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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.2012-11-08

3 Answers 3

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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, $\left|x+\frac1x\right|\ge2$ 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

$x^2-xy+1=0$

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

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

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

$f(y)=y^2-2$

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    solution is $x^2-2$2012-10-25
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$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2$ Let $x+\frac{1}{x}=z$. Then we get, $f(z)=z^2-2.$ Hence we put x on the place of z. And we get $f(x)=x^2-2$.

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    I have rolled back the improper edit and modified your original post to be properly formatted while preserving the original content. However, I should point out that your post adds nothing substantive to what has already been described over 4 years ago.2017-02-13