Question

在函数y=2^x中,当x2>x1>0时,f[(x1+x2)/2]

Answer 1

f[(x1+x2)/2]=2^[(x1+x2)/2]

[f(x1)+f(x2)]/2=(2^x1+2^x2)/2

由基本不等式(2^x1+2^x2)/2≧√[(2^x1)(2^x2)]=2^[(x1+x2)/2]

因为x1≠x2

所以,等号不成立

即(2^x1+2^x2)/2>2^[(x1+x2)/2]

即：当x2>x1>0时,f[(x1+x2)/2]