Question

若函数f(x)=x^3-3x在区间(a,6-a^2)上有最小值,则实数a的取值范围是

Answer 1

x=1时,f(x)min=-2.

f(x)=x^3-3x=-2时

x^3-3x+2=0x³-x-2x+2=0

x(x²-1)-2x+2=0

x(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)=0

x=1,x=-2

∴-2