已知函数f(x)=3x-b(2≤x≤4)的图象过点(2,1),则F(x)=[f-1(x)]2-f-1(x2)的值域为 ___ .
人气:382 ℃ 时间:2019-09-16 18:46:39
解答
由函数f(x)=3x-b(2≤x≤4)的图象过点(2,1),得32-b=1,解得b=2.则f(x)=3x-2,f-1(x)=log3x+2,F(x)=[f-1(x)]2-f-1(x2)=(log3x+2)2-(log3x2+2)=(log3x+1)2+1,由2≤x≤4得,f(x)∈[1,9].所以...
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