## Functions Modeling Change: A Preparation for Calculus, 5th Edition

Yes, $g(x)=-\frac{1}{6}x^9$
Use the fact that $(-x^3)^3=(-1 \cdot x^3)^3$, $(ab)^n=a^n(b^n)$, and $(a^m)^n=a^{mn}$ to get $$g(x)=\frac{(-1)^3(x^3)^3}{6}=-\frac{x^9}{6}=-\frac{1}{6}x^9$$ Therefore, this is a power function with $k=-\frac{1}{6}$ and $p=9$.