Question

高等数学的问题 来啊直线L与X轴平行且与曲线y=x-e^x相切,求切点坐标?要过程设f（x）在0可导,且f’(0)=6,

Answer 1

1、设切点P(x0,y0),则y0＝x0－e^(x0)

y＝x－e^x,y'＝1－e^x

曲线在点P处的切线的斜率k＝y'＝1－e^(x0)＝0,所以x0＝0,y0＝－1

所以,切点(0,-1)

2、lim(x→0) f(x)/x存在,则lim(x→0) f(x)＝lim(x→0) [x×f(x)/x]＝0

f(x)在x＝0处可导,从而连续,所以f(0)＝lim(x→0) f(x)＝0

若F(x)在x＝0处连续,则lim(x→0) F(x)＝f(0)＝a

所以,a＝lim(x→0) F(x)＝lim(x→0) [sinx/x＋f(x)/x]＝1＋lim(x→0) [f(x)－f(0)]/x＝1＋f'(0)＝1＋6＝7