Given the function f (x) = sin (2x + π / 3) - root 3cos (2x + π / 3), write the monotone interval of F (x) and find the range of F (x) on (- π / 6, π / 3)

Given the function f (x) = sin (2x + π / 3) - root 3cos (2x + π / 3), write the monotone interval of F (x) and find the range of F (x) on (- π / 6, π / 3)

The solution is f (x) = sin (2x + π / 3) - √ 3cos (2x + π / 3)
=2(1/2sin(2x+π/3）-√3/2cos（2x+π/3）)
=2sin(2x+π/3-π/3)
=2sin2x
We know that when 2K π - π / 2 ≤ 2x ≤ 2K π + π / 2, K belongs to Z and Y is an increasing function
That is, when k π - π / 4 ≤ x ≤ K π + π / 4, K belongs to Z and Y is an increasing function
So the increasing interval of the function is [K π - π / 4, K π + π / 4], K belongs to Z
We know that when 2K π + π / 2 ≤ 2x ≤ 2K π + 3 π / 2, K belongs to Z and Y is a decreasing function
That is, when k π + π / 4 ≤ x ≤ K π + 3 π / 8, K belongs to Z and Y is a decreasing function
So the decreasing interval of the function is [K π + π / 4, K π + 3 π / 8], K belongs to Z
From X to (- π / 6, π / 3)
We know that 2x belongs to (- π / 3,2 π / 3)
That is, sin2x belongs to (- √ 3 / 2,1]
That is, 2sin2x belongs to (- √ 3,2]
So function
The range of F (x) (- π / 6, π / 3) (- 3,2)]

Let a = [1, b] (b > 1), F & nbsp; (x) = 12 (x-1) 2 + 1 & nbsp; (x ∈ a). If the range of F & nbsp; (x) is also a, then the value of B is a______

If the function F & nbsp; (x) = 12 (x-1) 2 + 1 & nbsp; (x ∈ a), then the axis of symmetry x = 1, the function a = [1, b] is an increasing function, the value range of F & nbsp; (x) is also a, 12 (B-1) 2 + 1 = B, that is b2-4b + 3 = 0, the solution is b = 1 or B = 3, because b > 1, so B = 3

Let a = [1, b] (b > 1), f (x) = 0.5 * (x-1) &# 178; + 1, when x ∈ a, the range of F (x) is also a, try to find the value of B

F (x) = 0.5 (x-1) & sup2; + 1, so the symmetry axis of F (x) is x = 1
When x belongs to [1, b], f (x) is a monotone increasing function
So when x = B, f (x) is the maximum
0.5(b-1)²+1=b
0.5(b-1)²=b-1
Because B-1 is not equal to 0
So 0.5 (B-1) = 1
b-1=2,b=3

For the odd function f (x) defined on R, when x > 0, f (x) = 2, then the range of the odd function f (x) is______ ．

∵ the odd function f (x), ∵ f (- x) = - f (x), f (0) = 0 defined on R. let x < 0, then f (- x) = - f (x) = - 2 ∵ f (x) = 2 & nbsp; & nbsp; & nbsp; x > 00 & nbsp; & nbsp; & nbsp; X = 0 − 2 & nbsp; & nbsp; & nbsp; x < 0 ∵ the value field of odd function f (x) is: {- 2

The known function f (x) = (x + a) \ (x ^ 2 + b) is an odd function defined on R, and its range is [- 1,4,1,4]
(1) Try to find the value of a and B

Answer: if f (x) = (x + a) / (x ^ 2 + b) is an odd function defined on R, then it satisfies the following two formulas: F (- x) = - f (x) f (0) = 0: F (0) = A / b = 0. So: a = 0f (x) = x / (x ^ 2 + b) the definition field is r, which means that x ^ 2 + b > 0 is constant. So: b > 0 (when x = 0) f (x) = x / (x ^ 2 + b) = y ∈ [- 1 / 4,1 / 4]

Let the definition field of odd function f (x) be (T, T ^ 2-3t-8) and the range be (2t, T ^ 2 + 3T + 6), then what is the range of function y = f (x + 2011) + 1
A.(-2,2)
B.(-4,4)
C.(-3,5)
D.(-5,5)
The = = error of finding y = f (x + 1) + 1

(t,t^2-3t-8),
Because it's an odd function, so
t+t^2-3t-8=0
t^2-2t-8=0
(t+2)(t-4)=0
t

Let the definition field of odd function f (x) be (T.T ^ 2-3t-8) and the value field be (2t, T ^ 2 + 3T + 6), then the value field of function y = f (x + 1) + 1 is
A.(-2,2)
B.(-4,4)
C.(-3,5)
D.(-5,5)

Because f (x) is an odd function, its domain is symmetric, that is, t = T ^ 2-3t-8, and its range is also positive and negative symmetric, that is, 2T = T ^ 2 + 3T + 6. According to these two conditions, t can be obtained, so it is easy to find the range of y = f (x + 1) + 1

Given that f (x) is an odd function defined on [- 1,1] and an increasing function on [0,1], then the range of y = √ f (X & # 178; - 3) + F (x + 1) is

1,-1

Find the range f (x) = x / x ^ 2-x + 3, f (x) = x ^ 2 / x ^ 2-x + 3

F (x) = x / (x ^ 2-x + 3) let a = x ^ 2-x + 3 a > 0f '(x) = (3-x ^ 2) / A ^ 2 F' (x) = 0, then x = ± √ 3, so f (x) in (- infinity, √ 3) and (√ 3, + infinity) is minus (- √ 3, √ 3) is increasing 〈 f (x) min = f (- √ 3) = - 1 / (2 √ 3-1) f (x) max = 1 / (2 √ 3-1) 〉 range is [- 1 / (2 √ 3-1), 1 / (2 √ 3 -

The function f (x) = the square of X - 2aX + 3, X belongs to [- 1,2], if the maximum value of F (x) is m (a), find the analytic expression of F (x) and the range of M (0)!

The analytic formula of F (x) can not be obtained from the known condition