Question

函数f(x)=(x^2-1)^2 + 2的极值点

Answer 1

方法一：f'(x)=(x^2-1)'[(x^2-1)^2]'=(2x)*[2(x^2-1)]=4x(x^2-1)=0,x=0或x=-1或x=1.方法二：f(x)=x^4-2x^2+3,f'(x)=4x^3-4x=4x(x^2-1)=0,x=0或x=-1或x=1.设f'(x)=4x(x^2-1)>0,则x>0且x^2-1>0,或x