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$f(x+1/x)=x^2+1/x^2$, find f(x) In the above question,I get to solve that f(x)=x^2-2 but what i miss in my answer is that absolute value of x is greater than 2 always. I have two questions,first,how do we come to the statement that absolute value of x is greater than 2 here(In the question in my textbook,its given that x belongs to R-{0}. Why do we meed to do the absolhte value thing?)

And second is how to prepare myself whenever I encounter such questions so that I know what I have to look for,for finding range and domain of the expression.

Ethan Bolker
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Dilin Finn
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  • Think about the range of the function $g(x)=x+\frac{1}{x}$ with domain $\Bbb R^*$. That determines the domain of $f$. For example, is it possible to get $g(x)=1$? Sketching a graph will help – Shuri2060 Aug 22 '17 at 13:41
  • It cant be brought to 1 – Dilin Finn Aug 22 '17 at 13:46
  • So does including $1$ in the domain of the final answer $f(x)=x^2-2$ make sense? (and here it is important to remember the '$x$' here is different from that of the original question) – Shuri2060 Aug 22 '17 at 13:47
  • It will be -1,the explaination that i can give myself is that x^2-2 gives positive value for x so 1 cant be included in its domain.....but then it means that i have to modify the statement in the ques R-{0} – Dilin Finn Aug 22 '17 at 13:51

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Assuming you're reporting the problem exactly as written, you're puzzled about finding the "solution" to an appallingly badly stated problem.

For instance, if I tell you I have a function $f$, and that I know that for any real $x$, it's true that $$ f(x^2) = x^2 $$ you cannot conclude anything about the domain of $f$ except that it must include all nonnegative real numbers.

For instance, I might have been thinking about the function from the reals to the reals defined by $x \mapsto x$; that function has the property stated.

I might also have been thinking about $f(x) = |x|$, or thinking about $$ f(x) = \begin{cases} x & x \ge 0 \\ -13 & x < 0\end{cases}, $$ or about $$ f : \Bbb R^{+} \cup \{0\} \to \Bbb R^{+} \cup \{0\}: x \mapsto x. $$

All these functions have the property that I specified, but they have varying domains, and varying images and even have varying codomains.

So I guess my recommendation would be "Try not to sweat too much about solving problems like these; instead, watch carefully for them and run away from anyone who asks you to 'solve' such things."

John Hughes
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  • It was in my textbook,and the writer is one of the most reputed one in my country. Not that this is my homework or something,I just wanted to know what I can learn from this question. – Dilin Finn Aug 22 '17 at 14:06
  • To each their own. I've advised you on something you can learn, which is to sometimes ask "is this a question that has an answer?" Even great and respected textbook authors sometimes make errors. (And I say that as a not-so-great-and-respected textbook author who found many errors in other books while writing a new one.) – John Hughes Aug 22 '17 at 14:09
  • Thank You. Just one more clarification,how did domain of the function change on changing x+1/x to x, x being a member of R-{0} – Dilin Finn Aug 22 '17 at 14:13
  • You're given that $f(x + \frac{1}{x}) = x^2 + \frac{1}{x^2}$. The author assumes that this means the equality holds for all values of $x$ for this it makes sense, which means $x \in \Bbb R, x \ne 0$. (I personally think the author should say this, but ...) That means that every number of the form $x + \frac{1}{x}$ (for nonzero real $x$) must be in the domain of $f$. (It also appears to mean, to your author, that only those numbers are in the domain; here the author and I differ.) Suppose that $y = x + \frac{1}{x}$. What can you say about $y$? – John Hughes Aug 22 '17 at 14:25
  • Well, it can be anything of absolutely value at least $2$. And that's because $y = x + \frac{1}{x}$ (for $x \ne 0$) is the same as $$0 = x^2 -xy + 1,$$ whose solutions are $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$; such solutions exist if and only if $|y| \ge 2$. So the domain of $f$ (using the author's goofy rule) is that $f$ is defined on $(-\infty, -2] \cup [2, \infty)$. – John Hughes Aug 22 '17 at 14:27
  • Guess its hit it and leave it sometimes.... And i guess that this Question in its own needs to learn a bit – Dilin Finn Aug 22 '17 at 14:33