已知f(x)=2+log3x,x∈[1,9],求函数y=[f(x)]2+f(x2)的值域.
人气:428 ℃ 时间:2020-04-15 23:22:31
解答
∵f(x)=2+log
3x,x∈[1,9],
∴y=[f(x)]
2+f(x
2)的定义域为
;
∴即定义域为[1,3],
∴0≤log
3x≤1,
∴y=[f(x)]
2+f(x
2)=
(2+log3x)2+(2+log
3x
2)=
(log3x+3)2−3∴6≤y≤13;
∴函数y的值域是[6,13].
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