It is known that the definition domain of odd function f (x) is (- ∞, 0) ∪ (0, + ∞), and f (x) is an increasing function on the interval (0, + ∞), and f (1) = 0. The function g (x) = - x ^ 2 + MX + 1-2m (1) It is proved that the function f (x) is also an increasing function on the interval (- ∞, 0); (2) Solving the inequality f (x) about X

It is known that the definition domain of odd function f (x) is (- ∞, 0) ∪ (0, + ∞), and f (x) is an increasing function on the interval (0, + ∞), and f (1) = 0. The function g (x) = - x ^ 2 + MX + 1-2m (1) It is proved that the function f (x) is also an increasing function on the interval (- ∞, 0); (2) Solving the inequality f (x) about X

1, odd function is symmetric about the origin, 1 question is obviously true
If you have to process
Take any x1, x2x2, then - x1, - x2 > 0, - x1f-x1,
By the property of odd function, F-X = - FX, so there is - fx2fx1
So the question is set up
2, because F1 = 0, so F-1 = 0
The monotonicity of function is proved by 1 small question, obviously - 1

Senior high school mathematics: known function f (x) = cos square x + sinxcosx (x? R), the problem is in the supplement (1) Find the value of F (3 Pai / 8); Pai is the letter of pi / is the fractional line (2). Find the monotonic increasing interval of F (x)

f[x]=(1+cos2x)/2+1/2·sin2x
=1/2+1/2(cos2x+sin2x)
=1/2+√2/2·sin(2x+π/4)
(1)f(3π/8)=1/2+√2/2·sin(3π/4+π/4)
=1/2.
(2) 2K π - π / 2 ≤ 2x + π / 4 ≤ 2K π + π / 2
The solution
kπ-3π/8≤x≤kπ+π/8,
The monotonic increasing interval of F (x) is [K π - 3 π / 8, K π + π / 8]

Finding the minimum positive period y = cos square (π x + 2)

Y = cos squared (π x + 2)
=(1+cos2πx)/2
therefore
Period = 2 π / 2 π = 1

Let f (x) = √ 3sin2x-2sin ^ 2x. If x ∈ [- π / 6, π / 3], find the maximum and minimum of F (x) 2 sin (2x + π / 6) - 1 is obtained That's how to find the maximum and minimum

For example, if f (x) = 2Sin (2x + π / 6) - 1, it is correct. I will only explain the final result: F (x) = 2Sin (2x + π / 6) - 1, because x ∈ [- π / 6, π / 3], so, - π / 3 ≤ 2x ≤ 2 π / 3, - π / 6 ≤ 2x + π / 6 ≤ 5 π / 6, y = SiNx, on X ∈ [- π / 6,5 π / 6] - 1 / 2 ≤ y ≤ 1

The known function f (x) = 2Sin (2x - π / 3) + 1 φ x ∈ [1 / 4 π, 1 / 2 π] If the inequality | f (x) - M | 2 holds on X ∈ [1 / 4 π, 1 / 2 π], find the value range of real number M

The title is "known function f (x) = 2Sin (2x - π / 3) + 1"
When x = 5 / 12 π, the maximum value of F (x) is 2
When x = 1 / 4 π, the minimum value of F (x) is 1 / 2
"Inequality |f (x) -m| < 2" can be transformed into -2

Find the maximum value of the function f (x) = (COS ^ 2) x + SiNx That's cos to the power of X

Let t = SiNx - 1 ≤ t ≤ 1, so there is f (x) = - t + T + 1 = - (t-1 / 2) 2 + 5 / 4, because the opening of the function is downward, the maximum value of the function is f (x) max = 5 / 4, and T = SiNx = 1 /

Find the value range of the function y = sin? X + 3sinxcosx + 5cos? X

Find the value range of the function y = sin? X + 3sinxcosx + 5cos? X
y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+1
=2(1+cos2x)+(3/2)sin2x+1=2cos2x+(3/2)sin2x+3
=2 [cos2x + (3 / 4) sin2x] + 3 [let Tan φ = 3 / 4, φ∊ (0, π / 2); sin φ = 3 / 5, cos φ = 4 / 5)]
=2[cos2x+tanφsin2x]+3
=(2/cosφ)[cos2xcosφ+sin2xsinφ]+3
=[2/(4/5)]cos(2x-φ)+3
=(5/2)cos(2x-φ）+3
Therefore, Ymin = - 5 / 2 + 3 = - 1 / 2; ymax = 5 / 2 + 3 = 11 / 2;
That is, y ∊ [- 1 / 2,11 / 2] is the range of the function

Given the function y = 3sin? X-2 √ 3sinxcosx + 5cos? X, find the following (1) The range and period of function (2) monotone interval (3) state the image relation with function y = SiNx There is an analysis of the title said that the known trigonometric function into sine function y = asin (ω x + ψ). But I still don't know how to do it after reading the analysis. Thank you!

Y = 3sin? X-2 √ 3sinxcosx + 3cos? X + 2cos? X = 1 - √ 3sin2x + cos2x-1 + 1 = cos2x - √ 3sin2x + 2 = 2 (COS π / 3 * cos2x sin π / 3 sin2x) = 2cos (2x + π / 3) (the idea of reduction, the application of trigonometric formula) function value range [- 2,2] period T = π monotonically increasing interval

Find the minimum value of ω (2x) in the interval of ω (2x) (ω (2x)) (ω (x) is the minimum value of π (2x) of π (x)

f(x)=(1+cos2ωx)/2+√3sin2ωx
=sin(2ωx+π/6)+1/2
T=2π/(2ω)=π,ω=1,f(x)=sin(2x+π/6)+1/2
（1）f(2π/3)=sin(4π/3+π/6)+1/2=-1/2
（2）2x+π/6∈[2kπ-π/2,2kπ+π/2],k∈Z
The results show that x ∈ [K π - π / 3, K π + π / 6], K ∈ Z increases monotonically with increasing interval of (0, π / 6] and [2 π / 3, π)
The minus interval is [π / 6,2 π / 3]

The minimum positive period of the function FX = cos Λ 2 ω x + √ 3sin ω xcos ω x (ω > 0) is known. 1) the value of F (2 / 3 π) is obtained. 2) the monotone interval of the function FX and the symmetric axis equation of the image are obtained

What is the minimum positive period