Let the function f x satisfy f x 2 f x , and f 2 2, find f 2 4 100 20

<> haven't done it for a long time, it's all rusty, hope.

Because f(2)=2 so f(4)=f(2+2)=-f(2)=-2 because f(x+2)=-f(x) so f(x+4)=f(x+2+2)=-f(x+2)=-(-f(x))=f(x)=f(x).

So f(x) is a periodic function of period t=4, so f(-2)=f(2)=2f(100)=f(4*25)=f(0)=f(4)=-2

Solution: f(0+2) = f(0)=f(2)=2 So f(0)= 2f(-2+2)= f(-2)=f(0)=-2 So f(-2)=2f(2+2)= f(2)=f(4)=-2 So f(4)= 2, and so on, we can know f(100)=f(4)=-2

f（x）=2f（-x）+x

f(trace-x)=2f(x)-x

Multiply 2 on both sides to get it.

2f(-x)=4f(x)-2x

Belt pants with tassels into the original style:

f(x)=4f(x)-2x+x

3f(x)=x

f(x)=x/3

Let f(x)=ax+b, then the friend is defeated.

2f(x)+f(-x)

2(ax+b)+(ax+b)

2ax+2b-ax+b

ax+3b3x+2

So do implicit a=3, b=2 3

So f(x)=3x+2 is good to tremble 3

There is a question to scatter the silver meaning: f(x)=x+2f(1 x) 1So:f(1 x)=1 x+2f(x) 2 substitute 2 formulas into 1 formula: f(x)=x+2[1 x+2f(x)]f(x)=x+2 New Year's banquet x+4f(x).

3f(x)=x+2 finches sell x

f(x)=-x+2/x)/3

f(x)+2f(-x)=2x-1 (1) Use -x instead of x in the above equation to get :

f(-x)+2f(x)=-2x-1 (2)(2) 2 get:

2f(-x)+4f(x)=-4x-2 (3)(3)-(1):

3f(x)=-6x-1

f(x)=-2x-1/3

f（x）+2f（-x）=2x-1………1) Then f(-x)+2f(x)=2(-x)-1=-2x-1.........2）

Then (2)*2-1, we can find f(x).

f（x)=-2x-1/3

Let x=1 x, then f(1 x)+2f(x)=1 x. Return to the problem and this formula can be obtained f(x)=2 (3x)-x leak brother spring 3

<> solution: (1) f(x) can be known from the question

e?x，?1＜x≤0

4x?4x+1，0＜x≤1

From f(x+1)=-f(x), we can see that f(x+2)=f(x), that is, the function f(x) is a periodic function with 2 as the minimum positive period.

Therefore, the image of the function is shown in the figure on the right

As can be seen from the graph, the monotonically decreasing interval of the function f(x) is (2k?

k z), with an increasing interval of [2k+1

2k+1](k∈z)

6 points) (2) The number of zeros of the function g(x) at x [0,5] can be obtained from the image of the function.

This is the number of f(x)=kx roots, i.e., the number of intersections between the f(x) image of the function and the y=kx image.

Then when k e, the function f(x) image and y=kx image have an intersection point at x [0,5], so g(x) has a zero point at x [0,5];

Then when 1 k e, the function f(x) image and y=kx image have two intersection points at x [0,5], so g(x) has two zero points at x [0,5]. then when e

When k 1, the image of the function f(x) and the image of y=kx have three intersections at x [0,5], so g(x) has three zero points at x [0,5]. then when 1

At k e, the image of the function f(x) and the image of y=kx have four intersections at x [0,5], so g(x) has four zeros at x [0,5]. then when 1

At k 1, the function f(x) image and y=kx image have five intersections at x [0,5], so g(x) has five zeros at x [0,5].

then when 0 k 1

, the function f(x) image and y=kx image have six intersections at x [0,5], so g(x) has six zero points at x [0,5].

Solution: 1 x makes sense, x ≠ 0.

x, 1 x is also taken on the definition field, and x is replaced by 1 x2f(1 x)+f(x)=10 (1 x) (1).

f(1/x)+2f(x)=10^x (2)

The analytic formula for 3f(x)=2 10 x -10 (1 x)f(x)=(2 3) 10 x -10 (1 x) 3 is f(x)=(2 3) 10 x -10 (1 x) 3 (x≠0).

Note: x≠0 must be determined, and a defined field is the complete analytic formula.

Let y=1 x, then there is 2f(1 y)+f(y)=10 (1 y)=1 10 y, that is: 2f(1 x) + f(x)=1 10 x are combined, and this formula and the original formula, eliminate f(1 x).

It can be obtained: -3f(x)=1 10 x - 2*10 x so f(x) = ( 2*10 x - 1 10 x ) 3