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annavarenich
@annavarenich
July 2022
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Помогите пожалуйста: Найдите наименьшее значение функции y=16+sin^2x(всё под корнем) на отрезке [-п/3;п/3].
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sangers1959
Verified answer
Y=16+sin²x
y`= sinx*cosx/√(16+sin²x)=0
sinx*cosx=0
sinx=0
x=πn
-π3<πn<π/3
-1/3<n<1/3
n=0 ⇒ x=0
cosx=0
x=π/2+πn
-π/3<π/2+πn<π/3
-1/3<1/2+n<1/3
-1/6<n<1/6
n=0 ⇒ x=0
y(0)=√(16+0²)=4
y(-π/3)=√(16+(-√3/2)²=√16,75
y(π/3)=√(16+(∛√3/2)=√16,75
ymin=4.
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Answers & Comments
Verified answer
Y=16+sin²xy`= sinx*cosx/√(16+sin²x)=0
sinx*cosx=0
sinx=0
x=πn
-π3<πn<π/3
-1/3<n<1/3
n=0 ⇒ x=0
cosx=0
x=π/2+πn
-π/3<π/2+πn<π/3
-1/3<1/2+n<1/3
-1/6<n<1/6
n=0 ⇒ x=0
y(0)=√(16+0²)=4
y(-π/3)=√(16+(-√3/2)²=√16,75
y(π/3)=√(16+(∛√3/2)=√16,75
ymin=4.