高一數(shù)學(xué)...

2022-04-29 05:31

已知函數(shù)f(x)=根號(hào)(12x-9)/x, x屬于[12/13,4]1.x取何值時(shí),f(x)有最大值,最大值是多少?2.確定f(x)的單調(diào)遞增區(qū)間.并加以證明.help me!,.thanks!有沒(méi)有人啊..暈~help~..
2022-04-29 09:12
(1)
f(x)=根號(hào)(12/x-9/(x^2))=根號(hào)[-9(1/x-2/3)^2+4]
顯然1/x=2/3即x=3/2時(shí)有最大值2

(2)記g(x)=[f(x)]^2=-9(1/x-2/3)^2+4
只需判斷g(x)的單調(diào)性
而當(dāng)13/12<=x<=3/2時(shí),1/x-2/3遞減且大于0
故-9(1/x-2/3)^2+4在[13/12,3/2]上遞增
而當(dāng)3/2故-9(1/x-2/3)^2+4在(3/2,4]上遞減
這恰好與x=3/2時(shí)取最大值相吻合
更多回答
f(x)=根號(hào)(12x-9)/x=根號(hào)(12-9/x) 明顯x越大,f(x)越大,故x=4時(shí),f(x)最大為(根號(hào)39)/2
等我有空。好就沒(méi)有做了。
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