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 $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
$\begingroup$
$\endgroup$
1
-
$\begingroup$ 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. $\endgroup$– Hagen von EitzenNov 8, 2012 at 22:43
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
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 $f(x)=f(t+\frac1t)=t^2+\frac1{t^2}=x^2-2$.
answered Oct 25, 2012 at 16:07
$\begingroup$
$\endgroup$
2
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 $f$.
$$f(y)=\left(\frac{y \pm \sqrt{y^2 -4}}{2}\right)^2+\left(\frac{2}{y \pm \sqrt{y^2 -4}}\right)^2$$
$$f(y)=y^2-2$$
-
$\begingroup$ 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 :) $\endgroup$ Oct 25, 2012 at 17:33
-
$\begingroup$
$\endgroup$
2
$$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2$$ 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 $f(x)=x^2-2$.
-
1$\begingroup$ 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. $\endgroup$– heropupFeb 13, 2017 at 17:50
-
$\begingroup$ 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. $\endgroup$– heropupFeb 13, 2017 at 18:05