Answer
$f(-5)=-15$
$f(0)=1$
$f(1)=2$
$f(2)=3$
$f(5)=9$
Work Step by Step
Because x=-5 is in between $(-\infty,0)$ of $f(x)=3x$
so $f(-5)=3(-5)=-15$
x=0, 1, 2 are in between $[0,2]$ of $f(x)=x+1$
So $f(0)=0+1=1$
$f(1)=1+1=2$
$f(2)=2+1=3$
x=5 is in between $(2,+\infty)$ of $f(x)=(x-2)^{2}$
$f(5)=(5-2)^{2}=9$
