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annavarenich
@annavarenich
August 2022
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Помогите пожалуйста: Найдите 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) [-π/3;π/3]
y`=((16+sin²x)^1/2)`
y`=1/2*2*cos^*sinx*(16+sin²x)^(-1/2)
y`=sinx*cosx/√(16+sin²x)
sinx*cosx/√(16+sin²x)=0
sinx=0 x₁=0
cosx=0 x₂=π/2 x₂∉
y(0)=√16=+/-4
y(-π/3)=√(16+3/4)=√(67/4)≈+/-4,1
y(π/3)=√(67/4)≈+/-4,1
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annavarenich
Наименьшее значение функции, исправляюсь)
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Answers & Comments
Verified answer
Y=√(16+sin²x) [-π/3;π/3]y`=((16+sin²x)^1/2)`
y`=1/2*2*cos^*sinx*(16+sin²x)^(-1/2)
y`=sinx*cosx/√(16+sin²x)
sinx*cosx/√(16+sin²x)=0
sinx=0 x₁=0
cosx=0 x₂=π/2 x₂∉
y(0)=√16=+/-4
y(-π/3)=√(16+3/4)=√(67/4)≈+/-4,1
y(π/3)=√(67/4)≈+/-4,1