Find the maximum and minimum values of the function f (x) = [(SiNx) ^ 4 + (cosx) ^ 4 + (sinxcosx) ^ 2] / (2-2sin2x), and point out the value of X when the maximum value is taken?

Find the maximum and minimum values of the function f (x) = [(SiNx) ^ 4 + (cosx) ^ 4 + (sinxcosx) ^ 2] / (2-2sin2x), and point out the value of X when the maximum value is taken?

The form of cosx = [(SiNx) ^ 4 + (cosx) ^ 4 + (sinxcosx) ^ 2] / (2-4sinxcosx) where 1 = [(SiNx) ^ 2 + (cosx) ^ 2] ^ 2 = s = (SiNx) ^ 4 + (cosx) ^ 4 + 2 (cosxsinx) ^ 2, we get [(SiNx) ^ 4 + (cosx) ^ 4 + (sinxcosx) ^ 2] / (2-4sinxcosx) = [1-2 (cosxsi

When x → 0, the infinitesimal 1-cos2x is?
A. Higher order infinitesimal B lower order infinitesimal C same order non equal order infinitesimal D equal order infinitesimal

When x goes to zero,
1-cos2x is equivalent to 0.5 * (2x) ^ 2, that is, 2x ^ 2,
Therefore, compared with x ^ 2, it should be a non equivalent infinitesimal of the same order
Choose answer C

When x is equal to 0, what is the order of the following infinitesimals compared with x
1、x+sinx²
2、√x+sinx
3、4x²+6x³-5x^5
4、ln(1+x)=ln(1-x)
5. √ X & sup2; + & sup3; √ X & sup3; √ x under the first radical

Because x + SiNx & sup2. / X tends to be one, infinitesimal of the same order
Because √ x + SiNx / X tends to infinity, the infinitesimal of the bottom order
Infinitesimal of higher order 4x & sup2; + 6x & sup3; - 5x ^ 5 / X
Ln (1 + x) = ln (1-x) / x, which tends to be infinitesimal of the same order by using the law of lobita
Infinitesimal of lower order of √ X & sup2; + & sup3; √ x