## College Algebra 7th Edition

$f(2)=-1$ $f(-2)=\frac{5}{3}$ $f(\frac{1}{2})=0$ $f(a)=\frac{1-2a}{3}$ $f(-a)= \frac{1+2a}{3}$ $f(a-1)=\frac{3-2a}{3}$
We are given: $f(x)= \frac{1-2x}{3}$ We evaluate: $f(2)=\frac{1-2(2)}{3}=\frac{1-4}{3}=-1$ $f(-2)=\frac{1-2(-2)}{3}=\frac{5}{3}$ $f(\frac{1}{2})=\frac{1-2(\frac{1}{2})}{3}=0$ $f(a)=\frac{1-2a}{3}$ $f(-a)= \frac{1-2(-a)}{3}=\frac{1+2a}{3}$ $f(a-1)=\frac{1-2(a-1)}{3}=\frac{1-2a+2}{3}=\frac{3-2a}{3}$