f'(x)=3x^2+2ax+b
f'(1)=3+2a+b=0
f(1)=1+a+b+a^2=10
两式相减得 a^2-a-2=10,解得a=4,或者a=-3
带入3+2a+b=0
分别解得b=3,或者b=-11
f(1)＝1+a+b+a^2=10 (1)
f'(1)=3+2a+b=0 (2)
两式联立得
a^2-a-12=0 a1=4 a2=-3
分别代入（2）得，b1＝-11 b2=3
f'(x) = 3x^2 + 2ax + b
f'(1) = 3 + 2a + b = 0 => a = -1/2 (b+3)
f(1) = 1+a+b+a^2 = 10
=> 1+b-1/2(b+3) + 1/4 (b+3)^2 = 10
b = 3
由题意可得：f(1)=10
f'(1)=0
1+a+b+a^2=10
3+2a+b=0
b=3或者b=-11
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