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

$\pm r^2$
Use the fact that $\sqrt{a}=a^\frac{1}{2}$ along with the fact that $(a^b)^n=a^{bn}$ to get $$\sqrt{r^4}=r^{\frac{4}{2}}=r^2$$ However, $(-r^2)(-r^2)=r^4$ also. So, $\pm r^2$ is the result.