设f(x)，g(x)在x=x
0
某邻域有二阶连续导数，曲线y=f(x)和y=g(x)有相同的凹凸性．求证：曲线y=f(x)和y=g(x)在点(x
0
，y
0
)处相交、相切且有相同曲率的充要条件是：f(x)-g(x)=o((x-x
0
)
2
)(x→x
0
)．
【正确答案】正确答案：相交与相切即f(x
0
)=g(x
0
)，f'(x
0
)=g'(x
0
)．若又有曲率相同，即
亦即｜f''(x
0
)｜=｜g''(x
0
)｜． 由二阶导数的连续性及相同的凹凸性得，或f''(x
0
)=g''(x
0
)=0或f''(x
0
)与g''(x
0
)同号，于是f''(x
0
)=g''(x
0
)．因此，在所设条件下，曲线y=f(x)，y=g(x)在(x
0
，y
0
)处相交、相切且有相同曲率
f(x
0
)-g(x
0
)=0，f'(x
0
)-g'(x
0
)=0，f''(x
0
)-g''(x
0
)=0．
f(x)-g(x)=f(x
0
)-g(x
0
)+[f(x)-g(x)]'｜
x=x0
(x-x
0
)+
亦即｜f''(x
0
)｜=｜g''(x
0
)｜． 由二阶导数的连续性及相同的凹凸性得，或f''(x
0
)=g''(x
0
)=0或f''(x
0
)与g''(x
0
)同号，于是f''(x
0
)=g''(x
0
)．因此，在所设条件下，曲线y=f(x)，y=g(x)在(x
0
，y
0
)处相交、相切且有相同曲率
f(x
0
)-g(x
0
)=0，f'(x
0
)-g'(x
0
)=0，f''(x
0
)-g''(x
0
)=0．
f(x)-g(x)=f(x
0
)-g(x
0
)+[f(x)-g(x)]'｜
x=x0
(x-x
0
)+
【答案解析】