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

f'(x)=0

3x^2-3=0

x^2= 1

x 1= 1 x2=- 1

f(-3)=- 18

f(- 1)=2

f(-3/2)=9/8

f( 1)=-2

f(x)max=2

f(x)min=- 18

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

Let the tangent point be (x, x 3-3x)

(x^3-3x-6)/(x-2)=3x^2-3

x 1=x2=0 x3=3

Tangent point (0,0) or (3, 18)

k 1=-3 k2=24

l 1:3x+y- 12=0

l2:24x-y-48=0