Maximal domain for composite functions. 
Question $\mathbf 5$
If $f:(-\infty,1)\to R$, $f(x)=2\log_{\,e}(1-x)$ and $g:[-1,\infty)\to R$, $g(x)=3\sqrt{x+1}$, then the maximal domain of the function $f+g$ is
$\textbf{A.}\quad[-1,1)$ 
$\textbf{B.}\quad(1,\infty)$
$\textbf{C.}\quad(-1,1]$
$\textbf{D.}\quad(-\infty,-1]$
$\textbf{E.}\quad R$

It it correct to write the new domain as $[-1,1)$ because the values of $x$ are common to both domains? 
 A: Yes. In general, if $f: A \to B$ and $g : C \to D$, then $f + g$ has domain $A\cap C$.
To see why this is the case, recall that the function $f + g$ is defined by $(f+g)(x) := f(x) + g(x)$. So $(f+g)(x)$ is defined only when both $f(x)$ and $g(x)$ are defined, so $x \in A$ and $x \in C$.
A: Yes, you are completely correct. 
In addition to what you said, any value which does not satisfy the domain is not defined in either $f$ or $g$.
A: In this question there is no mention that x is an element of the Real Numbers. 
I suggest you check if the preamble says something like " x is a Real Number" 
If not mentioned, it is possibly an oversight or sloppiness.
f goes from the domain, the set of Real numbers less than 1, to the range, the set of Real numbers. 
g goes from the set of Real Numbers greater than or equal to negative one, to the range, the set of Real Numbers.
The word "maximal" adds an extra bit to the question, and indeed causes confusion. Sloppiness in the teaching, but don't say that to your teacher.
The domain is what you choose. It is a choice. The maximal domain is the set of which all other possible domains are subsets. 
The answer is a) if x is Real
