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Function notation says that any operations applied to a variable inside the parenthesis are applied to the variable before it enters the function, and anything applied to the function as a whole is applied to the entire function or to the result of the function. So you could say that...

If $f(x)=x^2$ Then $f(3x-2)=(3x-2)^2$, and $3f(x)-3=3x^{2}-3$

I understand that $f(x)^{-1}$ is used to denote the inverse of a function so...

$f(x)^{-1}=\pm\sqrt{x}$

But putting something to the -1 power gives you its reciprocal like...

$x^{-1}=\frac{x^{-1}}{1}=\frac{1}{x^1}= \frac{1}{can}$

So would it not make sense that...

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

My question is why $f(x)^{-1}$ is considered the inverse of $f(x)$? Was this notation chosen randomly, or is there some logical explanation for why it is that way.

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    $\begingroup$ I've never seen that notation for the inverse of $f$, but rather $f^{-1}$ [or, evaluated at $x$, $f^{-1}(x)$]. $\endgroup$ – Clarinetist Oct 2 '14 at 2:46
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    $\begingroup$ I usually see $f^{-1}(x)$ for the inverse function (assuming it exists). If I see $(f(x))^{-1}$, then to me that is the reciprocal of $f(x)$. I haven't seen $f(x)^{-1}$ before. $\endgroup$ – layman Oct 2 '14 at 2:47
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    $\begingroup$ That makes sense I totally missed the placement of the -1, thanks. Do you know why the superscript -1 is used though? $\endgroup$ – o.uinn Oct 2 '14 at 2:50
  • $\begingroup$ Yes, that's not the usual notation. $\endgroup$ – Thomas Andrews Oct 2 '14 at 2:56
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The notation $f^n$ was adopted in set theory to denote iterated functions: functions composed with themselves numerous times. $f^n(x) = f\circ f^{n-1}(x) = f(f^{n-1}(x))$

It is from this we get the definition of the inverse of a function; being that function which when composed with the function $f$ produces the identity function. That is: $f\circ f^{-1}(x) = f^0(x) = x$


Meanwhile the notation $f^n$ was adopted independently in trigonometric theory to denote exponentiated functions: when function are multiplied with themselves numerous times. Here we obtain the more familiar use, $f^n(x) = f\cdot f^{n-1}(x) = f(x)\cdot f^{n-1}(x)$

While this meaning of the notation is more commonly encountered, both uses remain in practice, depending entirely on context to identify the meaning. Although sometimes mathematicians do adopt the notation $f^{\circ n}$ to clarify when they mean iteration rather than exponentiation.


Confusingly even when exponentiation is otherwise used it has become standard practice for the $f^{-1}$ notation to always mean the iterated inverse of functions.

It's a historic anomaly.

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  • $\begingroup$ "Historic anomaly"--that pretty much sums it up. Unfortunately, I think this anomaly is what most people end up experiencing, not the more consistent notations of higher-level math. $\endgroup$ – David K Oct 2 '14 at 4:31
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As others have pointed out, the notation for the inverse function is $f^{-1}(x)$. $f(x)^{-1}$ probably would denote the reciprocal of $f(x)$.

The operation $\circ$ of composition of functions is in some ways analogous to multiplication.

Let $\mathrm{id}$ be the identity mapping, $\mathrm{id}(x) = x$. You have for example $f \circ f^{-1} = \mathrm{id}$, which corresponds to $xx^{-1} = 1$ for numbers.

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That notation, in my opinion, is not normal.

We always let $f^{-1}(x)$ or $f^{-1}$ denote the inverse of a function.

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See, Inverse of $a$ i.e. $a^{-1}$ (where $a$ is any element of some group) means that $a*a^{-1}=e$ where $e$ is the identity w.r.t. binary operation $*$. I guess you don't understand Group theory and all i.e. abstract algebra. But only thing to notice here is that in Functions binary operation is not multiplication but composition denoted by $\circ$ and identity is not 1, rather this 1 denotes identity function i.e. $e$ s.t. $e(x)=x$ for all $x$. So inverse of $f$ here means $f^{-1}$ s.t. $f\circ f^{-1}=e$. So if $f(x)=x^2$ and what will you compose $x^2$ with to get $e$? Replace $f(x)^{-1}=\pm\sqrt{x}$ insted of $x$ in $f(x)=x^2$, you will see, $f\circ f^{-1}(x)$=($\pm\sqrt{x})^2=x$ i.e. $f\circ f^{-1}=e$ i.e. identity function.

Just make sure you understand composition of functions, that is the binary operation here, not multiplication. $2^{-1}=1/2$ because operation is Multiplication in $\mathbb{R}$. To your wonder $2^{-1}=-2$ in $\mathbb{Z}$ as operation there is addition and identity is $0$ and it can also be $2$ itself if we consider $2$ as an element of another group say {$1,2$} and put operation as multiplication modulo 3 (means remainder of $2\times 2$ when divided by $3$, isn't it $1$? so isn't $2^{-1}=2$? but don't worry about that now.

I hope I didn't confused you. Just worry about composition of function for now, and understand that

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  • $\begingroup$ Thanks for the extensive answer I am slightly confused, but also intrigued. I will probably look more into these group theory, and abstract algebra. :) $\endgroup$ – o.uinn Oct 2 '14 at 3:12
  • $\begingroup$ wait till you pass high school, then you will be introduced, in 1st year of college. $\endgroup$ – Bhaskar Vashishth Oct 2 '14 at 3:29

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