解(1)f'(x)=2x^2+ax+1
Δ=a^2-8
令Δ=0得a=±2√2
当-2√22√2或者a0,解f'(x)=0得x=[-a±√(a^2-8)]/4
[-a-√(a^2-8)]/40,f(x)单增
(2)-2√2≤a≤2√2