定义在区间［0.,1］的函数g(x)=2^x-1 ,若X1≥ 0,X2≥ 0,x1+x2≤ 1,证明g(x1+x2)≥ g(x1)+g(x2)
问题描述：
定义在区间［0.,1］的函数g(x)=2^x-1 ,若X1≥ 0,X2≥ 0,x1+x2≤ 1,证明g(x1+x2)≥ g(x1)+g(x2)
答
0g(x1+x2)- [g(x1)+g(x2)]
=2^（X1+X2）-1-[2^X1-1+2^X2-1]
=2^（X1+X2）+1-2^X1-2^X2
=2^X1*2^X2+1-2^X1-2^X2
=(2^X1-1)*(2^X2-1)
因为0=0
所以g(x1+x2)- [g(x1)+g(x2)]>=0
所以g(x1+x2)≥ g(x1)+g(x2)
答
显然,x≥ 0时,2^x≥1,即g(x)≥ 0,
因为x1,x2≥ 0,所以2^x1-1≥ 0,2^x2-1≥ 0,
(2^x1-1)(2^x2-1)≥ 0,
2^x1*2^x2-2^x1-2^x2+1≥ 0,
2^(x1+x2)-1≥2^x1-2^x2-2,
即2^(x1+x2)-1≥ (2^x1-1)+(2^x2-1),
亦即g(x1+x2)≥ g(x1)+g(x2).