# Math homework challenge

1. Jun 5, 2008

1. The problem statement, all variables and given/known data
suppose f(x-1/x+1)+f(-1/x)+f(1+x/1-x)=x then find f(x)

2. Relevant equations

3. The attempt at a solution

2. Jun 5, 2008

### dirk_mec1

What's your attempt at finding a solution?

Last edited: Jun 5, 2008
3. Jun 5, 2008

change x to tan(x)

4. Jun 5, 2008

### dirk_mec1

Are the brackets in the right place? It seems to me that you're missing some. Furthermore is this the entire question?

5. Jun 5, 2008

would you like to know the answer

6. Jun 5, 2008

### dirk_mec1

7. Jun 5, 2008

f(x)=2xxxx+3xx-1/3x-3xxx it was a question of the iranian mathematical olympiad note xx means x power two

8. Jun 5, 2008

### dirk_mec1

So you mean: $$f(x)=2x^4+3x^2- \frac{1}{3} x-3x^3$$
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: Jun 5, 2008
9. Jun 5, 2008

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
prove that: Arctan(1)+Arctan(2)+Arctan(3)=Pi

10. Jun 5, 2008

### dirk_mec1

Is there a website or so where I can see the question and the answers of the previous question?

11. Jun 5, 2008

### rock.freak667

I can actually do this one...haha

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

12. Jun 5, 2008