设f(x)在(a,b)四次可导,x0∈(a,b)使得f’’(x0)=f’’’(x0)=0,又设f(4)(x)>0(x∈(a,b)),求证f(x)在(a,b)为凹函数.

admin2018-06-27  31

问题 设f(x)在(a,b)四次可导,x0∈(a,b)使得f’’(x0)=f’’’(x0)=0,又设f(4)(x)>0(x∈(a,b)),求证f(x)在(a,b)为凹函数.

选项

答案由f(4)(x)>0(x∈(a,b)),知f’’’(x)在(a,b)单调上升.又因f’’’(x0)=0,故[*]从而f’’(x)在(a,x0]单调下降,在[x0,b)单调上升.又f’’(x0)=0,故f’’(x)>0(x∈(a,b),x≠x0),因此f(x)在(a,b)为凹函数.

解析
转载请注明原文地址:https://kaotiyun.com/show/Pek4777K
0

最新回复(0)