(3)判断函数f(x)在R上的单调性,并证明: (4)设x1 x2∈R,试比较(f(x1)+f(x2))/2与f((x1+x2)/2)的大小 不好意思,昨天忘打了。
追答(3)
设x11
f(x1) = f[x2+x1-x2] = f[x2] f[x1-x2]
f[x1]/f[x2]=f[x1-x2]>1
所以f(x1)>f(x2)
所以函数f(x)在R上是减函数。
(4)
f(x1)+f(x2)- 2f((x1+x2)/2)
= [ f(x1/2) ]^2 + [ f(x2/2) ]^2 - 2f(x1/2) f(x1/2)
=[ f(x1/2) - f(x1/2) ]^2 ≥0
即 (f(x1)+f(x2))/2 ≥ f((x1+x2)/2)
第4问是什么啊?
追答作差比较大小
f(x1)=f(x1/2 + x1/2)= [ f(x1/2) ]^2
f(x2)=f(x2/2 + x2/2)= [ f(x2/2) ]^2
2f((x1+x2)/2)= 2f(x1/2) f(x1/2)
所以
f(x1)+f(x2)- 2f((x1+x2)/2)
= [ f(x1/2) ]^2 + [ f(x2/2) ]^2 - 2f(x1/2) f(x1/2)
=[ f(x1/2) - f(x1/2) ]^2 ≥0
即 (f(x1)+f(x2))/2 ≥ f((x1+x2)/2)