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 $(x^a)^b=x^{ab}$, $$g(x)=(x^2)^j$$ Use $x=-1$. This gives $$g(-1)=((-1)^2)^j=1^j$$ One to any positive integer is one. Therefore, all power functions of that form pass through $(-1,1)$.
