Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 192: 32

$f(-3)=5$ $f(0)=5$ $f(2)=5$ $f(3)=3$ $f(5)=7$

Because x=-3, x=0, x=2 are in between $(-\infty,2]$ of $f(x)=5$ so $f(-3)=5$ $f(0)=5$ $f(2)=5$ x=5, x=3 are in between $(2,+\infty)$ of $f(x)=2x-3$ So $f(3)=2(3)-3=3$ $f(5)=2(5)-3=7$