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已知x²-2x-8≤0,则函数y=log3(x+3)log3(3x+9)的值域?
人气:391 ℃ 时间:2020-03-22 23:34:52
解答
y=log3(x+3)log3(3x+9)=log3(x+3)*(log3(x+3)+1)=(log3(x+3))^2+log3(x+3)
=(log3(x+3)+1/2)^2-1/4
x²-2x-8≤0,-2≤x≤4
0≤log3(x+3)≤log3(7)
因此值域为0≤y≤[log3(7)]^2+log3(7)
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