设函数f(x)满足关系式f’’(x)+[f’(x)]2=x,且f’(0)=0,则 ( )

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问题 设函数f(x)满足关系式f’’(x)+[f(x)]2=x,且f(0)=0,则    (    )

选项 A、f(0)是f(x)的极大值
B、f(0)是f(x)的极小值
C、点(0,f(0))是曲线y=f(x)的拐点
D、f(0)不是f(x)的极值,点(0,f(0))也不是曲线y=f(x)的拐点

答案C

解析 由f(0)=0及f’’(x)+[f(x)]2=x知f’’(0)=0且f’’(x)=x一[f(x)]2,又x,f(x)可导,所以f’’(x)可导,于是f’’’(x)=1—2f(x)f’’(x),f’’’(0)=1>0,而f’’’(0)=,故f’’(x)在x=0左、右两侧异号,故选C.
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