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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?

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    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 $y=\ln x$?2011-07-07
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    The inverse is just another function, who argument is denoted by $x$. I don't find that counter-intuitive.2011-07-07
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    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.2011-07-07
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    I don't understand the downvote to the question. It is a genuine question, well written and future students might have the same question.2011-07-07
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    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?"2011-07-07
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    The independent variable of a single variable real function may be any letter, $x,t,u,y$, etc.2011-07-07
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    Mathematics is invariant under permutations of the alphabet2011-07-08
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    If you write $y$ as a function of $x$, in your case, $y = \sqrt{2x+1}$ for $x > \frac{1}{2}$, then it makes perfect sense to write $x = \frac{y^2 -1}{2}.$ But if you want to write the inverse of a function, then the emphasis changes to the function and its inverse function, so the variables used become of minor significance.2011-07-08
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    @tomcuchta That seems like a plausible explanation.2011-07-08

2 Answers 2

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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.

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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 $x$.