Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?

How about using calculus?
$$f(x)=\frac{x+2}{(x+1)^2}=\frac{1}{x+1}+\frac{1}{(x+1)^2} \\ f’(x) =\frac{-1}{(x+1)^2}-\frac{2}{(x+1)^3} =0 \\ \implies x=-3$$ which you can see clearly corresponds to a minimum. We see $f(-3) =-\frac 14$. Now, $\lim_{x\to\pm\infty}f(x)=0$ and also $\lim_{x\to -1} f(x) =+\infty$ which means that $-\frac 14$ is the global minimum. There is no upper bound and hence the range is $$\left[-\frac 14,\infty\right)$$
