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annavarenich
@annavarenich
July 2022
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Найдите наибольшее значение функции y=24+sin^2x(всё под корнем) на отрезке [п/6; 5п/6].
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sangers1959
Verified answer
Y`=(√(24+sin²x))` [-π/6;5π/6]
y`=sinx*cosx/√(24+sinx)=0
sinx*cosx=0
sinx=0
x=πn
π/6<πn<5π/6
1/6<n<5/6 ⇒n∉
cosx=0
x=π/2+πn
π/6<π/2+πn<5π/6
1/6<1/2+n<5/6
-2/3<n<1/3
n=0 ⇒x=π/2
y(π/2)=√(24+sin²(π/2))=√25=5
y(π/6)=√(25+sin²(π/6))=√25,25
y(5π/6)=√(25+sin²(5π/6))=√25,25
ymax=√25,25.
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Answers & Comments
Verified answer
Y`=(√(24+sin²x))` [-π/6;5π/6]y`=sinx*cosx/√(24+sinx)=0
sinx*cosx=0
sinx=0
x=πn
π/6<πn<5π/6
1/6<n<5/6 ⇒n∉
cosx=0
x=π/2+πn
π/6<π/2+πn<5π/6
1/6<1/2+n<5/6
-2/3<n<1/3
n=0 ⇒x=π/2
y(π/2)=√(24+sin²(π/2))=√25=5
y(π/6)=√(25+sin²(π/6))=√25,25
y(5π/6)=√(25+sin²(5π/6))=√25,25
ymax=√25,25.