## College Algebra (6th Edition)

I plug in the respective values of $x$ into $f(x)=(-1)^{x+1}(\frac{1}{2})^x$ For $x=1$: $f(x)=(-1)^{1+1}(\frac{1}{2})^1=1\cdot\frac{1}{2}=\frac{1}{2}$ For $x=2$: $f(x)=(-1)^{2+1}(\frac{1}{2})^2=-1\cdot\frac{1}{4}=-\frac{1}{4}$ For $x=3$: $f(x)=(-1)^{3+1}(\frac{1}{2})^3=1\cdot\frac{1}{8}=\frac{1}{8}$ For $x=4$: $f(x)=(-1)^{4+1}(\frac{1}{2})^4=-1\cdot\frac{1}{16}=-\frac{1}{16}$