(1)
f'(x)=3x^2-1
f'(t)=3t^2-1
y=(3t^2-1)(x-t)+t^3-t
=(3t^2-1)x-2t^3
(2)
由(1)知可设切线方程为:
y=(3t^2-1)x-2t^3=-2t^3+3xt^2-x
切线过点(a,b),有:
b=-2t^3+3at^2-a
可作三条,上面方程有三实数根
即:
g(t)=-2t^3+3at^2-a-b有三零点
结合图象知其极小值<0,极大值>0
g'(t)=-6t^2+6at=0,t=0或a
a>0,所以g(t)在t=0时取极小值,t=a时取极大值
g(0)=-a-b<0,
-b<a
g(a)=-2a^3+3a^3-a-b=a^3-a-b>0,
b<a^3-a=f(a)
