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Lenyaaaaa5
@Lenyaaaaa5
July 2022
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Найти наибольшее и наименьшее значения функции без использования производной
y=cos^2x-sinx
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sedinalana
Verified answer
Y=cos²x-sinx=1-sin²x-sinx=1-(sin²x+sinx+1/4)+1/4=5/4-(sinx+1/2)²
sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z⇒y=5/4-(-1/2+1/2)²=5/4
sinx∈[-1;1]
sinx+1/2∈[-1/2;3/2]
(sinx+1/2)∈[1/4;9/4]
5/4-(sinx+1/2)²∈[-1;1]
Ответ у наиб=5/4 ;унаим=-1
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Answers & Comments
Verified answer
Y=cos²x-sinx=1-sin²x-sinx=1-(sin²x+sinx+1/4)+1/4=5/4-(sinx+1/2)²sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z⇒y=5/4-(-1/2+1/2)²=5/4
sinx∈[-1;1]
sinx+1/2∈[-1/2;3/2]
(sinx+1/2)∈[1/4;9/4]
5/4-(sinx+1/2)²∈[-1;1]
Ответ у наиб=5/4 ;унаим=-1