Let f (x) = (- 2's x power + b) / (2's x + 1 power + a) be odd functions, find the values of a and B

Let f (x) = (- 2's x power + b) / (2's x + 1 power + a) be odd functions, find the values of a and B

First of all, since f (x) is an odd function, so f (0) = 0, substituting x = 0 to get b = 1 (1). If this is a choice filling problem, then according to the property of odd function f (- x) = - f (x), let x = 1, then f (- 1) = (1 / 2) / (a + 1), - f (1) = - [(- 1) / A + 4]. According to the properties of odd function, f (- 1) = - f (1), so a = 2 (2

It is known that the function f (x) = (B-2's x power) \ \ (2's x + 1 power + a) is an odd function. (1) finding the value of real numbers a and B (2) judging the function f (x) with the definition field R is odd And third, it's meaningless whether you can answer If t of any function belongs to R, inequality: F (square of t-2t) + square-k of F (2 (T)) (2) Monotonicity of judgment function f (x)

By substituting the odd function f (0) = 0, we get b = 1F (x) = (the x power of 1-2) \ \ (2's x + 1 power + a) f (1) = - f (- 1) into the solution, a = 2F (x) = (1-2's x power) \ \ (2's x + 1 + 2) = 1 / 2 ((1-2's x power) / (1 + 2's x power)) (1-2's x-th power) = 2 / (1 + 2 ^ x) - 12 ^ x increases monotonically, so 2 / (1 + 2 ^ x) - 12 ^ x increases monotonically

It is known that the function f (x) = - 2 x power + B / 2 x + 1 power + A is an odd function. Find the value of a B

Do you mean (- 2) ^ X or - 2 ^ x
But I can tell you what to do
Because f (x) is known to be an odd function whose domain is r
The two equations of F (0) = 0 and f (- 1) = f (1) are established
We can solve a, B
If it is - 2 ^ x
That is f (0) = - 1 + B / 2 + a = 0
f(-1)=-1/2+b+a=-f(1)=2-b/4+a
a=0,b=2
I hope you can be satisfied with my answer,

It is known that the function f (x) = - 2's x power + b) / (2's x + 1 power + 2) is an odd function (1) (2) it is proved that the function is a minus function on (- ∞, + ∞)

(1) Since f (x) is an odd function and its domain is r, f (0) = 0;
By introducing x = 0 and f (0) = 0 into the original formula, we get (B-1) / 4 = 0, that is, B = 1;
(2) As shown in the figure

It is known that a and B are two adjacent intersection points of the line y = 0 and the function f (x) = 2cos2 (Wx) / 2 + cos (Wx + π / 3) - 1, and ab = π / 2. (2) in the acute angle triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. If f (a) = - 3 / 2, C = 3, the area of triangle ABC is 3 times the root sign 3, then find the value of A

One
f(x)=2cos^2 wx/2+cos(wx+π/3)-1=coswx+cos(wx+π/3)=2cos(wx+π/6)•cos(π/6)=cos(wx+π/6)
|AB|=|x1-x2|=π/w=π/2
W=2
f(x)=cos(2x+Pai/6)
Two
f(A)=-3/2
cos(2A+π/6)=-√3/2
2A+Pai/6=5Pai/6
A=Pai/3
S = 1 / 2bcina = 1 / 2B * 3sinpai / 3 = 3 root number 3
Therefore, B = 4
a^2=b^2+c^2-2bccosA=16+9-2*4*3*1/2=13
A = radical 13

Given that the function f (x) = sin (Wx + a) + cos (Wx + a) (W > 0 | a | 0 | a | Pie / 2, the distance between two adjacent symmetry axes of an image is Pie / 2 and f (x) = f (- x), then the analytic expression of F (x) after simplification is?

F (x) = sin (Wx + a) + cos (Wx + a) = radical 2 [cos π / 4 * sin (Wx + a) + sin π / 4cos (Wx + a)] = = the distance between two adjacent symmetry axes of Radix 2Sin (Wx + A + π / 4) is π / 2, t = 2 * π / 2 = π
W = 2 π / T = 2 F (x) = f (- x) f (x) is an even function
F (x) = radical 2cos (2x)

It is known that the distance between the adjacent symmetry axes of the function f (x) = sin 4 power Wx + cos 4 power Wx is π / 2. Find the value of positive number W

Because f (x) = [(sinwx) ^ 2 + (coswx) ^ 2] ^ 2-2 (sinwx) ^ 2 * (coswx) ^ 2 = 1 - [(sin2wx) ^ 2] / 2 = 1 - (1-cos4wx) / 4 = 3 / 4 + (cos4wx) / 4. The period is t = 2 π / 4W = π / 2W

If the minimum positive period of the function y = 2sinwxcoswx (W > 0) is a pie, then a monotone increasing interval of the function f (x) = 2Sin (Wx + Pai / 2) is 2sinwxcoswx = sin2wx, so 2 π / 2W = π, w = 1,

f（x）=2sin（x+π/2）=2cosx
The monotone increasing interval of this function is [2K π - π, 2K π]

Given the function f (x) = 2Sin (Wx + φ), X belongs to R, w > 0, - π

2 π / w = 6 π, so w = 1 / 3
X / 3 + φ = π / 2 + 2K π or X / 3 + φ = - π / 2 + 2K π (k belongs to Z)
φ = π / 3 + 2K π or φ = - 5 π / 6 + 2K π
Again - π

It is proved that the cubic power of function f (x) = x + 3x is an increasing function on (negative infinity, positive infinity)

It is proved that the function f (x) = the cubic power of X + 3x is an increasing function on (negative infinity, positive infinity). Proof: two methods: Method 1: derivation, if you learn the derivative. If f (x) = x ^ 3 + 3x, then f '(x) = 3x ^ 2 + 3 > 0, then f (x) must be the increasing function on R; method 2: the definition of monotonicity: let x2 > X1 have: F