I'm reviewing functions for my Calculus $1$ course, and I'm stuck on why the algebraic method for finding a function's range works. (I'm aware of functions and inverse functions, which I learned in precalculus.)

I understand that when a function $f$ is defined in terms of variable $x$, finding the domain involves, by the domain convention, finding any potential restrictions on the real number values of $x$ (such as a zero in denominator, etc.). Intuitively, this makes sense. But why does this kind of process, of solving for $x$ and finding restrictions on $y$, work for finding the range algebraically? What's the intuition? Is there a particular way of thinking about it that can provide more understanding? Am I just overthinking this?


  • 1
    $\begingroup$ Rotate the graph of the function so that the y axis becomes the x axis and vice versa. You'll see the range becomes the domain (although if the original function was not 1-1, you won't actually have a function anymore) $\endgroup$
    – Aphyd
    Commented Jun 6, 2022 at 23:41
  • $\begingroup$ For example, consider y = x^2. Rotate it and you'll get a (multivalued) function x = sqrt(y). The domain of this function is the nonnegative reals, which corresponds to the range of x^2 $\endgroup$
    – Aphyd
    Commented Jun 6, 2022 at 23:43
  • 1
    $\begingroup$ By rotate I should have said flip $\endgroup$
    – Aphyd
    Commented Jun 6, 2022 at 23:48

1 Answer 1


So, if you'll recall, the range of a function $f : X \to Y$ is given by

$$\text{range}(f) = \{ y \in Y \mid \text{there is an $x \in X$ such that $f(x) = y$} \}$$

That is, $\text{range}(f)$ is the set of values $f$ maps something to.

If we take some function $f(x)$ and set it equal to $y$, such as

$$\frac{x^2 - 1}{x^2 + x + 1} = y$$

solving for $x$ will give us an $x$ which maps to that arbitrary $y$. (We only need one such $x$, even if multiple may exist, and we are constructing such an $x$.)

As a simple and explicit example, consider the function $y=f(x)$ where

$$f(x) = \frac{1}{x-2}$$

Then we set this equal to $y$. What $y$ is, is arbitrary. It just represents something arbitrary to be mapped to.

$$\frac{1}{x-2} = y$$

Well, if we solve this for $x$, we get

$$x = \frac 1 y + 2$$

What does this mean? For our particular function $f$, this means that

$$f \left( \frac 1 y + 2 \right) = y$$

whenever anything involved is defined. For instance, if I want to show that $2$ is in $\text{range}(f)$, then I let $y=2$, and we would have

$$\frac{1}{x-2} = 2 \text{ and } x = \frac 1 2 + 2 = \frac 5 2 \text{ and } f \left( \frac 5 2 \right) = 2$$

Hence, there is an $x$ for which $f(x)=2$, and thus $2$ is in the range.

Framed differently, you're looking at the domain of a (suitably defined) inverse function. In the ideal case, $f$ itself has an inverse function, call it $f^{-1}$. Then $\text{domain}(f) = \text{range}(f^{-1})$ and $\text{range}(f) = \text{domain}(f^{-1})$. The idea generalizes slightly more to noninvertible functions as well: if $f : X \to Y$, then $f^{-1} : Y \to X$.

That is essentially where the whole restrictions come in, because you're looking at the domain of a different function, in some loose sense.


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