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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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 $$x^2-xy+1=0$$ $$x=\frac{y \pm \sqrt{y^2 -4}}{2}$$ 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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