Even function f (x) is monotone on the interval [0, a] (a > 0), and f (0) * f (a)

Even function f (x) is monotone on the interval [0, a] (a > 0), and f (0) * f (a)

One
F (0) * f (a) 0) are monotone functions
Therefore, there is an intersection point with X axis on [0, a]
If the function is even, then there is an intersection point with X axis on [- A, 0]!
Therefore, in the interval [- A, a], there are two intersections between F (x) and X axis, that is, f (x) = 0 has two roots

Construct a f (x) =? With period pie, range [1 / 2,3 / 2] and a decreasing function on [0, Pie / 2]?

f(x)=1/2cos(2x+1)

How to construct an even function f (x) with period u, value range [1 / 2,3 / 2] and a minus function on [0, Wu / 2]=

1+1/2·cos2x

If f (x) = cos (arcsinx), then f (x): 1 is an even function, 2 is a periodic function, 3 is defined as [- π / 2, π / 2], and 4 is [0,1] A1，2 B3,4 C1,4 D 2,3

1.1 is obviously not. 3 for. 2. The definition field is the whole real number. Value range: [- 1,1]

Let g (x) be a function defined on R with period 1. If f (x) = x + G (x) on [0, 1] has a range of [- 2,5], then the value range of F (x) on the interval [0,3] is () A. [-2，7] B. [-2，5] C. [0，8] D. [-3，7]

If G (x) is a function of period 1 on R, then G (x) = g (x + 1)
The value range of the function f (x) = x + G (x) in the interval [0,1] (exactly the length of a periodic interval) is [- 2,5]
Let x + 1 = t, when x ∈ [0,1], t = x + 1 ∈ [1,2]
In this case, f (T) = t + G (T) = (x + 1) + G (x + 1) = (x + 1) + G (x) = [x + G (x)] + 1
Therefore, when t ∈ [1,2], f (T) ∈ [- 1,6] （1）
Similarly, let x + 2 = t, when x ∈ [0,1], t = x + 2 ∈ [2,3]
In this case, f (T) = t + G (T) = (x + 2) + G (x + 2) = (x + 2) + G (x) = [x + G (x)] + 2
Therefore, when t ∈ [2,3], f (T) ∈ [0,7] （2）
According to the known conditions and (1) (2), the value range of F (x) on the interval [0,3] is [- 2,7]
Therefore, a

Let f (x) be a function defined on R with a period of 1. If G (x) = f (x) - 2x has a value range of (- 2,6) in the interval "2,3", then G is at (- 12,12)

Your question is not finished, but I'll try to help you with the title,
Let f (x) be a function defined on R with a period of 1. If G (x) = f (x) - 2x has a range of (- 2,6) on the interval [2,3], then what is the range of G (x) on [- 12,12]?
∵ f (x) is a function defined on R with a period of 1,
∴f(x+1)=f(x)
Then G (x + 1) = f (x + 1) - 2 (x + 1) = f (x + 1) - 2x-2 = f (x) - 2x-2 = g (x) - 2
And ∵ g (x) on the interval [2,3] is (- 2,6), that is, when 2 ≤ x ≤ 3, there is - 2 < g (x) < 6
Let t = x + 1, then 3 ≤ t ≤ 4, so g (T) = g (x + 1) = g (x) - 2
∴-4＜g(t)＜4
That is, the value range of G (x) on the interval [3,4] is (- 4,4)
.
Similarly, the value range of G (x) on the interval [2 + N, 3 + n] is (- 2-2n, 6-2n). (n is an integer)
The value range of G (x) on [- 12,12] is the union of the ranges of values on [- 12, - 11], [- 11, - 10]... [11,12], i.e. (26,34) ∪ (24,32) ∪ (- 20, - 12) = - 20,34)

Let g (x) be a function defined on R with period 1. If the value range of function f (x) = x + G (x) on the interval [3,4] is [- 2,5], then the value range of F (x) on the interval [- 10, 10] is______ .

Method 1: ∵ g (x) is a function of period 1 on R, then G (x) = g (x + 1)
The value range of the function f (x) = x + G (x) in [3,4] is [- 2,5]
Let x + 6 = t, when x ∈ [3,4], t = x + 6 ∈ [9,10]
In this case, f (T) = t + G (T) = (x + 6) + G (x + 6) = (x + 6) + G (x) = [x + G (x)] + 6
Therefore, when t ∈ [9,10], f (T) ∈ [4,11] （1）
Similarly, let X-13 = t, when x ∈ [3,4], t = X-13 ∈ [- 10, - 9]
In this case, f (T) = t + G (T) = (X-13) + G (X-13) = (X-13) + G (x) = [x + G (x)] - 13
Therefore, when t ∈ [- 10, - 9], f (T) ∈ [- 15, - 8] （2）
...
By (1) (2) It is found that the value range of F (x) on [- 10, 10] is [- 15, 11]
So the answer is: [- 15, 11]
The second method is that f (x) - x = g (x) holds on R
So f (x + 1) - (x + 1) = g (x + 1)
So f (x + 1) - f (x) = 1
Therefore, if the independent variable increases by 1, the function value also increases by 1
Therefore, the value range of F (x) on [- 10, 10] is [- 15, 11]
So the answer is: [- 15, 11]

Let g (x) be a function defined on R with period 1. If the value range of function f (x) = x + G (x) on interval [0,1] is [- 2,5], then the value range of F (x) on interval [0,3] is___ .

If G (x) is a function of period 1 on R, then G (x) = g (x + 1)
The value range of the function f (x) = x + G (x) in the interval [0,1] (exactly the length of a periodic interval) is [- 2,5] （1）
Let x + 1 = t,
When x ∈ [0,1], t = x + 1 ∈ [1,2]
In this case, f (T) = t + G (T) = (x + 1) + G (x + 1) = (x + 1) + G (x) = [x + G (x)] + 1
Therefore, when t ∈ [1,2], f (T) ∈ [- 1,6] （2）
Let x + 2 = t,
When x ∈ [0,1], t = x + 2 ∈ [2,3]
In this case, f (T) = t + G (T) = (x + 2) + G (x + 2) = (x + 2) + G (x) = [x + G (x)] + 2
Therefore, when t ∈ [2,3], f (T) ∈ [0,7] （3）
We obtain the condition that (2, 2) is a given value of (2, 2) on the interval
So the answer is: [- 2, 7]

Let g (x) be a function defined on R with period 1. If f (x) = 2x + G (x) has a range of [- 1,3] on [0, 1], then the value range of F (x) on the interval [0, 3] is______ .

Let x ∈ [1,2], then X-1 ∈ [0,1], then f (x) = 2x + G (x) = 2 (x-1) + G (x-1) + 2 = f (x-1) + 2 ①, ? x ∈ [0,1], f (x) ∈ [- 1,3], ? for formula ①, f (x-1)) ∈ [- 1,3], ? f (x) = f (x-1) + 2 ∈ [1,5]. Similarly, when x ∈ [2,3], then X-2 ∈

Construct an even function whose domain is [- 1,1] and range is [- 2,5]

f(x)=7x^2-2