найдите производную функции 1) f (x) = (-2/3) x³+2x²-x

2) g (x) = (3+2x) / x-2 и вычислите g' (1)

3) решите неравенство g' (x) ≤0, если g (x) = 3x-5x²+x³

1) f (x) = (-2/3) x³+2x²-x

f' (x) = (-2/3) * 3x²+2*2x-1 = - 2x²+4x-1

2) g (x) = (3+2x) / x-2 = 3/x + 2 - 2 = 3/x

g' (x) = 3 * (-1) / x^2 = - 3/x^2

g' (1) = - 3/1^2 = - 3/1 = - 3

3) g (x) = 3x-5x²+x³ = x (3-5x+x²) = x * (x-x1) * (x-x2)

x1 = (5+D) / 2 x2 = (5-D) / 2 D=13^ (1/2)

ответ отрезек [х2; x1]