Background: I was trying to solving this question:

Find $\frac{1}{x}$, if $x \in(-1,3)-\{0\}$.

I know that I can solve the question by drawing the graph of the function, $\frac{1}{x}$, then I should find the values of the function at $-1$ and $3$, then I should "look" for the values the function assumes between the two values. But this method is impractical; what if the function is not easily "graphable"? How should one proceed then? That's why I tried to find algebraic method to solve this question and hence this type of questions.

I tried to solve it, algebraically, like this:

I know that domain of a inverse function is equal to the range of the function. So to find the range of $f(x)\frac{1}{x}$, with the restriction, I should first find the inverse of the function.

The inverse of the function is $f^{-1}(x)=\frac{1}{x}$, with $x, f^{-1}(x)\not= 0 $ and $f^{-1}(x) \in(-1,3)$.

Now all I have to do is to find Domain of this new function. But I don't know how to proceed from here. And this method is also impractical, because not every function is one to one function.

Question: The question is how to find range of a function with restricted domain, algebraically? Note that I am not asking how to solve that particular question of finding $\frac{1}{x}…$, but what I am asking is methods to solve this type of questions, algebraically.


Not exactly algebraically but you might find it useful.

You can look at the behaviour of the function, for example, for a continuous function, look at its maximum and minimum. And a particular case of the above, if the function is increasing (or decreasing) simply evaluate it at the extremes of the domain. (In both cases I assume the domain is an interval, but you can generalise for example to a union of several intervals or a function defined by parts).

  • $\begingroup$ I didn't understand your answer. Illustration(s) will be usefull. Thanks for your effort. $\endgroup$
    – user663117
    Jun 19 at 3:07
  • $\begingroup$ You seem to be new on Stack Exchange, please have a look on this tour and this and this post. $\endgroup$
    – user663117
    Jun 19 at 3:16

This is a problem that requires that you stretch your intuition.

Let $f(x) = (1/x).$

Then, $f(-1) = (-1).$

Further, as $x$ goes from $(-1)$ towards $0$ [i.e. approaches $0$ from below], $f(x)$ goes to $-\infty.$

Similarly, $f(3) = (1/3)$

and as $x$ goes from $(3)$ towards $0$, $f(x)$ goes to $+\infty.$

So, the range of the function, in the domain $\{x: -1 < x < 3, x \neq 0\}$, is

$(-\infty, -1) \bigcup (1/3, \infty)$.

That is, the range is $\{x < -1\} \bigcup \{1/3 < x\}.$

This should help you graph the function.

  • $\begingroup$ In this particular cases the answer was easy to find, but what if the question was something like: find $\frac{\sqrt{3x-5}+2}{x^{3}+22 }$, if $x \in (9,28)$? I don't have single clue about what the graph would look like; how the graph will behave between the points, how then shall I solve this question (without using graphing tools)? $\endgroup$
    – user663117
    Jun 19 at 5:44
  • $\begingroup$ But you may use calculator, though. $\endgroup$
    – user663117
    Jun 19 at 5:59
  • $\begingroup$ @HitlerDean In your comment/question, the first thing to notice is that for $x \in (9, 28)$, the denominator ranges from $751$ to $21974$, and thus the denominator is always positive.The next things to notice are [1] the numerator will (also) always be positive, throughout that range, and [2] the numerator increases much more slowly than the denominator. Again, your primary weapon is your intuition. Clearly, the function will therefore be strictly decreasing, so the obvious thing to do is to eval the function at (for example) $x=9$, $x=18,$ and $x=28$, and then rough out the graph. $\endgroup$ Jun 19 at 9:03
  • $\begingroup$ This is last question. How will you find the value of $\sqrt{x^2-4}$? Thanks. $\endgroup$
    – user663117
    Jul 1 at 12:15
  • $\begingroup$ @Διαισθητικός I am unsure what you are asking here. I interpret your question to be if $f(x) = \sqrt{x^2 - 4}$, then what is the domain of $f(x)$. Please advise if I have misinterpreted your question. Anyway, since negative square roots are disallowed, you have to identify all $x$ such that $x^2 - 4 \geq 0.$ For $0 \leq x$, this is clearly the subset $\{x ~: ~2 \leq x\}.$ Since $x^2$ is an even function, you also have the subset $\{x ~: ~x \leq -2\}.$ Therefore the domain is $(-\infty, -2] \cup [2, + \infty).$ $\endgroup$ Jul 1 at 12:23

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