第1个回答 2020-01-06
解:(1)函数y=x2的定义域是R,是凹函数.
证明如下:
∀x1、x2∈(0,+∞),x1≠x2,
f(x1+x22)=(x1+x22)2,12[f(x1)+f(x2)]=12[x12+x22].
∵f(x1+x22)-12[f(x1)+f(x2)]=(x1+x22)2-12[x12+x22]=-14(x1-x2)2<0,
所以f(x1+x22)<12[f(x1)+f(x2)],即函数f(x)=x2是凹函数.
(2)函数f(x)=log2x的定义域是(0,+∞),函数是凸函数.
证明如下:
∀x1、x2∈(0,+∞),x1≠x2,
f(x1+x22)=log2(x1+x22),12[f(x1)+f(x2)]=12log2x1+12log2 x2=12log2x1x2
∵f(x1+x22)-12[f(x1)+f(x2)]
=log2(x1+x22)-12 log2 x1 x2=log2(x1+x22)-log2 x1x2=log2(x1+x22x1x2)
而x1+x2-2x1x2=(x1-x2)2>0,所以x1+x22x1x2>1,log2(x1+x22x1x2)>0,
所以f(x1+x22)>12[f(x1)+f(x2)],即函数f(x)=log2x是凸函数.…(16分)