问题描述: 函数f(x)=2^x,X1,X2属于R,且X1≠X2,证明:1/2(f(X1)+f(X2))>f((X1+X2)/2) 1个回答 分类:数学 2014-12-16 问题解答: 我来补答 X1≠X2f(x1)≠f(x2)1/2(f(X1)+f(X2))-f((X1+X2)/2)=1/2(2^x1+2^x2)-2^[(x1+x2)/2]=1/2{2^x1+x^x2-2*2^[(x1+x2)/2]}=1/2[2^(x1/2)-2^(x2/2)]^2>01/2(f(X1)+f(X2))>f((X1+X2)/2) 展开全文阅读