Question

Suppose f is a real valued function with domain, all nonzero reals. Suppose further that
f(x)+2f(1/x)=x for all x in the domain. Find a formula for f(x).

Answer 1

Really not sure how to do this, Wolfram gives a solution, but will not show steps, on student pro version.

Answer 2

The solution is a function of x not involving f.

Answer 3

@ganeshie8 @Directrix @freckles @hartnn

Answer 4

@Kainui

Answer 5

@mathmale

Answer 6

\[f(x) + f(1/x) = x\]

replace \(x\) by \(1/x\) everywhere in above equation, what do you get ?

Answer 7

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

Answer 8

you could simplify 1/(1/x)

Answer 9

if you mean take the inverse of each input it would x

Answer 10

right, so you have two equations with two unknowns f(x) and f(1/x)
you can eliminate f(1/x)

Answer 11

\(\color{red}{f(x)} + 2\color{blue}{f(1/x)} = x\tag{1}\)

\(\color{blue}{f(1/x)} + 2\color{red}{f(x)} = 1/x\tag{2}\)

Answer 12

you may multiply second equation by 2, then subtract the first equation

Answer 13

I liked this equation so much that I decided to solve a general case of it and it has a beautiful general solution.

\[af(x)+bf(\tfrac{1}{x}) = g(x)\]
where g(x) is given. Then we can express f(x) as:
\[f(x) = \frac{ag(x)-bg(\tfrac{1}{x})}{a^2-b^2}\]

I found this to be sorta odd to look at and interesting. I didn't show the steps, because it's exactly the same steps @ganeshie8 is using. Just thought I'd share. :D

Answer 14

I got 2/(3x^2) which can't be right

Answer 15

Show your steps, and we can find exactly where your reasoning breaks.

Answer 16

following Ganeshie

Answer 17

\[f(x)+2f(1/x)=x \rightarrow f(1/x)+2f(x)=1\rightarrow (2f(1/x)+4f(x)=2/x)\]
\[-2f(1/x)-f(x)=x \rightarrow 3f(x)=(2/x)+x \rightarrow f(x)=(2+x^2)/3x\]

Answer 18

\(3f(x)=(2/x)\color{red}{-}x \)

Answer 19

Thanks guys!

Answer 20

That formula though...