Решение: k = f'(x0)

1) f(x) = x²-4x+2, x0 = -1

f'(x) = (x²-4x+2)' = 2x - 4

k = f'(x0) = 2(-1)-4 = -2-4 = -6

Ответ: -6.

2) f(x) = 2x⁴-x+3, x0 = 0

f'(x) = (2x⁴-x+3)' = 2•4x³-1 = 8x³-1

k = f'(x0) = 8•0³-1 = 0-1 = -1

Ответ: -1.

3) f(x) = 3x³-x²+4, x0 =-2

f'(x) = (3x³-x²+4)' = 3•3x²-2x = 9x²-2x

k = f'(x0) = 9•(-2)²-2(-2) = 9•4+4 = 36+4 = 40

Ответ: 40