Two function issues (Inverse and behavior) I encountered two functions lately while studying calculus and I either can't find my mistakes or I'm not fully understanding the definitions. I could use some help around here.


*

*Does $\:f\left(x\right)=\frac{e^{-x}}{x^2+1}$ have an inverse function? The function is monotone decreasing (Thus injective) and surjective but it appears as if it does not have any inverse function.

*$f\left(x\right)=\left(x-2\right)\sqrt{\frac{1+x}{1-x}}$ is a product of two monotone increasing functions. However, f(x) is monotone decreasing. How is that possible? 
 A: The first function
$$
f(x)=\frac{e^{-x}}{x^2+1}
$$
satisfies
$$
\lim_{x\to-\infty}f(x)=\infty,\qquad
\lim_{x\to\infty}f(x)=0
$$
Moreover
$$
f'(x)=\frac{-e^{-x}(x^2+1)-2xe^{-x}}{(x^2+1)^2}=
-\frac{e^{-x}(x+1)^2}{(x^2+1)^2}
$$
that vanishes only at $-1$ and is otherwise negative. So $f$ is monotone decreasing and hence has an inverse function defined over the interval $(0,\infty)$.
The fact that this inverse function cannot be expressed in terms of “elementary” functions is of no concern.

The second function
$$
f(x)=(x-2)\sqrt{\frac{1+x}{1-x}}
$$
is defined over $[-1,1)$. We can investigate on it being monotone by considering
$$
g(x)=\frac{1+x}{1-x}=\frac{2}{1-x}-1
$$
again over $[-1,1)$. Since $x\mapsto 1-x$ is decreasing, we have that $x\mapsto 2/(1-x)$ is increasing and subtracting $-1$ doesn't change this.
Is the function $f$ increasing, being the product of two increasing functions? Not necessarily: $x\mapsto x$ is increasing, but $x\mapsto x^2$ isn't.
However, $x\mapsto x-2$ only takes on negative values on $[-1,1)$. If $-1\le x<y<1$, we have
$$
x-2<y-2,\qquad \sqrt{\frac{1+x}{1-x}}<\sqrt{\frac{1+y}{1-y}}
$$
but we can't multiply those inequalities, because the first one is between negative numbers. For instance, it is true that
$$
-2<-1,\qquad 1/2<4
$$
but it's false that $-1<-4$.
You can rewrite
$$
f(x)=-\sqrt{(x-2)^2\frac{1+x}{1-x}}
$$
and so look at
$$
g(x)=\frac{(x-2)^2(1+x)}{1-x}=\frac{x^3-3x^2+4}{1-x}
$$
and check this function is increasing. This will prove $f$ is decreasing.
