曲线Y=2X^3-X在点(1,1)处的切线斜率为? 法线斜率为?

y对x求导,得y'=6x²-1
所以在点(1,1)处的切线斜率k = y'|x=1 = 6-1 = 5
法线斜率k' = -1/k = -0.2
在点(-1,-1)处的切线斜率k = y'|x=-1 = 6-1 = 5
由点斜式y= kx+b,代入(-1,-1)和k=5可得b=4,故切线方程为y=5x+4
法线斜率k' = -1/k = -0.2
由点斜式y= k'x+b,代入(-1,-1)和k'=-0.2可得b=-1.2,故法线方程为y=-0.2x-1.2