Show if the function $\frac{x+2}{x^2+2x+1}$ is injective $$\frac{x+2}{x^2+2x+1}=\frac{y+2}{y^2+2y+1} $$
The domain of the function is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
If $y=x$ the function is injective:
$$\frac{x+2}{(x+1)^2}=\frac{y+2}{(y+1)^2}$$
$$(x+2)(y+1)^2=(x+1)^2(y+2)$$
Here it stops. I would like to know if this can be simplified to $y=x$, or if it can't be (because it's of course not injective), how would you show it in the same way in condition that all you know about injectivity is, that $f(x)=f(y)$ and $x\neq y$ then it isn't injective, or if $x=y$ then it is.
Bellow is the image of the function $\frac{x+2}{x^2+2x+1}$.
Let's say that now we leave that condition in bold print above behind and focus on my next question. What is the best way to show for such a function that it is injective but only if we look at the left side of the assymptote, so you can't use the right side to find a value to be the same as a value on the left, which is very easy in this case. And you don't know it's graph, so how do you know where it has two different points in domain that $f(x)=f(y)$ and $x\neq y$ then it isn't injective, or if $x=y$ then it is. How to even find it without visualising.
I hope there aren't too many questions.

 A: The natural domain of this function is $\Bbb{R} \setminus\{-1\}$. Assuming you want to check it's injectivity (one-one) over the natural domain. From the last step you have in your work:
\begin{align*}
(x+2)(y+1)^2 & =(x+1)^2(y+2)\\
(x+2)(y^2+2y+1)&=(x^2+2x+1)(y+2)\\
xy^2+2xy+x+2y^2+4y+2&=x^2y+2xy+y+2x^2+4x+2\\
xy(y-x)+3(y-x)+2(y^2-x^2)&=0\\
(y-x)\color{red}{[xy+3+2y+2x]}&=0.
\end{align*}
Thus we see that $x=y$ is NOT the only conclusion the can be drawn because we can also have $\color{red}{2x+2y+xy+3=0}$ on $\Bbb{R} \setminus \{-1\}$. For example, we can take $x=0$ and $y=-3/2$. Then $f(0)=2=f(-3/2)$. So $f$ is not one-one to on the natural domain. In fact, you can solve for $y$ and get $y=-\frac{3+2x}{x+2}$. Thus for every $x \neq -2$, you can find a corresponding $y$ such that $f(x)=f(y)$. This shows how to find two possibly distinct values of the input at which the function will have the same value.
Observe that if we were to focus on for example $(0, \infty)$. Then $2x+2y+xy+3 > 0$. In which case the last condition won't be satisfied and hence $x=y$ would be the only conclusion and we can say that $f$ is one-one on $(0,\infty)$.
A: You can "sketch" rational functions. In this case for $f(x)=\frac{x+2}{(x+1)^2}$ we have:
$\bullet f(x)\to 0^-$ as $x\to -\infty$
$\bullet f(-2)=0$, i.e. $-2$ is a root.
and
$\bullet f(x)\to \infty$ as $x\to -1$
So, if you try to sketch your continuous function just using the data I mentioned above, you will end up with a graph like the one you got. In particular your function is not injective.
