \\ \\ y = -2 + 6(x-2) = -2 + 6x - 12 = 6x -14$