y'=x²1.斜率k=f‘(2)=4,∴切线方程为：y-4=4(x-2),即：y=4x-42.设切点是(m,1/3m^3+4/3)则k=f'(m)=m²∴切线方程为：y-(1/3m^3+4/3)=m²(x-m)∵点(2,4)在切线上,∴4-(1/3m^3+4/3)=m²(2-m)整理得,m^...