设H(x)在x=0处二阶导数连续,且H(0)=0,H'(0)不等于0,证明:曲线y=f(x)=(1—cosx)H(x)在x=0证明f(x)在x=0处必有拐点
问题描述:
设H(x)在x=0处二阶导数连续,且H(0)=0,H'(0)不等于0,证明:曲线y=f(x)=(1—cosx)H(x)在x=0
证明f(x)在x=0处必有拐点
答
y'(x)=sinx * H(x) + (1-cosx)H'(x)y''(x)=cosx * H(x) + 2sinx* H'(x)+ (1-cosx)H''(x)当 x --> 0 时,y''(x)/x = cosx * H(x)/x + 2sin(x)/x * H'(x)+ (1-cosx)/x * H''(x) ---> 1 * H'(0) + 2 * H'(0) + 0 * H''...