⑴f'(x)=3x?+2ax f'(1)=3+2a=1 a=-1
f(x)=x?-x?+b f(1)=b
∴b=1+1=2
a=-1 b=2
∴f(x)=x?-x?+2
⑵f'(x)=3x?-2x=0 x1=0 x2=2/3
x∈[-1,0) f'(x)>0 f(x)增函数
x∈(0,1/2] f'(x)
⑴f'(x)=3x?+2ax f'(1)=3+2a=1 a=-1
f(x)=x?-x?+b f(1)=b
∴b=1+1=2
a=-1 b=2
∴f(x)=x?-x?+2
⑵f'(x)=3x?-2x=0 x1=0 x2=2/3
x∈[-1,0) f'(x)>0 f(x)增函数
x∈(0,1/2] f'(x)