I want to find the inverse function of $f(x)=\ln(x)+x$, for $x>0$, but I heard it can't be done. Is there a way to find the inverse or to show it can't be found?
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Something incredibly important whenever we talk about functions that is sadly and frustratingly left out in many introductory classes... is that we must specify what the domain and codomain of the function are. The function from ${e}$ to ${e+1}$ given as $f(x)=\ln(x)+x$ is indeed invertible. The "function" from $\Bbb R$ to $\Bbb R$ given as $f(x)=\ln(x)+x$ does not exist since $f(x)$ is undefined for elements in its supposed domain such as $f(-1)$. – JMoravitz Jun 13 '18 at 17:36
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If you wish to talk about a specific domain and a specific codomain, we can then discuss whether it is invertible on that domain and codomain. The answer of whether or not it is invertible will depend heavily on those choices. Until you specify those, we can only guess as to what you mean. – JMoravitz Jun 13 '18 at 17:38
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So, if we were to set the domain as $\Bbb R^+$, we can learn that the range will be $\Bbb R$, and so by setting the codomain as $\Bbb R$ we can now actually begin the problem. The question of whether $f$ a function from $\Bbb R^+\to \Bbb R$ given as $f(x)=\ln(x)+x$ has an inverse, the answer is yes. One can reason using calculus that the function is strictly monotone increasing over the domain, is continuous, and is surjective and is hence bijective (i.e. an inverse exists). As for what the inverse looks like, it likely isn't a form you are familiar with. – JMoravitz Jun 13 '18 at 17:44
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Over that domain/codomain the inverse will be $f^{-1}(x)=W(e^x)$ where $W$ represents the Lambert W Function. – JMoravitz Jun 13 '18 at 17:45
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First of all, one shall start by finding the domain of the function and the domain of the inverse.
The function $f(x)$ is defined for such $x$, as : $$D_f = \{x\in \mathbb R : x>0\} = \mathbb R^+$$
Taking the first and second derivatives, we have :
$$f'(x) = \frac{1}{x} +1 >0, \; \; x>0$$
$$f''(x) = -\frac{1}{x^2} <0, \; \; x>0$$
Thus, the function $f$ is strictly increasing in its domain $D_f$ which also means that is $"1-1"$, which means that there exists an inverse function $f^{-1}(x)$ for $x>0$. Also take note that :
$$\lim_{x \to 0^+}f(x) = -\infty \; \; \text{and} \; \; \lim_{x \to \infty}f(x) = +\infty$$
Thus the range is $\mathbb R$ and the function $f$ is defined as $f: \mathbb R^+ \to \mathbb R$.
The inverse operation shall be carried out as finding an expression for $x$ involving the function family of values $y=f(x)$. Thus :
$$y=f(x) \Rightarrow y = \ln x + x $$
Now, take note that there is no possible way of solving this equation with respect to $x$. One cannot express $x$ simply and strictly in terms of $y$ and constants.
There are numerical methods and special function that handle such cases and for this particular example, the solution to the equation yielded is given as an expression of the product log function $\rm W$ :
$$y=\ln x + x \Leftrightarrow e^y = e^{\ln x + x} \Leftrightarrow e^y=xe^x \Leftrightarrow f^{-1}(x) = \rm W(e^x)$$
where $\rm W$ represents the Lambert $\rm W$ function. For more information about this special function, take a look here.
Finally, a graph showing both $f(x)$ and its inverse :
Rebellos
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