Answer
True.
Work Step by Step
Let $g(x)=x^p$ where $p$ is even. Therefore, $p=2j$ for some $j \in \Bbb{Z}$, meaning that $g(x)=x^{2j}$. Since $(x^a)^b=x^{ab}$, $$g(x)=(x^2)^j$$ Change $x$ to $-x$ to get $$g(-x)=((-x)^2)^j=((-1)^2(x^2))^j=(x^2)^j=x^{2j}$$ Therefore, $g(x)=g(-x)$ and the function must be symmetric about the $x$ axis. Thus, the statement is true.
