Transforming a function by taking the absolute value of the input Say we have a function $f(x)$ defined on the closed interval $x \in [-3, 4]$. Now, perform the transformation $g(x)=f(|x|)$. 
My question is: if we graph $g(x)$ over all defined outputs of $x$, would we include the interval $[-4, -3)$ on this graph? We do know the values $g(x)$ would take over this interval independent of the values of $f(x)$ from $[-4, -3
)$; but my line of reasoning is that we cannot "plug" these inputs from $[-4, -3)$ into $g(x)$ to obtain these outputs (since $f(x)$ only exists on the interval $[-3, 4]$), so $g(x)$ should only be graphed on the interval $[-3, 4]$. However, I graphed it over this interval on my math exam and it was marked as a mistake. Is my teacher correct? Thanks for the help.
 A: When asked for the natural domain of $f \circ g$, this is usually understood to be $g^{-1}(\mathrm{domain}(f))$. In your case:
\begin{align}
f &: \mathbb [-3,-4] \rightarrow \mathbb R \\
\\
g &: \mathbb R \rightarrow \mathbb R\\
  &: x \mapsto |x|
\end{align}
Then $g^{-1}([-3,4]) = [-4,4]$ as your teacher says. 
However, if your teacher was nonspecific about the domain of $g$, then they have no right to say you were wrong, as you were left to guess its domain, and you are correct provided $g$ was defined with the same domain as $f$:
$$g : [-3,4] \rightarrow \mathbb R$$
My personal opinion:
In my experience of school, though it is just mine in Australia, teachers know very little about their subject areas-- less than an enthusiastic student who has done some research after school. My advice is either to argue with them until they concede that they are incorrect (provided of course you have good reason to argue), or to put your head down and give them the answer they want, rather than the one that is right. By asking this question on Math.SE, you have probably already put more effort into it than your teacher did.
A: 
Say we have a function $f(x)$ defined on the closed interval $x \in [-3, 4]$.

So $\;f : [-3,4] \to \mathbb{R}\,$.

Now, perform the transformation $g(x)=f(|x|)$. 

So $\;g = f \circ h\,$, where $\,h(x)=|x|\,$.

My question is: if we graph $g(x)$ over all defined outputs of $x$

The modulus function is defined as $\,|\cdot| : \mathbb{R} \to \mathbb{R}\,$, but $h$ is restricted by the condition that $\,h(x) = |x| \in [-3,4] \iff x \in h^{-1}\left([-3,4]\right)= [-4,4]\,$. Since the question asks about "all defined outputs of $x$", this presumably means that $h$ is to be taken as $\,h : [-4,4] \to \mathbb{R}\,$.

would we include the interval $[-4, -3)$ on this graph?

Yes, per the above.

We do know the values $g(x)$ would take over this interval independent of the values of $f(x)$ from $[-4, -3
)$; but my line of reasoning is that we cannot "plug" these inputs from $[-4, -3)$ into $g(x)$ to obtain these outputs (since $f(x)$ only exists on the interval $[-3, 4]$), so $g(x)$ should only be graphed on the interval $[-3, 4]$.

By the same logic, if $g$ was defined as $\,g(x)=f(x-100)\,$ for example, then you would argue that the graph is empty, since there is no overlap between $\,[-3,4]\,$ and $\,[97,104]\,$. But that's a very tenuous, unlikely  interpretation of what "perform the transfomation" was supposed to mean.
