Examine the continuity of function $f(x)=\frac{2x^2-4x}{|x+1|+|x-3|-2}$

Question

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?

Answer 1

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