Find the local maximum and local minimum values for the function given by $g(x) = \frac{x}{2} + \frac{2}{x}, \quad x > 0$.

(A) Given $g(x) = \frac{x}{2} + \frac{2}{x}$ for $x > 0$.
First,find the derivative $g'(x)$:
$g'(x) = \frac{1}{2} - \frac{2}{x^2}$.
To find the critical points,set $g'(x) = 0$:
$\frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{2}{x^2} = \frac{1}{2} \implies x^2 = 4$.
Since $x > 0$,we have $x = 2$.
Now,find the second derivative $g''(x)$:
$g''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.
Evaluate $g''(x)$ at $x = 2$:
$g''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2}$.
Since $g''(2) > 0$,by the second derivative test,$x = 2$ is a point of local minima.
The local minimum value is $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$.
There is no local maximum value for this function as $x > 0$.