найдите точки экстремума функции Y=(2-x)^2/(3-x)^2. Created by Дмитрий2211. algebra-ru

(3-x)^2...

Дмитрий2211 @Дмитрий2211

August 2022 1 16 Report

найдите точки экстремума функции Y=(2-x)^2/(3-x)^2

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sedinalana

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Y`=[-2(2-x)*(3-x)²+2(3-x)(2-x)²]/(3-x)^4=
=(2-x)(3-x)*(-6+2x+4-2x)/(3-x)^4=-2(2-x)(3-x)/(3-x)^4=(2x-4)/(3-x)³=0
x=2
_ +
------------------(2)--------------------
min

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