Answer
$5$ is a critical number of the function $g$.
However, $g$ does not have a local extreme value at $x = 5$
Work Step by Step
$g(x) = 2+(x-5)^3$
$g'(x) = 3(x-5)^2$
$g''(x) = 6(x-5)$
When $x=5$, then $g'(x) = 0$
Thus, $5$ is a critical number of the function.
When $x \lt 5$, then $g''(x) \lt 0$
When $x \gt 5$, then $g''(x) \gt 0$
The graph is concave downward when $x \lt 5$, but the graph is concave upward when $x \gt 5$. Therefore, $g$ does not have a local extreme value at $x = 5$
