y=x^4-2x^2+2y'=x^3-4x=x(x^2-4) 令y'=0得x=0,x=-2,x=2. 所以f(x)在1和3上取极值。min=1max=65
y=x^4-2x^2+2 =(x^2-1)^2+1在[-3,3]上,-1<=x^2-1<=8则1<=(x^2-1)^2+1<=65y max=65y min=1
