Find the minimum positive period, maximum value and minimum value of the function y = (3-sinxcosx) / (3 + sinxcosx), and point out the independent variable x when the function obtains the maximum value and minimum value

Find the minimum positive period, maximum value and minimum value of the function y = (3-sinxcosx) / (3 + sinxcosx), and point out the independent variable x when the function obtains the maximum value and minimum value

The original formula is 9 - (SiNx) square * cosx square, let SiNx square = y (0

What are sin2x, cos2x and tan2x?

Two fold angle formula
sin2x=2sinxcosx
cos2x=(cosx)^2-(sinx)^2=2(cosx)^2-1=1-2(sinx)^2
tan2x=2tanx/(1-(tanx)^2)
If you do not understand, please hi me, I wish you a happy study!

Find the monotone decreasing interval of function y = sin2x + cos2x

=Root 2 (root 2/2*sin2x + root 2/2*cos2x)
=Radical 2Sin (2x + π / 4)
π/2+2kπ≤2x+π/4≤3π/2+2kπ
π/4+2kπ≤2x≤5π/4+2kπ
π/8+kπ≤x≤5π/8+kπ
[π/8+kπ,5π/8+kπ]
Is monotone decreasing interval

The minimum positive period and the maximum positive value of the function y = sin (2x + Pai / 6) + cos (2x + 3 / PAI) are respectively?

y= √3/2 * sin2x + 1/2cos2x + 1/2 *cos2x - 3/2sin2x
= cos2x
T = 2π/|W| = 2π/2 = π
ymax = 1

It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the graph of function f (x) is drawn, and the analytic formula of function f (x) is obtained

∵ when x ≥ 0, f (x) = x (1 + x) = (x + 1)
2）2-1
4，
F (x) is an odd function defined on R,
When x < 0, - x > 0,
f（-x）=-x（1-x）=（x-1
2）2-1
4=-f（x），
∴f（x）=-（x-1
2）2+1
Four
∴f（x）=
(x+1
2) 2-1
4 x≥0
-(x-1
2) 2+1
4 x＜0

If the function y = sin (2x + φ) (0 ≤ φ ≤ π) is an even function of R, then the value of φ is I know that the translation transformation needs to transform the 2 of the function formula y = sin (2x + φ) into sin (2 (x + φ / 2)). Then, to become an even function in this way is to shift the π / 2 unit to the left, that is, φ / 2 = π / 2? Isn't φ just π? But the answer is π / 2!

Who told you that to become even is to shift π / 2 units to the left
It depends on which function it is!
Y = SiNx shifts π / 2 units to the left, and y = cosx is an even function,
This translation actually shifts T / 4
Then y = sin2x shifts right, t / 4 should be π / 4
That is, y = sin [2 (x + π / 4)], and φ = π / 2
In addition, translation is not recommended

If f (0.5) = 9, then f (8.5) is equal to () A. -9 B. 9 C. -3 D. 0

∵ f (x-1) is an odd function, so f (- x-1) = - f (x-1), that is, f (- x) = - f (X-2)
And ∵ f (x) is an even function, f (x) = - f (X-2),
If f (x-4) = f (x) holds for any x ∈ R, the minimum positive period of F (x) is 4,
∴f（0.5）=f（8.5）=9．
Therefore, B

If f (1) = 9, then f (9) =?

F (x-1) is an odd function
f(-x-1)=-f(x-1)
F (x) is even function
f(-x)=f(x)
f(9)=f(-9)=f(-8-1)
=-f(8-1)
=-f(7)
=-f(-7)
=-f(-6-1)
=f(6-1)
=f(5)
=f(-5)
=f(-4-1)
=-f(4-1)
=-f(3)
=-f(-3)
=-f(-2-1)
=f(2-1)
=f(1)
=9

The tangent equation at x = 1 is y = X-2, and the analytic formula of F (x) is obtained

Because f (x) = f (- x) = a (- x) ^ 4 + B (- x) ^ 3 + C (- x) ^ 3 + C (- x) ^ 2 + D (- x) + e = ax ^ 4-bx ^ 3 + CX ^ 2-dx + ef (x) is even function, so f (x) = f (- x) ax ^ 4 + BX ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = ax ^ 4-bx ^ 3 + CX ^ 3 + DX + e = ax ^ 4-bx ^ 3 + e2bx ^ 3 + 2DX ^ 3 + 2DX = 0, no matter what value of X is set up, so B = 0, d = 0, so f (x) = ax ^ 4 + CX ^ 2 + e image over P (P) P (P) image over the image is too P (x ^ 2 + e image over 0,1), so 1

If f (x) = ax ^ 2 + BX + 3A + B is an even function and its domain is [a-1,2a], then a = (), B = () ∵ the function f (x) is an even function The definition domain of the function is symmetric interval ∴a-1+2a=0 a=1/3 Can you explain this step`

Even function. F (x) = f (- x)
So the domain should be zero symmetric
So A-1 + 2A = 0
Do you understand