Fixed points of $ e^{x+1/x} = x$ in $\mathbb{R}$

Question

This particular question was one of the options in a question in amock test i am solving of masters in mathematics.

Does $f(x) = e^{x+1/x}$ has unique fixed point in $\mathbb{R}$?

Attempt: the equation can be simplified to $x+\frac{1}{x} = \log x$ and then $ x^2 +1 = x \log x $.

But I don't know the graph of $x \log x$ and hence can't solve the problem and don't know any other idea.

So, can you tell a elegant method to solve this question.

Thank you!!

Answer 1

For all $x\ne0$, $f(x)>0$ and there can be no fixed-point in the negatives. Then for $x>0$, $f(x)\ge 1+x+\dfrac1x>x$ and there can be no fixed point in the negatives.

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Note that if $x$ can take values in $\mathbb C$,

$$\overline{e^{x+1/x}}=e^{\bar x+1/\bar x}$$and the conjugate of a fixed point is also a fixed point !

Answer 2

Hint :

as stated in the comments by @kccu $e^{x+\frac{1}{x}}$ has no fixed points (in $\mathbb{R}$, so maybe it's in $\mathbb{C}$ as stated in the answer of @Yves Daoust :P) so I assume it is $e^{\frac{x+1}{x}}$ .

Surely you have to search for it when $x>0$ , if you study the derivative of $$f(x) = e^{\frac{x+1}{x}}-x$$

what can you say about the number of fixed points? Now you need just to find a point where the function is positive and a point where the function is negative so that you can apply Bolzano's theorem to conclude.