要使函数在开区间（a,6-a^2)上有最小值,则x=1必须包含于（a,6-a^2)有：
a答案是【-2,1）x=1时，f(x)min=-2.f(x)=x^3-3x=-2时x^3-3x+2=0x³-x-2x+2=0x(x²-1)-2x+2=0x(x+1)(x-1)-2(x-1)=0(x²+x)(x-1)-2(x-1)=0(x-1)(x²+x-2)=0(x-1)(x+2)(x-1)=0(x-1)²(x+2)=0x=1,x=-2∴-2
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