f(x)=2x^3-x+2f′(x)=6x^2-1设切点为（m,n)∴切线斜率为k=f′(m)=6m^2-1∵切线经过切点（m,n),n=f(m)=2m^3-m+2∴切线方程为y-n=(6m^2-1)(x-m)y-(2m^3-m+2)=(6m^2-1)(x-m)………①∵切线经过点M(0,-2)∴将点M代入切线...