What is the solution to this math homework challenge?

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Homework Help Overview

The original poster presents a problem involving a function f defined by an equation that includes multiple transformations of the variable x. The context appears to be related to algebraic manipulation and function analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various substitutions, such as changing x to tan(x), and question the placement of brackets in the original statement. There are attempts to derive the function f through algebraic transformations and polynomial expressions.

Discussion Status

Some participants have offered algebraic manipulations and transformations, while others express skepticism about the validity of proposed solutions. There is ongoing exploration of the problem without a clear consensus on the correct approach or solution.

Contextual Notes

Participants note potential issues with the function's definition at certain points, as well as the need for clarity regarding the completeness of the original problem statement.

hadi amiri 4
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Homework Statement


suppose f(x-1/x+1)+f(-1/x)+f(1+x/1-x)=x then find f(x)


Homework Equations





The Attempt at a Solution

 
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What's your attempt at finding a solution?
 
Last edited:
change x to tan(x)
 
Are the brackets in the right place? It seems to me that you're missing some. Furthermore is this the entire question?
 
would you like to know the answer
 
Yes please!
 
f(x)=2xxxx+3xx-1/3x-3xxx it was a question of the iranian mathematical olympiad note xx means x power two
 
So you mean: <br /> f(x)=2x^4+3x^2- \frac{1}{3} x-3x^3 <br />
This isn't possible the function is defined for 1, -1 and 0 and the exercise says that that isn't possible.

What method did they use to get the solution other than just trying different order of polynomials? Can you give me the website where you got this?
 
Last edited:
t=x-1/x+1 ------ > tx +t=x-1 -------- >x=1+t/1-t , -1/x=t-1/t+1 ,1+x/1-x=-1/t

------->f(t)+f(t-1/t+1)+f(-1/t)=1+t/-t -------->f(x)+f(x-1/x+1)+f(-1/x)=1+x/1-x

then
t=-1/x --------->x=-1/t , x-1/x+1=t+1/1-t , 1+x/1-x=t-1/t+1
------->f(t+1/1-t)+f(t)+f(t-1/t+1)=-1/t >f(x+1/1-x)+f(x)+f(x-1/x+1)=-1/x

then
t=1+x/1-x----------->t-tx=1+x ----->x=t-1/1+t------ > -1/x =t+1/1-t , x-1/x+1=-1/t
------>f(-1/t)+f(t+1/1-t)+f(t)=t-1/t+1------->f(-1/x)+f(x+1/1-x)+f(x)= x-1/x+1

two more step remains that i think you can do them yourself
what about this one
prove that: Arctan(1)+Arctan(2)+Arctan(3)=Pi
 
  • #10
Is there a website or so where I can see the question and the answers of the previous question?
 
  • #11
hadi amiri 4 said:
what about this one
prove that: Arctan(1)+Arctan(2)+Arctan(3)=Pi

I can actually do this one...haha

just use tan(A+B+C) where tanA=1,tanB=2,tanC=3

:biggrin:
 
  • #12
a geometric proof
 
  • #13
sorry!
 

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