Let f (x) be an odd function, G (x) be an even function, and f (x) - G (x) = x & # 178; - x find f (x)

Let f (x) be an odd function, G (x) be an even function, and f (x) - G (x) = x & # 178; - x find f (x)

F (x) - G (x) = x ^ 2-x equation 1
f(-X)-g(-X)=x^2+X
f(-X)=-f(X) g(-X)=g(X)
-F (x) - G (x) = x ^ 2 + X equation 2
Equation 1 - equation 2
f(X)-g(X)-[-f(X)-g(X)]=2f(X)=x^2-x-x^2-x=-2x
f(X)=-x

Given f (x) = x & # 178; + (a + 1) x + A & # 178; (a ∈ R), if f (x) can be expressed as the sum of an odd function g (x) and an even function H (x)
(1) Finding the analytic expressions of G (x) and H (x)
(20) if f (x) and G (x) are time functions in the interval (- ∞), (a + 1) & # 178;], the value range of F (1) is obtained
(2) If f (x) and G (x) are decreasing functions in the interval (- ∞), (a + 1) & # 178;], the value range of F (1) is obtained

(1)
f(x)=g(x)+h(x) ①
f(-x)=h(x)-g(x) ②
① + 2
h(x)=[f(x)+f(-x)]/2=x²+a²
g(x)=f(x)-h(x)=(a+1)x
(2) What is "time function"?

Given the function f (x) = 4sinxcos (x + Pie / 3). (1) find the minimum positive period of F (x). (2... When x belongs to [0,4]. Find the range of F (x)

(1) The simplest method is to use the formula of "integral sum difference" 2Sin α cos β = sin (α + β) + sin (α - β) original formula = 2 × 2sinxcos (x + π / 3) = 2 [sin (x + X + π / 3) + sin (x-x - π / 3)] = 2 [sin (2x + π / 3) + Si

Given the function f (x) = 4sinxcos (x-pie / 6) - 2, find the minimum positive period of F (x)

The integrable sum difference formula sin α · cos β = (1 / 2) [sin (α + β) + sin (α - β)]; f (x) = 2Sin (2x pi / 6) - 3 / 2; t = 2pi / W; w = 2; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi; t = pi;

Given the function f (x) = 4sinxcos (x + π / 3) + radical 3 (1), if x ∈ R, find the minimum positive period and monotonicity of F (x)
(1) If x ∈ R, find the minimum positive period and monotone increasing interval of F (x)
(2) If x ∈ [- π / 3, π / 4], find the maximum and minimum of function f (x)

As a result, 3 = 4sinx (1 / 2cosx - [3 / 2 / 2 SiNx (3 / 2cosx - [3 / 2 / 2 SiNx (3 / 2cosx - [3 / 2 / 2 / 2 SiNx (3 / 2 / 2 / 2 / 2 / 2 / 2 / 2 SiNx) + 3 = 4sinx (1 / 2cosx - [3 / 2 / 2 / 2 / 2 / 2 SiNx (2 / 2 / 2 / 2 / 2 / 2 SiNx) x (3 / 2 / 2 / 2 / 2 / 2 / 2 / 2 osx-2) cosx-2-2-2-2-2-2 \\\\\\\\\\\\\\\\\\\\\inthis paper, the author introduces the concept of the new system

Given that the minimum positive period of function f (x) = Tan (Wx Pie / 4) (W > 0) is Pie / 2 (1), find w (2)
It is known that the minimum positive period of function f (x) = Tan (Wx Pie / 4) (W > 0) is Pie / 2 (1) to find w (2) to find monotone interval and symmetry center

It is known that the minimum positive period of function f (x) = Tan (Wx Pie / 4) (W > 0) is Pie / 2 (1) to find w (2) to find monotone interval and symmetry center
Analysis: the minimum positive period of ∵ function f (x) = Tan (Wx Pie / 4) (W > 0) is Pie / 2
T=π/2==>w=π/T=2
∴f(x)=tan(2x-π/4)
Monotone increasing interval: K π - π / 2

Given the function f (x) = cos2x-2cosx + 2, then the range of this function is

Let cosx = a ∥ the original formula = 2cos & sup2; x-1-2cosx + 2
=2a²-2a+1
=2(a-1/2)²+1/2
∵-1≤a≤1
The value range is [1 / 2,5]

Given the vector a (2cos & sup2; X, √ 3), the vector b (1, sin2x) and the function f (x) = a * B, find the analytic expression and the minimum positive period of the function f (x)

f(x)＝2cos²2x+√3sin2x
＝cos2x+1+√3sin2x
＝2sin（2x+π/6）+1.
Minimum positive period = π

The period of the function y = 4sin (3x + Wu / 4) + 3cos (3x + Wu / 4) is
What is the maximum value

Y = 4sin (3x + Wu / 4) + 3cos (3x + Wu / 4) = 5 [4 / 5 * sin (3x + Wu / 4) + 3 / 5 * cos (3x + Wu / 4)] let cosa = 4 / 5, Sina = 3 / 5; then y = 5 * sin (3x + Wu / 4 + a); so the maximum value is 5 and the period is 2 Wu / 3

Y = 3cosx, X belongs to R; 3cos (x + 2 Wu) = 3cosx. How to calculate the period of the original function is 2 Wu!

The period is the minimum t of Y (x + T) = y (x),
y(x+T)=3cos(x+T)=y(x)=3cosx
We get cos (x + T) = cosx
T = 2 μ