Let f f x 1 x x 2 then f x detailed process...

Obviously, x+2 can be seen as (x+1)+1 then (1+x) (x+2) equals (x+1) [(x+1)+1] equals 1 [(x+1)+1] (x+1) (x+1) The idea of dividing 1 by its reciprocal result remains unchanged! is equal to 1 [1+1 (x+1)] So because f[f(x)]=1 [1+1 (x+1)], then f(x)=1 (1+x) is so tired. It looks better to write on paper with fractions.

It's tiring to look at it this way.

The logarithm is meaningful, (1-x) (1+x)>0

x-1)/(x+1)<0

10,x2+2>0, so (x1-x2) [(x1+2)(x2+2)]<0

0f(x2)-f(x1)<0

The f(x2) function f(x) decreases monotonically on (-1,1).

In this game, Tong Ge has to find a way to set up the term of f[f(x)] containing the unknown number x, and it is obvious that x+2 can be regarded as (x+1)+1 then (1+x) (x+2) is equal to (x+1) [x+1)+1] is equal to 1 [(x+1)+1] (x+1) Here we use the idea of dividing 1 by its reciprocal result unchanged! is equal to 1 [1+1 ..

Summary. Wait a minute, I'm writing.

Let f(x) x 1) 2(x 1)丨, find f'(x) and find the process.

Wait a minute, I'm writing.

Okay, thanks. You're welcome, you can come back to me if you have any questions

Here's how, please take the test:

If it helps, big and bold.

Please take the town.

The expression of the known function is.

f(x) Sakurasan x 2-4x.

Then the schizoid clan is the source of respect.

1) F (A2) (A2) 2A4 (A2).

2) f(x1) (x1) 2-4-4(x-1) x2-2x1-4x-4

x 2 a 2 x a 3.

Solution: Let x'=x+1, regimental sky x'=x+1, then collapse x=x'-1, substituting f(x+1)=x -2x to get :

f(x‘)=x'-1)²-2(x'-1)

x'²-2x'+1-2x'+2

x'²-4x'+3

x=x', substituting f(x') into the demolition raid yields: f(x)=x -4x+3f(x)=x -4x+3

Let x+1=t, then x=t 1.

f(x+1)=x 2x can be converted into f(t)=(t 1) 2(t 1), and the modulus is called f(t)=t 4t+3.

Then let t=x, then f(x) Dan Kaiming car = x 4x+3

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

So f(x)=x 4x+3

Method 2: Commutation method so that x+1=t then x=t-1f(t)=(t-1) 2-2(t-1)=t 2-4t+3, so f(x)=x 4x+3

This kind of topic swap method is the most commonly used version and also right.

is very effective. Pay attention to the usual summary and induction.

Solution: Let x'=x+1， x'=x+1, then x=x'-1, substitute f(x+1)=x -2x to get: back.

Answer: f(x')=(x'-1)²-2(x'-1)=x'²-2x'+1-2x'+2

x'²-4x'+3

x=x', substituting f(x') gets: f(x)=x -4x+3 f(x)=x -4x+3