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?
 A: Notice that
\begin{align*}
f(x) = \frac{x+2}{x^{2} + 2x + 1} = \frac{(x+1) + 1}{(x+1)^{2}} = \frac{1}{x+1} + \frac{1}{(x+1)^{2}}
\end{align*}
If we set $y = 1/(x+1)$ and consider $c\in\textbf{R}$, we arrive at the following equation
\begin{align*}
y + y^{2} = c \Longleftrightarrow y^{2} + y - c = 0\Longleftrightarrow y = \frac{-1\pm\sqrt{1+4c}}{2}
\end{align*}
which has real roots if and only if $c\geq-1/4$.
So the answer is $\text{Im}(f) = [-1/4,\infty)$.
A: 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 $$\left[-\frac 14,\infty\right)$$
A: Why? Just write the function as follows: 
$$f(x) = \frac{(x+1)+1}{(x+1)^2} = \frac{x+1}{(x+1)^2}+ \frac{1}{(x+1)^2}
$$
later, 
$$f(x)= \frac{1}{(x+1)}+\frac{1}{(x+1)^2}=u+u^2 = (u+0.5)^2-0.25 = \left(\frac{1}{x+1}+0.5\right)^2-0.25$$
Since here you should be able to continue, just see the variation of $x$ and transform to $f(x)$
A: $$f(x)=\frac{x+2}{(x+1)^2}$$
$$f'(x)=\frac{(x+1)-2(x+2)}{(x+1)^3}$$
$$=\frac{-x-3}{(x+1)^3}$$
so
$$f((-\infty,-1))=[f(-3),+\infty)$$
and
$$f((-1,+\infty))=(0,+\infty)$$
thus, the range is $$[-\frac 14,+\infty).$$
A: Alternate way to find the range : 
$$f(x) = y =\frac{x+2}{x^2+2x+1}$$
$$yx^2+(2y-1)x+(y-2)=0  $$
Now this quadratic has real roots (since real points exist belonging to the function for all values of x (we can remove the case of -1 later)) . So applying the condition for real roots : ($b^2-4ac \geq 0$)
$$(2y-1)^2-4(y)(y-2)\geq0$$
$$-4y+1 +8y\geq0$$
$$y\geq\frac{-1}{4}$$
$$y \in  \left[ -\frac{1}{4},\infty \right)$$
Now the value of $y$ at $x=-1$ $\to \infty$ so no need to remove it from the range
A: $$\text{You wrote: } x=\frac{y+2}{y^2+2y+1} $$
$$x(y^2+2y+1)=y+2$$
$$
xy^2 + ( 2x-1 )y +(x-2)=0 \tag 1
$$
$$
ay^2 + by + c = 0, \quad\text{where } a=x,\, b=(2x-1), \, c = x-2 \tag 2
$$
$$
y = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{1-2x\pm \sqrt{(2x-1)^2 -4x(x-2)}}{2x} \tag 3
$$
except that $(2)$ can be a valid solution of $(1)$ only when $a=x\ne 0.$ When $x=0,$ equation $(1)$ becomes
$$
y+2=0, \text{ so } y=-2.
$$
Thus the domain of the function defined on line $(3)$ excludes only those values of $x$ for which the expression under the radical is negative. So we want
$$
(2x-1)^2 - 4x(x-2)\ge0.
$$
$$
-4x + 1+8x\ge0
$$
$$
x \ge \frac {-1} 4.
$$
