Let $A=\{-1, 0, 1, 2\}$ and $B=\{-4, -2, 0, 2\}$. Let $f, g: A \rightarrow B$ be functions defined by $f(x)=x^{2}-x$ for $x \in A$ and $g(x)=2\left|x-\frac{1}{2}\right|-1$ for $x \in A$. Are $f$ and $g$ equal? Justify your answer.

Two functions $f$ and $g$ are equal if $f(a) = g(a)$ for all $a \in A$.
For $x = -1$:
$f(-1) = (-1)^{2} - (-1) = 1 + 1 = 2$
$g(-1) = 2\left|-1 - \frac{1}{2}\right| - 1 = 2\left|-\frac{3}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2$
So,$f(-1) = g(-1)$.
For $x = 0$:
$f(0) = (0)^{2} - 0 = 0$
$g(0) = 2\left|0 - \frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0$
So,$f(0) = g(0)$.
For $x = 1$:
$f(1) = (1)^{2} - 1 = 0$
$g(1) = 2\left|1 - \frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0$
So,$f(1) = g(1)$.
For $x = 2$:
$f(2) = (2)^{2} - 2 = 4 - 2 = 2$
$g(2) = 2\left|2 - \frac{1}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2$
So,$f(2) = g(2)$.
Since $f(a) = g(a)$ for all $a \in A$,the functions $f$ and $g$ are equal.