Question

若f(x)=9的x次方-(k+1)3的x次方+1>0恒成立,则k的范围是

Answer 1

f(x)=9^x-(k+1)3^x+1=(3^x)^2-(k+1)3^x+1=[3^x-(k+1)/2]^2+1-[(k+1)/2]^2

1、若(k+1)/2>0也即k>-1,则有

f(x)=[3^x-(k+1)/2]^2+1-[(k+1)/2]^2≥1-[(k+1)/2]^2

要想f(x)>0恒成立,必须有

1-[(k+1)/2]^2>0

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