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From what I understand invertible is if we can equate x = f(x) of some form.. Whereas Inverse is where the function can be reflected across the y=x axis. Are those related in any way?

Is it true that only $f(x)=\frac{1}{x}$ out of these below functions are invertible? If so, why?

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

nvs0000
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    A solution of $x=f(x)$ is a fixed-point of $f$. That has nothing to do with being invertible or not. – Martin R Aug 30 '21 at 11:41
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    Hello :) "Invertible" means, that for all $y$ in the range of $f$, we can find exactly one $x$ in the domain of $f$ such that $y=f(x)$. – Jochen Aug 30 '21 at 11:41
  • A sufficient (but not necessary condition) that a function is invertible within an interval $[a,b]$ is that it is differentiable everywhere in this interval and either $f'(x)>0$ holds for every $x$ in the interval or $f'(x)<0$ holds for every $x$ in the interval. Equality is allowed on isolated points. – Peter Aug 30 '21 at 11:46
  • For each of those functions, you should state the domain over which you are defining them. And a function is invertible if and only if it is one-to-one and onto, i.e. the function is a bijection. This is not necessarily a definition of invertible, but it a useful and quick way of deciding if a function is invertible. See: https://en.wikipedia.org/wiki/Inverse_function#Definitions – Adam Rubinson Aug 30 '21 at 11:46
  • In all but the $f(x)=1/x$ choice, the value of $f(x)$ is defined, and is the same, at $x=1$ as it is at $x=-1.$ So those functions are not invertible. – coffeemath Aug 30 '21 at 11:47
  • It seems that your understanding of invertibility is more "graphical" than by definition. Can you plot these graphs accurately? The criteria you've written for functional invertibility is almost correct (the reflection across the line $y=x$ should look like a function i.e. for each $x$ there should be only one $y$ such that $(x,y)$ lies on the graph) and I don't understand the $x=f(x)$ part, it'll need more clarity. Rather than focus on the definitions, you should focus on getting your graph, and the reflection correct. Same for the answers. – Sarvesh Ravichandran Iyer Aug 30 '21 at 11:57
  • sorry so only 1/x is invertible... correct?.. – nvs0000 Aug 30 '21 at 12:07

2 Answers2

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A function is invertible if and only if it is one-to-one. A one-to-one function is a function where no two inputs produce the same output, i.e. for all $a$ and $b$ in the domain of $f$, $$ f(a)=f(b)\implies a=b \, , $$ or, equivalently, $$ a\neq b\implies f(a)\neq f(b) \, . $$

As Martin R mentions in the comments, if $f(x)=x$ for some $x$, then $f$ is said to have a "fixed point" at $x$. This has nothing to do with whether $f$ is invertible.

So for each of your functions, you have to consider whether $f(a)=f(b)$ implies $a=b$. For instance, does $a^2=b^2$ imply that $a=b$?

Joe
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    Doesn't a function have to also be surjective in order for it to be invertible? You don't mention this, but wikipedia does say this: https://en.wikipedia.org/wiki/Inverse_function#Definitions – Adam Rubinson Aug 30 '21 at 11:59
  • @AdamRubinson: There's not a straightforward answer to that question. It depends on the definition of "function" (see here). The simplest definition (and the one I consider to be the best) is that a function $f$ is a set of ordered pairs such that if $(a,b)\in f$ and $(a,c)\in f$, then $b=c$. According to this definition, the functions $f:\Bbb R\to\Bbb R,,f(x)=\sin x$ and $g:\Bbb R\to[-1,1] ,,g(x)=\sin x$ are identical. Therefore, to call $f$ non-surjective and $g$ surjective is non-sensical. – Joe Aug 30 '21 at 12:21
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    A function can only be surjective onto a particular set. So $f$ and $g$ are both surjective onto $[-1,1]$, and not surjective onto $\Bbb R$. Note also that when we write $f:A\to B$, the set $B$ is just intended to be any superset of the range of $f$, and changing what $B$ is doesn't change the definition of $f$. However, if you define functions in a different way, then the answer is also very different. – Joe Aug 30 '21 at 12:21
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$f$ is invertible (or injective) if whenever $f(x)=f(y)$, we have $x=y$. Or equivalently for each $z$ in the range of $f$ there is only one $x$ such that $f(x)=z$. For $1,2$ and $4$ you should try to find $f(x)=f(y)$ where $x\neq y$.

Hint: For $1$ we have $1^2=(-1)^2$.

spinosarus123
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