1

I'd like to show that below function holds the inverse mapping:

$f: \;P(\Bbb R) \rightarrow \Bbb R $ where $f(x) = {1\over x^2}-x\cdot arctanx+{1 \over 2}log(1+x^2) $

To show the existence of inverse mapping, I want to use the property that every inverse mapping has a bijection between domain and range of its orignial function.

But each term of $f(x)$ holds different domain and range which makes this process cubersome.

Any brief approach to check the inversibility of the given function?

Daschin
  • 675

1 Answers1

1

Hints (Every monotone function has a inverse mapping):

Since $$f'(x)=-\frac{1}{x^3}-\arctan x,$$ we can conclude that $f'(x) > 0$ whenever $x<0$; $f'(x) < 0$ whenever $x>0$, which leads that $f(x)$ has a inverse mapping.

Red shoes
  • 6,948
Paul
  • 20,553
  • thx. it's not a hint but contains every reasoning what I need. Any reference that denotes well-proved statement of "Every monotone function has a inverse mapping"? – Daschin Jun 10 '17 at 08:25
  • @Daschin You are welcome! Every monotone function has a inverse mapping, because every monotone function is one to one. – Paul Jun 10 '17 at 08:40
  • note that function is Even, which means $f(x) = f (-x)$ for all $x \in R$ so $f: R \to R$ is not one-to-one! it is monotone on $(0 , + \infty)$ or $( - \infty , 0)$ but not both.. @Daschin – Red shoes Jun 10 '17 at 08:55