Domain Definition in Scaled Function Suppose the function $f(x)$ is defined on the domain $\{x_1,x_2,x_3\}$, so that the graph of $y=f(x)$ consists of just three points. Suppose those three points form a triangle of area $32$.
The graph of $y = 2f(2x)$ also consists of just three points. What is the area of the triangle formed by those three points?
If I look at the properties of functions using shifting / scaling, etc, then I can arrive at an answer for this question.
I have a different doubt though - I am confused about the domain of definition of $f$. Since it is explicitly defined only for three points, wouldn't it be incorrect to evaluate the function for points not in the domain. 
 A: If $f(x)$ is defined on $\{x_1,x_2,x_3\}$, then $f(2 x)$ is defined on $\{x_1/2,x_2/2,x_3/2\}$
More generally, $f(g(x))$ can only be defined on those points for which $g(x)$ lies in the domain of $f$.
A: 
It's eight.

The domain is scaled down by factor $2$: if $x_1$ belong to a domain $D$ of $f$, you need to have $x=x_1/2$ to get $2x = x_1\in D$.

So you have three points forming a triangle with sizes equal half of the original sizes. Its area is then $1/4$ of the original: $32/4 = 8$.


Edit
I have misunderstood the main part of the problem, as I thought $x_1, x_2, x_3$ are points in a plane and they form 'the triangle'. As a result, multiplying those $x$-es (points, i.e vectors in $\mathbb R^2$) by $2$ made my imagined triangle twice smaller in all directions, thus reducing its area by $1/4$.
That's also why I've neglected the importance of the other $2$ in my comment.
Now, when I read the problem again, I can see I was completely wrong.
The basic problem is defined as scaling the domain and co-domain of a specific real-valued function of a real-value parameter $f(x)$ by defining a new function $2f(2x)$. In this case multiplying the argument by $2$ shrinks the domain by a coefficient $1/2$, while multiplying the value by $2$ expands the co-domain by a factor $2$. Consequently the area of a figure defined by the function's graph is divided by $2$ and multiplied by $2$, hence stays unchanged.
As for the main question of OP:

wouldn't it be incorrect to evaluate the function for points not in the domain.

the answer is: NO, because we evaluate the function for points in the domain.
One needs carefully define, though, what exactly is the function being considered and what exactly is its domain.
To show it more clearly let's describe a new function as a composition:
$$y = 2f(u) \text{  where  } u=2x.\tag{*}$$
Also describe the three domain values with some neutral symbol, like $p$:
 $$f: \{p_1, p_2, p_3\} \to \mathbb R$$
Then we can easily see that for function ($*$) we must obviously have the argument of $f$ belonging to the domain of $f$:
$$u\in D = \{p_1, p_2, p_3\} $$
but that means $x$, the argument of ($*$) must belong to a shrinked domain:
$$x\in \left\{\frac{p_1}2, \frac{p_2}2, \frac{p_3}2\right\} $$
The point is, $f(x)$ and $2f(2x)$ are two different functions, hence their domains differ, too; however $f$ itself is evaluated in both cases on the same set of input values.
