Find the local maximum and local minimum values for the function given by $g(x) = x^{3} - 3x$.

(A) Given function: $g(x) = x^{3} - 3x$.
First,find the derivative: $g'(x) = 3x^{2} - 3$.
To find critical points,set $g'(x) = 0$:
$3x^{2} - 3 = 0 \Rightarrow 3(x^{2} - 1) = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$.
Now,find the second derivative: $g''(x) = 6x$.
Apply the second derivative test:
For $x = 1$: $g''(1) = 6(1) = 6 > 0$. Since $g''(1) > 0$,$x = 1$ is a point of local minima.
The local minimum value is $g(1) = (1)^{3} - 3(1) = 1 - 3 = -2$.
For $x = -1$: $g''(-1) = 6(-1) = -6 < 0$. Since $g''(-1) < 0$,$x = -1$ is a point of local maxima.
The local maximum value is $g(-1) = (-1)^{3} - 3(-1) = -1 + 3 = 2$.