Question

xf(x) + 2f(2/x) = x + 1 (x isn't 0)
I need to find f(x), answer is 1/x. I've tried everything but I just don't get it, can someone help me please?

Answer 1

man this is a pain, and i am getting stuck with the algebra but the gimmick is this:
you have one equation
\[xf(x)+2f(\frac{2}{x})=x+1\] replace \(x\) by \(\frac{2}{x}\) everywhere and you get
\[\frac{2}{x}f(\frac{2}{x})+2f(x)=\frac{2}{x}+1\] or
\[2f(x)+\frac{2}{x}f(\frac{2}{x})=\frac{2}{x}+1\]

Answer 2

multipliy the second equation by x and subtract from the first should do it

Answer 3

I'm getting nonsense answers again

Answer 4

\[xf(x)+2f(\frac{2}{x})=x+1\]
\[2xf(x)+2f(\frac{2}{x})=2+x\] subtract top from bottom and get
\[xf(x)=1\] and therefore \[f(x)=\frac{1}{x}\]

Answer 5

ok that works, now lets go slow
we start with \[xf(x)+2f(\frac{2}{x})=x+1\]

Answer 6

replace x by \(\frac{x}{2}\) everywhere

Answer 7

how come that 2f(2/x) isn't multiplied by x?

Answer 8

we get another equation involving \(f(x)\) and \(f(\frac{2}{x})\) namely
\[\frac{2}{x}f(\frac{2}{x})+2f(x)=\frac{2}{x}+1\]

Answer 9

multiply the second equation through by x to get
\[2f(\frac{2}{x})+2xf(x)=2+x\]

Answer 10

subtract the first equation from the last one to get
\[xf(x)=1\] and therefore
\[f(x)=\frac{1}{x}\]

Answer 11

I'm sorry I will let satellite show you

Answer 12

no you have it in one step, i wrote many

Answer 13

I can put it back up

Answer 14

only if you want

Answer 15

yeah put it back, i gotta run

Answer 16

Thank you both, please myini put it back so I can understand everything!

Answer 17

\[\text{ equation(1) :} x f(x)+2 f(\frac{2}{x})=x+1\] \[\text{ equation(2) : } 2f(x)+\frac{2}{x}f(\frac{2}{x})=\frac{2}{x}+1\] Do x*(eq2)+ -2*(eq1) . The result is this: \[2xf(x)+2f(\frac{2}{x})=2+x\] \[-2xf(x)-4f(\frac{2}{x})=-2x-2\] Now add. \[-2f(\frac{2}{x})=-x\] so we have \[f(\frac{2}{x})=\frac{x}{2}\] => We really don't need to do the the following but you can for more fun :) So I just used the firs equation and plugged in what i just found f(2/x) to be \[xf(x)+2 \cdot \frac{x}{2}=x+1 \] \[xf(x)+x=x+1 \] Subtracting x on both sides gives \[xf(x)=1\] Now dividing both sides gives us our pretty f(x) :)

Answer 18

how did you get equation 2?

Answer 19

I got it now, thank you guys ;)

Answer 20

Oh, why is it that I can't choose more than 1 better answer? Such a pity... Thank you both, anyway!