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[函数] 函数一题

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longma posted 2016-5-31 15:30 |Read mode
Last edited by hbghlyj 2025-4-6 03:36定义在正实数上的连续函数 $f(x)$ 满足:$f(1)=2$ ,且对于任意的正实数 $x$ ,均有 $f\left(x^2\right)$ $=\sqrt{x} f(x)$ ,则 $f(4)=$      

这题除了构造函数 $f(x)=2 \sqrt{x}$ ,其他方法怎么做?求教!|

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色k posted 2016-5-31 15:40
见《撸题集》第743页题目5.3.31

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original poster longma posted 2016-5-31 15:49
回复 2# 色k


    看到,谢谢!

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isee posted 2016-5-31 18:14
这次竟然没单调性的事。。。。

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游客 posted 2016-6-1 13:06
“连续”这个条件是关键,“连续”的概念就是要求极限。

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