Find the maximum value of the function y = sin (above is a 2) x + cosx + 3, and the set of X when the maximum value is obtained

Find the maximum value of the function y = sin (above is a 2) x + cosx + 3, and the set of X when the maximum value is obtained

y=sin²x+cosx+3
=1-cos²x+cosx+3
=-cos²x+cosx+4
=-(cosx-1/2)²+17/4
When cosx = 1 / 2, that is, when x = 2K π ± π / 3, K ∈ Z, ymax = 17 / 4
The maximum value of the function is 17 / 4, and the set of X is {x | x = 2K π ± π / 3, K ∈ Z}
y=(sinx)^2+cosx+3=1-(cosx)^2+cosx+3=-(cosx)^2+cosx+4=-(cosx-1/2)^2+17/4
Because - 1 ≤ cosx ≤ 1
So - 3 / 2 ≤ cosx-1 / 2 ≤ 1 / 2
So the maximum value of Y is 17 / 4
When cosx-1 / 2 = 0, that is, when cosx = 1 / 2, the maximum value is taken
Then x = ± π / 3 + 2K π (K ∈ z)
If SiN x * SiN x > cos x * cos x, then the value range of X is?
Number shape combination method: from the known | SiNx | > | cosx |, draw the images of y = | SiNx | and y = | cosx |, from which we can see (K π + π / 4, K π + 3 π / 4)
SiNx ^ 2-cosx ^ 2 > 0, then cos2x > 0, then - pi / 2 + 2npi
Monotone interval problem of logarithmic function!
Monotone decreasing interval of y = log0.8 (- x ^ 2 + 4x)
By the way, the monotone interval method of logax is better said to be popular~
Y = log0.8 (x) is decreasing, so find the increasing interval of y = (- x ^ 2 + 4x), y = (- x ^ 2 + 4x) = - (X-2) ^ 2 + 4
If domain: - x ^ 2 + 4x > 0, then 0
The monotone decreasing interval of function y = LG (1-x) + 1 / (x-1) is?
1-x > 0, x = 0, monotonically increasing
When 1
The local meaning of Y (x) is X
The value range of logarithmic function x is
The value range of lgx is x > 0, that is (0, + ∞)
The LG (2-ax) power of F (x) = a is a simple decreasing function on [0,1]?
f'(x)=a^lg(2-ax)*lna*(-a)/(2-ax)=-a*lna*[a^lg(2-ax)]/(2-ax)
If the function f (x) monotonically decreases on [0,1], then f '(x) 0,2-ax > 0 and a ^ LG (2-ax) > 0 will always exist
So a * LNA > 0 and a
Derivative, because the simple subtraction, so the derivative is less than 0, get the inequality, and then simplify the calculation
The value range of logarithmic function a, X
A greater than zero is not equal to one. X greater than zero
If f (x) is an odd function and x > 0, f (x) = LG (x + 1), the analytic expression of F (x) is obtained?
Odd function f (- x) = - f (x)
Let X
How to calculate the logarithm function with the base of 1 and the true number of 2?
Such as the title
meaningless
In the definition of logarithmic function, the value range of base a is greater than 0 and not equal to 1
It can't be calculated
Value does not exist
In other words, any power of 1 is 1
It doesn't make sense to base on 1
It doesn't count. It doesn't make sense
If the function f (x) = f (x) LG (x + √ (1 + x ^ 2)) (x ∈ R) is odd, is f (x) odd or even?
Let g (x) = LG (x + √ (1 + X & # 178;))
(x + √ (1 + X & # 178;) > 0 is always greater than 0 on R
F(x)=f(x)g(x)
g(x)=lg(x+√(1+x²))
g(-x)=lg(-x+√(1+x²))
g(x)+g(-x)=0
Ψ g (x) is an even function
∵ f (x) = f (x) LG (x + √ (1 + X & # 178;)) (x ∈ R) is an odd function
F (x) is an even function
Can't judge
F(-x)=f(-x)lg(-x+√(1+(-x)^2))=f(-x)lg(1/(x+√(1+x^2)))=-f(-x)lg(x+√(1+x^2))
F (x) is an odd function, so f (- x) = - f (x), so - f (- x) LG (x + √ (1 + x ^ 2)) = - f (x) LG (x + √ (1 + x ^ 2)), so f (- x) = f (x), so f (x) is an even function
Even function
Since f (x) is an odd function, f (- x) = - f (x)
That is, left = f (- x) LG (- x + √ (1 + x ^ 2)) = f (- x) LG [1 / (x + √ (1 + x ^ 2))] = - f (- x) LG (x + √ (1 + x ^ 2))
Right = - f (x) LG (x + √ (1 + x ^ 2))
Comparing the left with the right, f (- x) = f (x).
even
Even function
Since f (x) is an odd function, f (- x) = - f (x)
F(-x)=f(-x)lg(-x+√(1+(-x)^2))=f(-x)lg(1/(x+√(1+x^2)))=-f(-x)lg(x+√(1+x^2))
So - f (- x) LG (x + √ (1 + x ^ 2)) = - f (x) LG (x + √ (1 + x ^ 2)), so f (- x) = f (x), so f (x) is even function