m^2+1≥1
f'(x)=3x^2-3
f(x)在x=-1,x=1處有極點,
當(dāng)x＞1時,f(x)遞增,
所以 f(x)在[m^2+1,√2]內(nèi)單調(diào)遞增,
f(x)在x=m^2+1處有最小值,
f(m^2+1)=(m^2+1)^3-3(m^2+1)=m^2-2
設(shè)m^2+1=a
a^3-4a+3=0
(a-1)(a^2+a-3)=0
a1=1,a2=(-1+√13)/2,a3=(-1-√13)/2(a3舍去)
m^2+1=1或m^2+1=(-1+√13)/2
m=0或m^2=(-3+√13)/2
m=0或m=±(1/2)√(-6+2√13)