If possible, find the location of an absolute extremum of the function on the given interval.
\[
g(x)=2 x^{3}-3 x^{2}-72 x+19 ;(0,6)
\]

Step 1 ：Find the derivative of the function \(g(x) = 2x^3 - 3x^2 - 72x + 19\). The derivative is \(g'(x) = 6x^2 - 6x - 72\).

Step 2 ：Set the derivative equal to zero and solve for \(x\). The equation is \(0 = 6x^2 - 6x - 72\). Simplify it to \(0 = x^2 - x - 12\). The solutions are \(x = 4\) and \(x = -3\).

Step 3 ：Check if the solutions are in the given interval \((0,6)\). Only \(x = 4\) is in the interval. So, \(x = 4\) is the only critical point in the interval.

Step 4 ：Evaluate the function at the endpoints of the interval and at the critical point. The values are \(g(0) = 19\), \(g(4) = -85\), and \(g(6) = -79\).

Step 5 ：\(\boxed{\text{The absolute maximum value of the function on the interval (0,6) is 19 at x = 0.}}\)

Step 6 ：\(\boxed{\text{The absolute minimum value of the function on the interval (0,6) is -85 at x = 4.}}\)