Let f (x) = log4 (4 ^ x + 1) + KX be an even function. Let g (x) = log4 (A2 ^ x-4 / 3a) be a function F（

Let f (x) = log4 (4 ^ x + 1) + KX be an even function. Let g (x) = log4 (A2 ^ x-4 / 3a) be a function F（

f（x）=log4（4^x+1）+kx=x+0+kx=(k+1)x
A function of degree one is a straight line, not even function
From the question, the function is even function
Then K + 1 = 0, k = - 1
Is your topic not finished?

If f (x) is an odd function defined on R with 3 as its period and f (2) = 0, then the minimum number of solutions of the equation f (x) = 0 in the interval (0,6) is

∵ the period of the function f (x) is 3
∴f（2）=f（5）=0=f（-1）=f(-4)
∵ the function f (x) is an odd function
∴f（-x）=-f（x）
∴f(-1)=-f(1)=0
f(-4)=-f(4)=0
∵ f (x) is an odd function
∴f（0）=0
∵ the period of the function f (x) is 3
∴f（3）=f（0）=0
The minimum number of solutions of the equation f (x) = 0 in the interval (0,6) is 5 (1,2,3,4,5,)

If f (x) is an odd function defined on R with period 3 as its period, if f (2) = 0, then the number of solutions of the equation f (x) = 0 in the interval (0, 6) () A. It's three B. It's four C. It's five D. More than 5

∵ f (x) is an odd function with period 3 defined on R, f (2) = 0. If x ∈ (0, 6), then f (5) = f (2) = 0
If f (x) is an odd function, then f (- 2) = - f (2) = 0, then f (4) = f (1) = f (- 2) = 0
And the function f (x) is an odd function defined on R. it can be concluded that f (0) = 0, thus f (3) = f (0) = 0
In F (x + 3) = f (x), let x = - 3
2, then f (- 3)
2）=f（3
2) From the definition of odd function, we can get f (- 3)
2）=-f（3
2），∴f（3
2）=0．
Therefore, f (9
2）=f（3
2) = f (4) = f (1) = f (3) = f (5) = f (2) = 0,
Therefore, D

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 with period 1 on R, then G (x) = g (x + 1) and

It is known that f (x) is an even function defined on R and has a period of 2, then "f (x) is an increasing function on [0,1]" is a () A. Conditions that are neither sufficient nor necessary B. Sufficient but unnecessary conditions C. Necessary but insufficient conditions D. Necessary and sufficient conditions

∵ f (x) is an even function defined on R,
If f (x) is an increasing function on [0, 1], then f (x) is a decreasing function on [- 1, 0],
And ∵ f (x) is a function defined on R with period 2, and the difference between [3,4] and [- 1,0] is two periods,
The monotonicity of the two intervals is the same, so it can be concluded that f (x) is a decreasing function on [3,4], so the sufficiency holds
If f (x) is a decreasing function on [3,4], it can be concluded that f (x) is a decreasing function on [- 1,0] from the periodicity of the function, and then it can be concluded that f (x) is an increasing function on [0,1] if the function is even
In conclusion, "f (x) is an increasing function on [0,1]" is a necessary and sufficient condition for "f (x) to be a decreasing function on [3,4]"
Therefore, D

If the function f (x) is an even function with period U / 2 and f (Wu / 3) = 1, find the value of F (- 17 / 6) If the function f (x) is an even function with period U / 2 and f (Wu / 3) = 1, find the value of F (- 17 / 6 times of U)

If f (x) = ACOS ω x, ? t = π / 2, ? ω = 4,  f (x) = acos4x, ? f (π / 3) = 1,  acos4 π / 3 = 1, that is, ACOS (π + π / 3) = 1, 57368; a (- 1 / 2) = 1,  a = - a = - 2, then f (x) = - 2cos4x, - 17 π / 6 = - 2 π - 5 π / 6, (- 17 π / 6) = - 2cos (- 8 π - 10 π - 10 - 10 - 10 π - 10 π - 5 π - 5 π - 6, π / 2) = - 2cos (- 10 π / 2)

If the function f (x) = a ^ 2 + (A-2) x + B is defined as (B, A-1), it is even function? Sorry, it's wrong. The function f (x) = ax ^ 2 + (A-2) x + B is defined as (B, A-1) and is even function. Then the range of F (x) is? The answer is [- 1,1)

F (x) = ax ^ 2 + (A-2) x + B is an even function,
So A-2 = 0 (1)
The domain is symmetric about the origin, so B + A-1 = 0 (2)
From (1) (2), a = 2, B = - 1,
So f (x) = 2x ^ 2-1 (- 1)

If the function f (x) = ax ^ 2 + (a + 1) x + 2 is an even function on the domain [- 2,2], find the value range of this function

Because it's even,
f(x)=f(-x)
ax^2+(a+1)x+2=ax^2-(a+1)x+2
a+1=-1-a
a=-1
f(x)=-x^2+2
So the range is [- 2,2]

The function f (x) = ax ^ 2 + BX + 3A + B is even function, and its definition domain is [a-1.2a] (a, B belongs to R) to find the value range of F (x)

The definition domain of even function is symmetric about the origin
So A-1 and 2a are opposite numbers
a-1=-2a
a=1/3
Definition domain [- 2 / 3,2 / 3]
Even function
f(-x)=ax²-bx+3a+b=f(x)=ax²+bx+3a+b
Then - BX = BX
bx=0
This is an identity
So B = 0
f(x)=x²/3+1
-2/3

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]