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Дмитрий2211
@Дмитрий2211
August 2022
1
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найдите точки экстремума функции Y=(2-x)^2/(3-x)^2
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sedinalana
Verified answer
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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Answers & Comments
Verified answer
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