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I am trying to prove that this function is bijective, but I don't know how to do it.

$$f:x \mapsto \frac{e^x-1}{x}$$

For that, I try to use the fact that this function has necessarily a reciprocal if it is bijective.
So, I think I need to prove that :
$$f(x)=y \Leftrightarrow x = f^{-1}(y)$$
If I understand correctly, this means that :
$$\frac{e^x-1}{x} = y \Leftrightarrow x = \frac{y}{e^y-1}$$

But my problem is that I don't know how to do this, and I don't know if there is a more efficient way to solve the problem.
Can anyone help me solve this problem?

Gary
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Lucas
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    You won't find a reciprocal in closed form. Instead, show it is continuous, injective and unbounded. – Thomas Andrews Dec 16 '21 at 23:52
  • Ok, so to prove that the function is continuous, it is not very hard, I think. Then, to prove that it is injective, I suppose I need to introduce a new function g(x) = e^x and compute the derivative of g(x), right ? But finally, how can I prove that this is unbounded ? I mean, $0 \leq f(x)$, no ? – Lucas Dec 16 '21 at 23:52
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    And no, it is not true that $y=\frac{e^x-1}x$ iff $x=\frac{y}{e^y-1}.$ – Thomas Andrews Dec 16 '21 at 23:54
  • By the way, the function is undefined at $x=0,$ so you need to define it there caerfully to make the function continuous. – Thomas Andrews Dec 16 '21 at 23:55
  • Yes, my bad ^^" ! But how can I prove that the function is unbounded ? – Lucas Dec 16 '21 at 23:55
  • I don't know, is it unbounded? Does the question say what it is a bijection with? – Thomas Andrews Dec 16 '21 at 23:56
  • I need to prove that $f$ is a bijection from $\mathbb{R}$ to $\mathbb{R}^{+*}$, and that $f(0) = 1$ – Lucas Dec 16 '21 at 23:57
  • Then you don't need unbounded below, just that as $x\to -\infty,$ $f(x)\to0.$ – Thomas Andrews Dec 16 '21 at 23:59
  • So, if I prove that $x \rightarrow - \inf, f(x) \rightarrow 0$, and $x \rightarrow + \inf, f(x) \rightarrow + \inf$, and that $f(x)$ is continuous and that $f$ is injective, am I good ? – Lucas Dec 17 '21 at 00:00
  • The limit of $f(x)$ as $x$ gets to $0$ is just the definition of the derivative of $e^x$ at $x=0$. – Geoffrey Trang Dec 17 '21 at 00:18
  • @markvs It seems that the task is to prove that the function is a bijection between $\mathbb R$ and $\mathbb R^+$. Of course, in its current form this is not true either because $1$ is not in the range of $f$. OP, please write precisely in the question how you define $f(x)$ for $x=0$ and between what sets you have to prove bijectivity. – Gary Dec 17 '21 at 00:26
  • Your $\frac{e^x-1}{x} = y \Leftrightarrow x = \frac{y}{e^y-1}$ is wrong.Your function $f$ as bijection has a reciprocal function whose graphic is symmetric to $f$ respect to the diagonal $y=x$. You can have many points of $f^{-1}$. For example you have points for all natural $n$ $$\left(\ln(n),\frac{n-1}{\ln(n)}\right)\in f\iff \left(\frac{n-1}{\ln(n)},\ln(n)\right)\in f^{-1}$$ You can verify this way the error of your equivalence. – Piquito Dec 17 '21 at 00:35

3 Answers3

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$\newcommand{\R}{\mathbb{R}}$ Let us add some more structure. Consider the function $f: \R\setminus\{0\} \to \R_{+}\setminus \{1\}$ defined by $$ f(x) = \frac{e^x -1}{x} $$ and define its extension $g:\R \to \R_{+}$ where $g(0) = 1$. We can easily show that $g$ is continuous at $x = 0$ by L'Hopital's rule.

Now consider the derivative of $g$,

$$ g'(x) = \frac{e^x(x-1) +1}{x^2} $$

We have $$ \lim_{x \to 0} g'(x) = \frac{1}{2} $$

Hence from Hermis14, $g'(0) = 1/2$.

Furthurmore, $g'(x)$ is always positive and $g$ is surjective, which means $g$ is bijective.

Therefore, $f$ is bijective.

Hermis14
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It's useful to notice that $$\frac{e^x-1}{x} = \int_0^1 e^{x t} d t$$ so the function $f$ and all its derivatives are $>0$, with $\lim_{x\to -\infty} \frac{e^x-1}{x} = 0$, $\lim_{x\to \infty} \frac{e^x-1}{x} = \infty$.

Obs: The reciprocal $g(x) = \frac{1}{f(x)} = \frac{x}{e^x-1}$ is the generating function of the Bernoulli numbers

orangeskid
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Set $f(x)=\frac{e^x-1}{x}.$ Then, the claim follows immediately from the following easily verified facts:

$f$ is continuous on $\mathbb R\setminus \{0\},\ \underset {x\to \infty}\lim f(x)=\infty,\ \underset {x\to -\infty}\lim f(x)=0\ $ and $f'(x)=\dfrac{\left(x-1\right)e^x+1}{x^2}>0$ on $\mathbb R\setminus \{0\}.$

Matematleta
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