Domain of function $g$ question when $g(x) = f(x+3)$ Suppose $f$ is defined on $[0;4]$ and  $g(x) = f(x+3)$ what is the domain of $g$?

I'm having a tough time solving this problem. I know that $0 \leq x \leq 4$ for the domain of $f$. So would it be correct to say that for the function $g$, $0 \leq x +3\leq 4$ so:$$f(0) \leq f(x+3) \leq f(4)$$
$$ \Rightarrow f(0) \leq g(x) \leq f(4)$$

I'm not sure how to proceed. A little assistance would be greatly appreciated.
As you say, you require the argument of $f$ to be in $[0,4]$, so

$$x+3\in [0,4]$$
$$\iff 0\leq x+3\leq4$$
$$ \iff (0)-3\leq (x+3) - 3 \leq 4-3$$
$$ \iff -3 \leq x \leq 1$$
$$\iff x\in [-3,1]$$

So the domain of $g$ (the set of valid values of $x$) is $[-3,1]$.

Notice that my approach is "if I know where $x+3$ is, where must $x$ be?".
$$x\in\text{dom(g)}\iff x+3\in\text{dom(f)}\iff x+3\in[0,4]\iff x\in[-3,1].$$