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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 $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$

J. M. ain't a mathematician
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youdontknowme
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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 08 '12 at 22:43

3 Answers3

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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 $f(x)=f(t+\frac1t)=t^2+\frac1{t^2}=x^2-2$.

Hagen von Eitzen
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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 $f(x)=x^2-2$.

heropup
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user416006
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    Not only are you answering an old question that has already been answered adequately, your answer isn't even algebraically correct. Please explain why you think $$x^2 + \frac{1}{x^2} = \left(\frac{x+1}{x}\right)^2 - 2?$$ It appears that another user has improperly edited your original response. To this user: if you are going to edit someone else's post, you better be CERTAIN you have done it properly. – heropup Feb 13 '17 at 17:50
  • 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. – heropup Feb 13 '17 at 18:05
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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 $f$.

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

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

wnvl
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