Find the domain and range of $f(x)=\frac{2}{x+1}$ I am to find the domain and range of the function $f(x)=\frac{2}{x+1}$.
I can find the domain by ensuring the denominator is not 0:
$$(-\infty,-1)\cup(-1,\infty),$$  because if $-1$ is an input, the denominator is $0$.
I know to exclude $-1$ because I don't want my denominator to ever $= 0$:
$$x+1=0,$$
i. e., $x=-1$, so exclude $-1$ in the domain.
So far so good and my textbook answers section confirms this is correct for the domain, all real numbers except $-1$.
Where I'm confused is for the range. The solution provided is
$(-\infty,0)\cup(0,\infty)$.
So all values except $0$. How was this arrived at?
 A: Finding the range is usually considerably more difficult than finding the domain of a function. In this case, there is a naive method that works.
The range is all possible values of $y$ in the equation $y = \frac{2}{x + 1}$. If, for a given $y$, we can find a valid value for $x$, then $y$ is in the range. If $y \neq 0$, then
\begin{align*}
y = \frac{2}{x + 1} &\iff y(x + 1) = 2 \\
&\iff x + 1 = \frac{2}{y} &\text{remember: }y \neq 0 \\
&\iff x = \frac{2}{y} - 1.
\end{align*}
So, for $y \neq 0$, there is a value of $x$ so that the equation is satisfied. If $y = 0$, then we get
$$0 = \frac{2}{x + 1} \implies 0 = 0(x + 1) = 2,$$
which is absurd. So, no such $x$ exists, and $0$ is not part of the range. Therefore, the range is indeed $(-\infty, 0) \cup (0, \infty)$.
Note that this relies on the equation being solvable for $x$. For many problems, this is not feasible to do! Lucky that the problem is relatively simple.
A: These problems are often nice to do graphically. Start with the graph of $y=\frac1x$. You should know off the top of your head that this has range and domain $(-\infty,0)\cup(0,+\infty)$. Then the graph $y=\frac1{x+1}$ is just $y=\frac1x$ translated $1$ unit to the left, so it has the same range as before but now the discontinuity is at $x=-1$, not $x=0$.
Then the graph of $y=\frac2{x+1}$ is the graph of $y=\frac1{x+1}$ scaled vertically away from the $x$-axis by a factor of $2$. This doesn't affect anything horizontal, so these two graphs have the same domain. And any value of $\frac2{x+1}$ is the double of a value of $\frac1{x+1}$, so they also have the same range.
Putting it all together, the domain has changed its discontinuity from $x=0$ to $x=-1$ and the range is unchanged.
A: You are correct in that there is a vertical asymptote at $x=-1$. So, the domain of the function is all real numbers except $x=-1$.
For the range, you need to evaluate where $y=f(x)$ is undefined. This can be done by substituting $y$ with $x$
$$y = \frac{2}{x+1} ~\Rightarrow~ x = \frac{2}{y+1} ~\Rightarrow~ x(y+1)=2 ~\Rightarrow~ y+1=\frac{2}{x}~\Rightarrow~ y= \frac{2}{x}-1$$
and observing that 
$$y= \frac{2}{x}-1$$
is undefined if $x=0$. Therefore, the range of the function is all real numbers except $y=0$.
A: Let $$y = \frac2{x+1}\implies x+1 = \frac{2}{y}\implies x = \frac 2y-1$$
$$\text{As you've found that }-\infty<x<-1 \text{ or } -1<x<\infty$$
$$ \iff-\infty<\frac 2y-1<-1 \text{ or } -1<\frac 2y-1<\infty $$
$$\iff-\infty<\frac 2y<0 \text{ or } 0<\frac 2y<\infty$$
$$\iff -\infty <y<0 \text{ or } 0<y<\infty$$

$$R_f \in(-\infty,0) \ \cup \ (0,\infty)$$

A: To find the range of the given function, a more conventional method is already mentioned, however, you can also find the range of this particular function using limits.
For instance, $$\lim\limits_{x\to \infty} \frac{2}{x+1} = 0$$, 
So the range of $f(x)$ would be  $\mathbb{R}\setminus \{0\}$
A: Now, $f(x) = \frac{2}{x+1}$ is a rational function, and so is defined everywhere except for any $x$ that would cause division by $0$; in this case, $D_{f} = \mathbb{R} - \lbrace -1 \rbrace = (-\infty, -1) \cup (-1, \infty)$.
As for the range, it will contain every real number except $0$, as there is no $x$ for which $\frac{2}{x+1} = 0$. Hence, $R_{f} = (-\infty, 0) \cup (0, \infty)$. 
(This could be inferred by observing that the graph of $f(x)$ would be the graph of $y = \frac{2}{x}$ shifted one unit to the left. So, the range of $f(x)$ is the same as the range of $y$.)  
