My question is based on some questions about piecewise functions. The problem is when I "sum" two piecewise functions that don't share any interval. When I draw the graphic of this function in geogebra, for example, it does not exist. Because $ f(x) $ is a real number, then I thought that the problem was that adding some $ x $ real number with a number that does not exists, the sum is indeterminate. I just want to confirm that.
We got a function $f(x)=1$ for $0\leq x\leq1$ and $f(x)=2$ for $1<x\leq2$. We got also a function $g(x)=f(2x)$ We got also a function $h(x)=f(x-2)$ And finally we got $k(x)=g(x)=h(x)$, which graphic does not exists.
Formally, you can sum two functions only when they share the same domain and codomain. Since your $f:[0,2]\to\mathbb{R}$ is only defined for $0\leq x\leq 2$, it is easy to see that, in the best scenario (that is, the greatest domain possible), $\text{Dom}(g)=[0,1]$ and $\text{Dom}(h)=[2,4]$. So, what happens with $k(x)=g(x)+h(x)$? Well, $g$ and $h$ have different domains so it is undefined and, therefore, it won't have any graph. However, not only that... If we redefined the sum of two real functions $f_1:D_1\to\mathbb{R}$ and $f_2:D_2\to\mathbb{R}$ as $(f_1+f_2):D_1\cap D_2\to\mathbb{R}; (f_1+f_2)(x)=f_1(x)+f_2(x)$, $k(x)=g(x)+h(x)$ would be well defined but, since now $\text{Dom}(k)=\text{Dom}(f+g)=\text{Dom}(f)\cap\text{Dom}(g)=\emptyset$, it will simply be the empty function and, thus, it will have empty graph.