(1)
f(x)=x^3-ax^2+bx
f'(x)=3x^2-2ax+b
f'(-1)=3+2a+b=0 1
f'(3)=27-6a+b=02
1式-2式得
-24+8a=0
a=33
3式代入1式得
b=-9

(2)
f(x)=x^3-3x^2-9x
y(1)=1-3-9=-11
y'=3x^2-6x-9
k=y'(1)=3-6-9=-12
所以切线方程是
y+11=-12(x-1)
y=-12x+12-11
=-12x+1

(3)
f(x)=x^3-3x^2-9x
f'(x)=3x^2-6x-9=0
x^2-2x-3=0
(x-3)(x+1)=0
x=3x=-1
当x0
所以在区间(3,+∞)为增区间