We have the function \begin{equation*}f(x)=-\frac{(x-3)^2}{x+1}\end{equation*}
I want to determine the domain and the range of the function.
The root of the deniminator is $x=-1$. Therefore, the domain is $D_f=\mathbb{R}\setminus\{-1\}=(-\infty, -1)\cup (-1,+\infty )$.
Does it holds that \begin{equation*}W_f=f(D_f)=f\left ((-\infty, -1)\cup (-1,+\infty )\right )=f\left ((-\infty , -5)\cup (-5,-1) \cup (-1,3) \cup (3,+\infty)\right )\end{equation*} ?
If this is correct, then we have to know the monotonicity of $f$ at each of these intervalls.
We have that at $(-\infty , -5)$ the function is decreasing, at $(-5,-1)$ the function is incresing, at $(-1,3)$ the function is increasing and at (3,+\infty)$ the function is decreasing.
We have the following:
\begin{align*}f\left ((-\infty , -5)\right )= \left (16, +\infty\right ) \end{align*}
\begin{align*}f\left ((-5,-1)\right )= \left (16, +\infty\right )\end{align*}
\begin{align*}f\left ((-1,3)\right )= \left (-\infty, 0\right )\end{align*}
\begin{align*}f\left ((3,+\infty)\right ) = \left (-\infty, 0\right )\end{align*}
Therefore the range is \begin{equation*}W_f= \left (-\infty, 0\right )\cup \left (16, +\infty\right ) \end{equation*}
$$$$
Is everything correct? Or do we have to determine the range in an other way?