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.
So $\;f : [-3,4] \to \mathbb{R}\,$.
So $\;g = f \circ h\,$, where $\,h(x)=|x|\,$.
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 $\,h : [-4,4] \to \mathbb{R}\,$.
Yes, per the above.
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.