设f(x)具有二阶连续导数,f(0)=0,f'(0)=0,f''(0)>0.
展开全部
在曲线y=f(x)上任意一点(x,f(x))(x不等于0)处做此曲线的切线:Y-f(x)=f'(x)(X-x),交x轴于点(u,0),
∴u=x-f(x)/f'(x),
u'=1-[(f'(x)]^2-f(x)f''(x)]/[f'(x)]^2=f(x)f''(x)/[f'(x)]^2,
x→0时u→0-f'(x)/f''(x)→-f'(0)/f''(0)=0,
x...
∴u=x-f(x)/f'(x),
u'=1-[(f'(x)]^2-f(x)f''(x)]/[f'(x)]^2=f(x)f''(x)/[f'(x)]^2,
x→0时u→0-f'(x)/f''(x)→-f'(0)/f''(0)=0,
x...
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
