How to restrict a function $f(x)=\sqrt{\ln{\frac{1}{(x+1)^2}}}$ such that it becomes injective?

Question

We can easily determine the domain of the function : $D(f)\in[-2, 0]\backslash\{-1\}$. We can try to find inverse but will end up with this function: $f^{-1}(x)=-1\pm\sqrt{\frac{1}{e^{x^{2}}}}$. We can see that this function doesn't have unique inverse since it isn't injective.
My question is how can we, from this information, determine how to restrict this function such that it becomes injective i.e. has an unique inverse?

Answer 1

hint

For $ x\ne -1,$

$$f^2(x)=-\ln((x+1)^2)$$ $x$ is in the domain if $$0<(x+1)^2\le 1$$ or

$$0<x^2+2x+1\le 1$$

thus the domain is $$D(f)=[-2,-1)\cup (-1,0]=I\cup J$$

check that $f'(x)$ has the same sign than $$\frac{-2}{x+1}$$

hence, $f$ is continuous and strictly increasing at $[-2,-1)$. it is a bijection from $ I$ to $ [0,+\infty)$. its inverse is such that $$y^2=-\ln((x+1)^2)=-2\ln(-x-1)$$ and $$x=f^{-1}(y)=-1-e^{\frac{-y^2}{2}}$$