f(x)=x3+ax2+bx
f'=3x^2+2ax+b
x=2處有極值-6
f'(-2)=3*(-2)^2+2a*(-2)+b=12-4a+b=0.(1)
f(-2)=(-2)^3+a*(-2)^2+b=8+4a+b=-6.(2)
由（1）、（2）解得：
a=-1/4,b=13
f(x)=x^3-1/4x^2+13x
f'(x)=3x^2-1/2x+13=1/2(6x^2-x+26)=1/2(x+2)(6x-13)
x∈(-∞,-2）和（13/6,+∞）時(shí),f'(x)＞0,f(x)單調(diào)增
x∈（-2,13/6）時(shí),f'(x)＜0,f(x)單調(diào)減
f'(x)對(duì)x∈[-1,1]都有f'(x)≤2
3x^2+2ax+b≤2
即：g(x)=3x^2+2ax+b-2=3[x^2+2a/3 x+（b-2)/3]≤0
x∈[-1,1]都有f'(x)≤2
相當(dāng)于x^2+2a/3 x+（b-2)/3=(x+1)(x-1)=x^2-1
即：2a/3=0,（b-2)/3=-1
∴a=0,b=-1
b/(a-1)=-1/(0-1)=1