## Intermediate Algebra (12th Edition)

According to the definition of negative exponents $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, a. $(\frac{1}{3})^{-1}=\frac{1}{(\frac{1}{3})}=3$ b. $(-\frac{1}{3})^{-1}=\frac{1}{(-\frac{1}{3})}=-3$ c. $-(\frac{1}{3})^{-1}=-\frac{1}{(\frac{1}{3})}=-3$ d. $-(-\frac{1}{3})^{-1}=-(\frac{1}{-\frac{1}{3}})=-(-3)=3$