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 $%4$ \t $%5$ \t $%6$ \t $%7$ \t $%8$ \t $%9$ \t $%10$ \t $%11$ \t $%12$
The natural domain of this function consists only of the numbers $-4, -3, -2, -1, 0, 1, 2, 3, 4$, as those are the only values of $x$ for which $f(x)$ is defined.

For example, when $x = -2$, $f(x) = 2$. So,
    $f(-2) = 2$. (Mouse over to highlight corresponding table column.)
Similarly, when $x = 2$, $f(x) = 0.5$. So,
    $f(2) = 0.5$.
Also,
$f(-2+ 2)$ $= f(0) \qquad$ First evaluate the argument in parentheses.
$= 2$ Then look up the value of $f$.
Whereas
$f(-2) +f(2) $ $= 2 + 0.5$ First evaluate $f(-2)$ and $f(2)$ separately.
$= 2.5$ Then add the values.
Some for you:
$f(%0)$ = BOX
$f(%1)$ = BOX
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MESSAGE

$f(%3)$ = BOX
$f(%0) %2 f(%1)$ = BOX
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MESSAGE
RANDOMIZE