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The maximum value of the third power * cosx of y = SiNx
∵ Y & # 178; = (Sin & # 179; xcosx) & # 178; = cos & # 178; X (SiNx) ^ 6 ((SiNx) ^ 6 is the sixth power of SiNx)
And COS & # 178; X (SiNx) ^ 6
=(3cos²x)(sin²x)(sin²x)(sin²x)/3
≤[(3cos²x+sin²x+sin²x+sin²x)/4]⁴/3
=(3/4)⁴/3
=27/256
∴y²≤27/256
∴y≤3√3/16
That is, the maximum value of Y is 3 √ 3 / 16, and the equivalent condition is: 3cos & # 178; X = Sin & # 178; X
Find the sum of the maximum and minimum value of F (x) = ((x + 1) power 2 + SiNx) / (x power 2 + 1)
The function should be f (x) = [x ^ 2 + 1 + 2x + SiNx] / (x ^ 2 + 1)
f(x)=[x^2+1+2x+sinx]/(x^2+1)=1+(2x+sinx)/(x^2+1)
Let g (x) = (2x + SiNx) / (x ^ 2 + 1),
Then f (x) = 1 + G (x)
G (x) is an odd function if its maximum is g (x0) = a,
Then the minimum value is g (- x0) = - A,
They are opposite to each other, so m = 1 + A, M = 1-A, so m + M = 2
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