Examine the continuity of function $f(x)=\frac{2x^2-4x}{|x+1|+|x-3|-2}$ Using the definition of absolute value for
$$|x+1|=\begin{cases}
x+1, & x\ge -1\\
-x-1, & x>-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x>3
\end{cases}$$
and checking cases when $$x<0,x=0,x>0$$
I get this definition of a function $f(x)$:
$$f(x)=\begin{cases}
3, & x=3,x=-1\\
x(x-2), & x>0\\
0, & x=0\\
2-x, & x<0
\end{cases}$$
Is this definition of a function $f(x)$ correct, and if not what is wrong?

$$|x+1|=\begin{cases}
x+1& x\ge -1\\
-x-1, & x\color{red}<-1
\end{cases}$$
and
$$|x-3|=\begin{cases}
x-3, & x\ge 3\\
-x+3, & x\color{red}<3
\end{cases}$$
