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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.
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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).
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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 ..
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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
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Here's how, please take the test:
If it helps, big and bold.
Please take the town.
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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.
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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
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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
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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.
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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
Knowable. f(0)=0
f'(x)=e^x-1-2ax >>>More
Your teacher's unraveling method is right.
Because it's an equation, the left and right sides are equal, but in different forms, for example, if the numbers on both sides of the 2=2 equal sign are obviously equal, and square them to 4=4, isn't it the same? >>>More
If you first take the value of x as 1, then the left and right sides of the equation become: 1+2+1=a0+0+0+0+0+0, so a0=4, and you take the value of x to 0, then the equation becomes: 0+0+1=a0-a1+a2-a3+a4-a5, that is: >>>More
If you start with y as a distance, the equation may be easier to understand. d = root number (x 2+4x+13) + root number (x 2-2x+2) = root number ((x+2) 2+9) + root number ((x-1) 2+1) = root number ((x+2) 2+(0-3) 2) + root number ((x-1) 2+(0+1) 2). This is the sum of the distances between the point (x, 0) and the point (-2,3) and the point (1,-1). >>>More
Answer: Conclusion: x (x+1) (x+1) x
Proof is as follows: x (x+1) (x+1) x >>>More