Найти уравнение касательной. 1)f(x)=2x²+⅓x³ x₀=-3 2)f(x)=x+1/x-1 x₀=2 3)f(x)=⅓x²-2x x₀=3 4)f(x)=x²+1/x x₀=2 надо! заранее

1) f(x)=2x^2+1/3x^3

f'(x) = 4x+x^2

y=f(x0)+f'(x0)*(x-x0)

f(x0) = 18-9=9

f'(x0) = -12+9 = -3

y=9-3(x+3) = -3x

2)f(x)=(x+1) / (x-1)

f'(x) = x-1 -(x+1) / (x-1)^2 = -2/(x-1)^2

y=f(x0)+f'(x0)*(x-x0)

f(x0)=3

f'(x0) = -2

y=3-2(x-2) = -2x+7

3) f(x)=1/3x^2-2x

f'(x) = 2x/3 - 2

y=f(x0)+f'(x0)*(x-x0)

f(x0)=3-6= -3

f'(x0) = 0

y= -3

4) f(x)=x^2+1/x

f'(x)=2x-1/x^2

y=f(x0)+f'(x0)*(x-x0)

f(x0)=4,5

f'(x0)=3/4

y=4,5+3/4(x-2) = 3x/4 - 3