Let $f$ be the function specified by the following table:
#[Let ][Sea ]# $f: \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \to \mathbb{R}$ be the function specified by the following table:
$x$ \t $f(x)$ \\ $-4$ \t $%4$ \\ $-3$ \t $%5$ \\ $-2$ \t $%6$ \\ $-1$ \t $%7$ \\ $0$ \t $%8$ \\ $1$ \t $%9$ \\ $2$ \t $%10$ \\ $3$ \t $%11$ \\ $4$ \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 $x = -2$, $f(x) = %6$. So,
$f(-2) = %6$. (Mouse over to highlight corresponding table column.)
Similarly, when $x = 2$, $f(x) = %10$. So,
$f(2) = %10$.
Also,
 $f(-2+ 2)$ $= f(0) \qquad$ First evaluate the argument in parentheses. $= %8$ Then look up the value of $f$. Whereas $f(-2) +f(2)$ $= %6 + %10$ First evaluate $f(-2)$ and $f(2)$ separately. $= %16$ Then add the values.
Some for you:
$f(%0)$ = BOX
$f(%1)$ = BOX
BUTTONS
MESSAGE

$f(%3)$ = BOX
$f(%0) %2 f(%1)$ = BOX
$f\left[(%13[0])(%13[1])\right]$ = BOX
$f(%13[0])f(%13[1])$ = BOX
$f\left(%15\right)$ = BOX
$f(%14)^2$ = BOX
BUTTONS
MESSAGE
RANDOMIZE