Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $y = f(x) = \sqrt{2x + 1}$ for $x \geq -1/2$. Then, $f$ is injective on its domain and therefore its inverse is well-defined. To find the inverse, we simply invoke the necessary algebraic operations to solve for $x$ and determine that

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

and therefore

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

Now, I realize the name of the indeterminate has no effect on the validity of the expression but in every elementary text I see, the inverse is written instead as $$ f^{-1}(x) = \frac{x^2 -1}{2} $$ which is really counterintuitive. If our original function maps from the "x-axis" to the "y-axis" then it makes sense that the inverse would map from the "y-axis" to the "x-axis", not conversely.

So my question is, Is there a reason why most texts choose the latter representation instead of the former or is it just a convention that is followed without any apparent justification?

share|cite|improve this question
There is a strong geometric flavour here. For arbitrary functions from one set to another, that flavour may be muted or missing. Also, $f^{-1}$ is a function like any other, it should not be forever tied to its origins. For example, $\ln$ is the inverse of $\exp$. If we insisted on writing $\ln y$, how would we deal with the curve $y=\ln x$? – André Nicolas Jul 7 '11 at 22:05
Think about it this way: given two functions $f(x) = \sqrt{2x + 1}, x≥−1/2$ and $g(x) = \frac{X^2 -1}{2}, x≥0$ the choice of which is the forward and which is the inverse is arbitrary. They are inverses of each other. We therefore use $x$ as the argument for both. – Tpofofn Jul 7 '11 at 22:29
I don't understand the downvote to the question. It is a genuine question, well written and future students might have the same question. – Aryabhata Jul 7 '11 at 22:29
When this is taught in practice there is a convention that many people use to teach it as "switch x and y and solve for x". I think that's because students are "used to" solving for x and telling them to solve for y may seem counterintuitive. So it may just alleviate questions like: "Why are we solving for y here when we usually solve for x?" – tomcuchta Jul 7 '11 at 22:40
Mathematics is invariant under permutations of the alphabet – Thomas Rot Jul 8 '11 at 0:29
up vote 4 down vote accepted

First, the former representation is also commonly used; see, for example, Wikipedia's table here. The justification for the latter representation, however, is simply that functions are usually written in terms of $x$; see here for a concrete example.

share|cite|improve this answer

As long as you are considering $f$ and $f^{-1}$ at the same time, e.g. when discussing the continuity of $f^{-1}$ or deriving a formula for the derivative of $f^{-1}$ you should definitely keep $y$ as name for the independent variable of $f^{-1}$. But when you start discussing $f^{-1}$, say $\log$, in its own right then it might be helpful to consider it as a function of a ${\it new}\ $ horizontally scaled variable $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.