Find the local maximum and local minimum values for the function $g(x) = \frac{1}{x^{2}+2}$.

(A) Given function: $g(x) = \frac{1}{x^{2}+2}$.
To find the critical points,we find the derivative $g'(x)$ using the chain rule:
$g'(x) = \frac{d}{dx}(x^{2}+2)^{-1} = -1(x^{2}+2)^{-2} \cdot (2x) = \frac{-2x}{(x^{2}+2)^{2}}$.
Set $g'(x) = 0$:
$\frac{-2x}{(x^{2}+2)^{2}} = 0 \Rightarrow -2x = 0 \Rightarrow x = 0$.
Now,we apply the first derivative test:
For $x < 0$,let $x = -1$: $g'(-1) = \frac{-2(-1)}{((-1)^{2}+2)^{2}} = \frac{2}{9} > 0$.
For $x > 0$,let $x = 1$: $g'(1) = \frac{-2(1)}{(1^{2}+2)^{2}} = \frac{-2}{9} < 0$.
Since $g'(x)$ changes from positive to negative at $x = 0$,$x = 0$ is a point of local maxima.
The local maximum value is $g(0) = \frac{1}{0^{2}+2} = \frac{1}{2}$.
Since the denominator $(x^{2}+2)^{2}$ is always positive and the numerator $-2x$ does not change sign in a way that creates a local minimum,there is no local minimum value for this function.