急 设f(x)满足f''(x)+【f'(x)】^2=x,且f'(0)=0,则点(0,f(0))必为拐点.

问题描述:

急 设f(x)满足f''(x)+【f'(x)】^2=x,且f'(0)=0,则点(0,f(0))必为拐点.

f''(x)+【f'(x)】^2=x,且f'(0)=0
x=0代入上式,因为f'(0)=0得f''(0)=0
所以根据拐点定义,(0,f(0))必为拐点(x=0处一阶导数二阶导数都为0)