Joining two graphs Suppose I have $f_1(x)=x$
And i restrict its domain as $\color{blue}{(-\infty,0]}$ using $g_1(x)=\dfrac{x}{\frac{1}{2\left(x-0\right)}\left(x-0-\left|x-0\right|\right)}$
Resulting in :

Now, suppose i have $f_2(x)=\sin{(x)}$
And i restrict its domain to $\color{blue}{(0,\infty)}$ using $g_2(x)=\dfrac{\sin x}{\frac{1}{2\left(0-x\right)}\left(0-x-\left|0-x\right|\right)}$
Resulting in :



Now I want a new equation which can join $g_1(x)$ and $g_2(x)$ in one single equation. How should i achieve it?

I mean $\large g_{\text{joined 1+2}}(x)$ would look like :

Also i want a single equation which is NOT of the form of $\color{red}{
    f(x)= 
\begin{cases}
    x& \text{if } x\leq 0\\
    \sin{(x)}& \text{if }  x> 0
\end{cases}}$

My attempt :
I tried using $(y-g_1(x))(y-g_2(x))=0$ but they don't seem to work...
They seem to be working with $y=x,x \in (-\infty,0]$ and $y=2x, x \in (-\infty,0]$ as :



So how to proceed?
Please help. Thanks!


I found shorter ways of generalizing this. I am as old as you are so I am glad to know someone has a similar kind of interest as I do. I notice that when $f(0)\neq{0}$ and $g(0)\neq{0}$ my method could be quicker.
Basically you can take two functions and "fuse" them by the floor function with exponents (as long as the function does not have a restricted domain ex.$\sqrt{x}$). I found that using this I can bring $f(x)\in(-\infty,a)$ and $g(x)\in[a,\infty)$ to a single equation. 
 $${{f(x)}^{\frac{-\text{sgn}{\lceil(x-a-1)+.5}\rceil+1}{2}}} {g(x)}^{\frac{{\text{sgn}\lceil(x-a-1)+.5}\rceil+1}{2}}$$
To switch which domain of the function $x=a$ is included in, replace the "ceil" function with the "floor" on your computer. Also you may need to use "sign" instead of "sgn" for the signum function.
Unfortunately this fusion works best when were fusing functions that have no undefined intervals. With $f(x)\in(a,b)$ and $g(x)\in(c,d)$, where $a<c$ and $b<c$ if there is a gap between $b$ and $c$ one must use manipulations to get...
$$\left({{\left(\left(f(x)\right)^{-1}+1\right)}^{\frac{\text{sgn}{\lceil(x-a-1)+.5}\rceil-\text{sgn}\lceil(x-b-1)+.5\rceil}{2}}}{{\left(\left(g(x)\right)^{-1}+1\right)}^{\frac{\text{sgn}{\lceil(x-c-1)}+.5\rceil-\text{sgn}\lceil(x-d-1)+.5\rceil}{2}}}-1\right)^{-1}$$  
However this equation has an undefined region making it less artificial.
Now you can easily use this approach for more than one function. Here are the "switches" of my method as the exponents of functions.
$c_n(x)$ includes $f(x)$ and $g(x)$.
$\left(\left(\prod_{n=0}^{k}{\left({\left({c_n}(x)\right)}^{-1}+1\right)}^{{d_n}(x)}\right)-1\right)^{-1}\quad$ Union of all defined intervals ("undefining" unwanted intervals). 
$d_0(x)=\frac{-\text{sgn}\lceil(x-a-1)+.5\rceil+1}{2}$ $(-\infty,a)$
$d_1(x)=\frac{\text{sgn}\lceil(x-a-1)+.5\rceil+1}{2}$ $(a,\infty)$
$d_2(x)=\frac{\text{sgn}\lceil(x-a-1)+.5\rceil-\text{sgn}\lceil(x-b-1)+.5\rceil}{2}$ $(a,b)$
