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

Question

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

Answer 1

将x=0代入f''(x)+【f'(x)】^2=x
解得f''(0)=0
下面说明f"(x）在x=0两边是异号的
对等式f''(x)+【f'(x)】^2=x取极限x→0
得到limf"(x)=0
那么limf'(x)/x=limf"(x)=0 (x→0)
这说明(x→0)时f'(x)是比x更高阶的无穷小
当x<0时f''(x)+【f'(x)】^2=x右边是小于零的,左边【f'(x)】^2是大于零的,因此f''(x)小于零
当x>0时,等式右边是大于零的,而等式左边【f'(x)】^2是比x^2更高阶的无穷小,因此可以忽略不计,因此f''(x)大于零
综上所述f''(0)=0且在x=0两边异号,因此点(0,f(0))为拐点