What is the inverse of $f(x)=x+\arccos(\frac{x}{1-x^2})-\frac{1}{x^2}e^\frac{1}{2x^2}$

Question

I received the task to find the natural domain, injectivity, surjectivity and the inverse of the real function $f(x)=x+\arccos(\frac{x}{1-x^2})-\frac{1}{x^2}e^\frac{1}{2x^2}$ as part of my homework. I have been working on this for hours but I only know how to find the domain:

$D(f)=(-\infty,\frac{-1-\sqrt{5}}{2}]\cup[\frac{1-\sqrt{5}}{2},0)\cup(0,\frac{-1+\sqrt{5}}{2}]\cup[\frac{1+\sqrt{5}}{2},+\infty)$

I know the definitions for injectivity, surjectivity and the inverse, but I don't know how to apply them to this function. How do I do this? (without using calculus)

Answer 1

Without using calculus I can just give you a graph.

You can also see that in the intervals where $f(x)$ is not injective the inverse is not a function.

Hope this can be useful, but with no calculus little can be done...

Edit

The previous graph was wrong. I had to replot a part by hand because values drop so rapidly to $-\infty$ that even Mathematica made it wrong!

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