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.
Best Answer
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.