## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(0)=0$ and $f(2)=3$
$x=0$ lies in the interval $x\leq0$; therefore, $f(x)=x^2$ When $x=0$, then we have: $f(0)=0^2=0$ Since, $x=2$ lies in the interval $2\leq x\leq5$, therefore $f(x)=5-x$ When $x=2$, then we have: $f(2)=5-2=3$