Question

急急急！急求解答！函数问题！函数f(x)与y=a^x的图象关于y=x对称，记g(x)=f(x)[f(x)+2f(2)-1].若g(x)在[1/

函数f(x)与y=a^x的图象关于y=x对称，记g(x)=f(x)[f(x)+2f(2)-1].若g(x)在[1/2,2]上为增函数，求a的值域

Answer 1

函数f(x)与y=a^x的图象关于y=x对称
f(x)=log(a为底）x
g(x）=[log(a为底）x]*{[log(a为底）x]+[2log(a为底）2]-1}
令t=log(a为底）x
g(x)=t^2+{[2log(a为底)2]-1}t
对称轴1/2-[log(a为底)2]
开口向上
若g(x)在[1/2,2]上为增函数,
则对称轴<=1/2
1/2-[log(a为底)2]<=1/2
[log(a为底)2]>=0
得a>1