Let $f$ be the function specified by the following table:
$x\ $ \t $-4$ \t $-3$ \t $-2$ \t $-1$ \t $0$ \t $1$ \t $2$ \t $3$ \t $4$ \\ $f(x)$ \t $-2$ \t $1$ \t $2$ \t $4$ \t $2$ \t $1$ \t $0.5$ \t $-2.5$ \t $0$
Then $f(-2) = 2,\ f(1) = 1,$ and $f(2) = -2.$ Also,
$f(%0)$ = BOX and $f(%1)$ = BOX