Forming a conjecture about the relationship between $g(A \cup B)$ and $ g(A) \cup g(B)$ for an arbitrary function g. Question: Form and prove a conjecture about the relationship between $g(A \cup B)$  and $ g(A) \cup g(B)$ for an arbitrary function g. 

I've already shown this:
For an arbitrary function $g$: $\mathbb{R} \to \mathbb{R}$, it is always true that\ $g(A\cap B)\subseteq g(A) \cap g(B)$ for all sets  $A, B \subseteq \mathbb{R}$.

by this method:

Suppose $y \in g(A \cap B)$. Then there exists $x \in A \cap B$ such that $g(x)=y$. So, $x \in A$ and $x \in B$. Thus, $g(x) \in g(A)$ and $g(x) \in g(B)$. So, $y\in g(A)$ and $y \in g(B)$. Therefore, $y \in g(A) \cap g(B)$. Thus,  $g(A\cap B)\subseteq g(A) \cap g(B)$. 

-Really just need help forming the conjecture. I don't really know where to start.
Conjecture: Let $g(x)$ be a function. Then $g(A\cup B)=g(A)\cup g(B)$
For any $h \in g (A \cup B) $ then there is an $a \in A \cup B $ with $h = g (a) $ Since $a \in A \cup B$ then either $a \in A $ or $a \in B $ (or both) and $g(a) \in g (A) $ or $g (a) \in g (B) $ (or both) so $h=g(a) \in g (A) \cup g (B) $.  

So $g (A \cup B) \subset g (A) \cup g (B)$.

If $h \in g (A) \cup g (B) $ then either $h \in g (A) $ and $h = g (a)$ for some $a \in A $ or $h \in g (B);h = g (a);a \in B $ (or both).  Either way $h = g (a) $ for some $a \in A \cup B$ so $h \in g (A \cup B) $.

So $g (A)\cup g (B) \subset g (A \cup B) $

So $g (A)\cup g (B) = g (A \cup B) $