# 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.

• Yes you look totally correct. – SchrodingersCat Jan 11 '17 at 14:31
• Hem, no you look totally wrong. The inequations on $f$ have nothing to do with the domain, but with the codomain, which is irrelevant here. – Yves Daoust Jan 11 '17 at 14:34
• @SchrodingersCat : What?? – MPW Jan 11 '17 at 14:35
• @MPW the inequality $0 \leq x +3\leq 4$ is true, that's what I meant.. nothing else though.. – SchrodingersCat Jan 11 '17 at 14:37

## 3 Answers

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?".

• Thanks, but what is confusing me is why do we say that $0 \leq x + 3 \leq 4$ ? – Aspiring Mathlete Jan 11 '17 at 14:37
• @AspiringMathlete : Because whenever you see $f(\boxed{\;})$ you always must have that $\boxed{\;}$ is in the domain of $f$, i.e., $\boxed{\;} \in [0,4]$, which is another way to say $0\leq \boxed{\;}\leq 4$ – MPW Jan 11 '17 at 14:39

It is $g:[-3,1] \to\mathbb K$ because $f(0)=f(-3+3)=g(-3)$ and $f(4)=f(1+3)=g(1)$

$$x\in\text{dom(g)}\iff x+3\in\text{dom(f)}\iff x+3\in[0,4]\iff x\in[-3,1].$$

• Silent (and mislead) downvotes not welcome. – Yves Daoust Jan 11 '17 at 16:47