Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. Also indicate the $x$-value at which each extremum occurs. \[ f(x)=x^{2}+\frac{160}{x} ;(0, \infty) \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The absolute maximum value(s) is(are) $\square$ at $x=$ (Simplify your answers. Type integers or decimals rounded to three decimal places as needed. Use a comma to separate answers as needed.) B. There is no absolute maximum value.

Step 1 ：First, we need to find the derivative of the function \(f(x) = x^2 + \frac{160}{x}\). Using the power rule and the quotient rule, we get \(f'(x) = 2x - \frac{160}{x^2}\).

Step 2 ：Next, we set the derivative equal to zero and solve for \(x\) to find the critical points. So, \(2x - \frac{160}{x^2} = 0\) simplifies to \(2x^3 - 160 = 0\). Solving this equation gives us \(x = \sqrt[3]{80}\).

Step 3 ：Then, we need to check the endpoints of the interval and the critical point to find the absolute maximum and minimum values. The function is undefined at \(x = 0\), so we only need to check \(x = \sqrt[3]{80}\).

Step 4 ：Substituting \(x = \sqrt[3]{80}\) into the function, we get \(f(\sqrt[3]{80}) = (\sqrt[3]{80})^2 + \frac{160}{\sqrt[3]{80}} = 80 + 2\sqrt[3]{80}\).

Step 5 ：Since the function is defined for all \(x > 0\), and \(f(x)\) tends to infinity as \(x\) tends to infinity, there is no absolute maximum value.

Step 6 ：Thus, the absolute minimum value is \(80 + 2\sqrt[3]{80}\) at \(x = \sqrt[3]{80}\).