My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.
Page 56. Exercise 6. Let $f$ be defined as follows: $f(x)=1$ for $0\le x\le1$; $f(x)=2$ for $1\lt x\le2$. The function is not defined if $x\lt0$ or if $x>2$.
$a)$ Draw the graph of $f$.
$b)$ Let $g(x)=f(2x).$ Describe the domain of $g$ and draw its graph.
$c)$ Let $h(x)=f(x-2).$ Describe the domain of $h$ and draw its graph.
$d)$ Let $k(x)=f(2x)+f(x-2).$ Describe the domain of k and draw its graph.
The attempt at a solution: I drew graph of $f$, it would look like this (Except function is not defined when $x < 0$ or $x > 2$):
As for $b)$ and $c)$, since $g(x)=f(2x)$, on interval $0\le x\le1$, $g(x)=2$ as I understand, and on interval $1\lt x\le2$, $g(x)=4$, but since function $f$ is defined from $x=0$ to $x=2$, wouldn't same be true for $g(x)$? Wouldn't same be true for $c)$? What am I missing? because answer says that domain of function of $g(x)$ and $h(x)$ is different from domain of $f$, and domain of $k(x)=f(2x)+f(x-2)$ is empty.