令g(x)=f(x)+5则g(x)=ax^3+bx,为奇函数∴g(-2)+g(2)=0即f(-2)+5+f(2)+5=0∴f(2)=-20
f(2)+f(-2)=-10,f(2)=-20
由题得f (-2)=-2a -2b -5=10,即2a +2b =-15,且f (2)=2a +2b -5 所以f (2)=-20