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Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.

The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$

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You've been given two answers a while ago (with the same tenor), but you have accepted none. Do you still require any clarification? Or need to discuss how the textbook answer could be wrong? The expressions are definitely not equivalent, hence the textbook is wrong or that function refers to something else. –  Hagen von Eitzen Nov 8 '12 at 22:43
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2 Answers

Note that $(x+\frac1x)^2=x^2+2+\frac1{x^2}$. Hence it looks like $f(x)=x^2-2$ is a good candidate. Of course, $x+\frac1x\ge2$ implies that we cannot say anything about $f(x)$ if $x<2$. But for $x\ge 2$, we can find a real number $t$ such that $t^2-xt+1=0$ (and hence $t+\frac1t=x$), namely $t=\frac{x\pm \sqrt{ x^2-4}}2$, and then see that indeed $f(x)=f(t+\frac1t)=t^2+\frac1{t^2}=x^2-2$.

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Let $y=x+\frac{1}{x}$, try now to express $x$ as a function of $y$.

We have


$$x=\frac{y \pm \sqrt{y^2 -4}}{2}$$

Substitute this value for x in your expression for $f$.

$$f(y)=\left(\frac{y \pm \sqrt{y^2 -4}}{2}\right)^2+\left(\frac{2}{y \pm \sqrt{y^2 -4}}\right)^2$$


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I wonder if I finish that,will it end up like 1+x^2+x^4/x*sqrt(1-x^2) but thank u anyway :) –  youdontknowme Oct 25 '12 at 17:33
solution is $x^2-2$ –  wnvl Oct 25 '12 at 18:02
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