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The function $f(x)$ is not defined when $x=0$. This function has the property that $f(x) + 2f\left(\frac 1x\right) = 3x$. Find all such values of $y$ such that $f(y) = f(-y)$. (This means it is an even function!).

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  • $\begingroup$ The equation only concerns values of the function on the same side. I think you can make any $y$ so that $f(y)=f(-y)$. $\endgroup$ – user1537366 Dec 14 '14 at 2:02
  • $\begingroup$ user1537366, I don't think that is the case. The problem asks for us to find a value that satisfy the equation f(x)+2f(1/x)= 3x. $\endgroup$ – Derek Zhou Dec 14 '14 at 2:12
  • $\begingroup$ OK, I didn't realise $f$ is fully determined. $\endgroup$ – user1537366 Dec 14 '14 at 2:14
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First of all, we can find $f$ explicitly. Indeed, we have $$\begin{align} f(x)+2f(1/x) &= 3x \\ f(1/x)+2f(x)&=3/x\end{align}$$ which is a simple system of linear equations whose solution gives $$f(x)=\frac{2}{x}-x.$$ Hence, $f(y)=f(-y)$ if and only if $y=\pm \sqrt{2}$.

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    $\begingroup$ Nice solution! I'll definitely keep this trick in mind for the future. $\endgroup$ – Cameron Williams Dec 14 '14 at 2:19
  • $\begingroup$ Thanks man! Smooth solution. I was worried that something went wrong because I thought you couldn't substitute 1/x for x. But, nope you are right though. $\endgroup$ – Derek Zhou Dec 14 '14 at 2:30

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