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?

$$f(x)=\frac{x+2}{(x+1)^2}$$
$$f'(x)=\frac{(x+1)-2(x+2)}{(x+1)^3}$$
$$=\frac{-x-3}{(x+1)^3}$$
so
$$f((-\infty,-1))=[f(-3),+\infty)$$
and
$$f((-1,+\infty))=(0,+\infty)$$
thus, the range is $$[-\frac 14,+\infty).$$
