## Precalculus (10th Edition)

$f(0)=0,f(2)=3.$
$x=0$ is in the interval $x\leq0$, thus $f(x)=x^2,f(0)=0^2=0.$ $x=2$ is in the interval $2\leq x\leq5$, thus $f(x)=5-x,f(2)5-2=3.$