0
$\begingroup$

Consider $$ f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} + 5$$ Find the relation of $f(x)$ in $(-2\ \ \ 2)$ interval.

We discussed previously about domain of $f(x)$ (find $f(x)$ and its domain given $f(x+\frac{1}{x})$) and the result was this :

$f(x) = \begin{cases} \begin{align} x^2 + 3 & \quad \quad x \in (-\infty,-2] \cup [2,\infty) \\ \text{arbitrary} & \quad \quad x \in (-2,2) \end{align} \end{cases}$

But I want to know what's the meaning of "arbitrary" in this context ? And also how we can find relation of $f(x)$ in $(-2\ \ \ 2)$ interval?

2 Answers 2

1

$\left| x + \frac 1x \right| \ge 2$, therefore no matter which value we give to the function on the interval $(-2,2)$, the equation will be satisfied.

  • 0
    Okay , but I'm really confused about it.Why we include this interval ? – 2017-01-29
  • 0
    @S.H.W Because f is defined on $\mathbb{R}$, so we have to give $f$ values in that interval. If the function was $f:\mathbb{R} - (2,2) \to \mathbb{R}$, then we shoudn't define $f$ on that interval. – 2017-01-29
  • 0
    But I think domain of $f(x)$ should be $\mathbb{R} - (2,2)$. – 2017-01-29
  • 0
    Suppose $g(x) = x + \frac{1}{x}$ . The range of $g(x)$ equals to domain of $f(x)$. Am I right ? – 2017-01-29
  • 0
    @S.H.W The domain of f isn't stated, so usually it's assumed to be $\mathbb{R}$. – 2017-01-29
  • 0
    Okay , So we can assume any relation for $f$ in $(-2,2)$ . For example $e^x$. Am I right ? – 2017-01-29
  • 0
    @S.H.W yeah, you are right – 2017-01-29
  • 0
    Okay , thank you a lot . – 2017-01-29
  • 0
    @S.H.W you are welcome – 2017-01-29
0

It means that you can define $f$ as you want on $(-2,2)$ : the relation gives absolutely no information on $f$ on this interval

  • 0
    Why we define domain of $f(x)$ is $\mathbb{R}$ ? If we say $g(x) = x + \frac{1}{x}$ , can we conclude that range of $g(x)$ is equal to domain of $f(x)$ ? – 2017-01-29