# Why does solving for $x$ in order to find range algebraically work? What's the intuition?

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?

Thanks.

• 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) Commented Jun 6, 2022 at 23:41
• 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 Commented Jun 6, 2022 at 23:43
• By rotate I should have said flip Commented Jun 6, 2022 at 23:48

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.