The range of the function y = 3 + cosx is______ .

The range of the function y = 3 + cosx is______ .

∵ - 1 ≤ cosx ≤ 1, ∵ 2 ≤ 3 + cosx ≤ 4, ∵ the range of function y = 3 + cosx is: [2,4], so the answer is: [2,4]
Given the function y = loga (x ^ 2 + mx-m) (a > 0, a ≠ 1), we can find the range of real number m under the following conditions: (1) when the range is R;
If the value range is r, then all the positive numbers x ^ 2 + mx-m must contain (0, positive infinity), and the value range of x ^ 2 + mx-m itself is (0, positive infinity). Then why is there a discriminant greater than zero (that is, y < 0)
The range is r
Then the true number gets all positive numbers
So the minimum value of true number is less than or equal to 0
Then the discriminant is greater than or equal to 0
So m ^ 2 + 4m > = 0
M=0
The domain is r
Then the true number is always greater than 0
So the discriminant is less than 0
m^2+4m
It is known that the definition domain of the function f (x) = sin (x + θ) + cos (x + θ) is R. (1) when θ = π / 2, find the monotone increasing interval of F (x)
RT
f(x)=sin(x+θ）+cos（x+θ)
= √2[√2/2sin(x+θ)+√2/2cos(x+θ)]
=√2[sin(x+θ)cosπ/4+cos(x+θ)sinπ/4]
=√2sin(x+θ+π/4)
When θ = π / 2, f (x) = √ 2Sin (x + θ + π / 4)
=√2sin(x+3π/4)
When x + 3 π / 4 ∈ [2K π - π / 2,2k π + π / 2], f (x) increases monotonically;
Then x ∈ [2K π - 5 π / 4,2k π - π / 4] where k ∈ Z
Did you learn derivative? First, we obtain a new function f '(x) by deriving f (x). Let f '(x) > 0 solve x, then the monotone increasing interval of F (x) is obtained. If I remember correctly, f '(x) should be f' (x) = cos (x + θ) - sin (x + θ)
f(x)=sin(x+π/2）+cos（x+π/2)
=cos(x）-sin（x)
=√2cos(x+π/4)
X ∈ [2K π - 5 π / 4, 2K π - π / 4] where k ∈ Z.
The known points a (a, Y1), B (2a, Y2), C (3a, Y3) are all on the parabola y = 5x * x + 12x
(1) When a = 1, find the area of triangle ABC
(2) Is there any formula which contains Y1, Y2, Y3 and has nothing to do with a
To be specific, I'll copy it down directly and see it clearly. I won't. don't take up the position
(1) By substituting points a (1, Y1), B (2, Y2), C (3, Y3) into y = 5x * x + 12xy1 = 5 + 12 = 17y2 = 20 + 24 = 44y3 = 45 + 36 = 81, we can get the following formula: triangle area * 2 = (81-17) * (3-1) - (81-17) * (2-1) - (44-17) * (3-1) = 64-54 = 10 triangle area = 10 / 2 = 5 (2) Y1 = 5A ^ 2 + 12ay2 = 20A ^ 2 + 24ay3
It is known that the domain of definition of the function f (x) = sin (x + θ) + cos (x - θ) is R. when θ ∈ (0, π) and X ≠ 0, what is the value of θ, f (x) is an even function
f(x)=sin(x+θ)+cos(x-θ)
=sinxcosθ+cosxsinθ+cosxcosθ+sinxsinθ
F (x) is an even function,
Then f (- x) = f (x)
∴ sin(-x)cosθ+cos(-x)sinθ+cos(-x)cosθ+sin(-x)sinθ=sinxcosθ+cosxsinθ+cosxcosθ+sinxsinθ
∴ - sinxcosθ+cosxsinθ+cosxcosθ-sinxsinθ=sinxcosθ+cosxsinθ+cosxcosθ+sinxsinθ
∴ sinxcosθ+sinxsinθ=0
It can only be cos θ + sin θ = 0
∴ tanθ=-1
∵ θ∈（0,π）
∴ θ=3π/4
It is known that the image of the line Y1 = x, y2 = 13X + 1, Y3 = - 45x + 5 is as shown in the figure. If no matter what the value of X is, y always takes the minimum value of Y1, Y2, Y3, then the maximum value of Y is______ .
As shown in the figure, the coordinates a (32, 32); B (259259); C (60173717) of the intersection of Y1, Y2 and Y3 are obtained respectively, when x < 32, y = Y1; when 32 ≤ x < 259, y = Y2; when 259 ≤ x < 6017, y = Y2; when x ≥ 6017, y = Y3. ∵ y always takes the minimum value of Y1, Y2 and Y3, ∵ y max = 3717
Given that the domain of definition of function y = AX-1 / ax ^ 2 + 4ax + 1 is r, the value range of real number a is obtained
When a = 0,
When a is not zero, there are 16A ^ 2-4a
If Y1 < Y2, then the value range of the independent variable x is______ .
Simultaneous y = X2Y = − 12x + 3, the solution is X1 = − 2y1 = 4, X2 = 32y2 = 94, so, Y1 < Y2, then the value range of independent variable x is - 2 < x < 32. So the answer is: - 2 < x < 32
The function f (x) = log2 ˇ (2 + x) + log2 ˇ (2-x) is known to be an odd function-
Domain of definition
2+x>0,2-x>0
-2
F [x] = [1-2x] [x + 1] f [x] = [radical x-4] [radical x + 2] f [x] = [radical x + 1] [1 / 2-x] to find the definition field of function
The first one is over R, which is from negative infinity to positive infinity
The second x > = 4 if you mean the root (x-4), otherwise it is x > = 0 (root x) - 4
Third - 1
X belongs to R; x > = 0; 0=