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. \[ f(x)=x^{2}+\frac{160}{x} ;(0, \infty) \]

Step 1 ：The function is \(f(x)=x^{2}+\frac{160}{x}\) and we are looking for its absolute maximum and minimum values over the interval \((0, \infty)\).

Step 2 ：To find the extrema of a function, we first need to find its critical points. Critical points occur where the derivative of the function is either zero or undefined.

Step 3 ：First, we find the derivative of the function \(f(x)\). Using the power rule and the quotient rule, we get \(f'(x)=2x-\frac{160}{x^{2}}\).

Step 4 ：Next, we set the derivative equal to zero and solve for \(x\). This gives us the equation \(2x-\frac{160}{x^{2}}=0\).

Step 5 ：Multiplying through by \(x^{2}\) to clear the fraction gives us \(2x^{3}-160=0\).

Step 6 ：Solving for \(x\) gives us \(x=\sqrt[3]{80}\).

Step 7 ：We also need to check where the derivative is undefined. However, the derivative is defined for all \(x\) in the interval \((0, \infty)\), so we don't have any additional critical points.

Step 8 ：Now we need to determine whether the critical point is a maximum, minimum, or neither. We do this by using the second derivative test.

Step 9 ：The second derivative of the function is \(f''(x)=2+\frac{320}{x^{3}}\).

Step 10 ：Substituting the critical point into the second derivative gives us \(f''(\sqrt[3]{80})=2+\frac{320}{(\sqrt[3]{80})^{3}}=2+4=6\), which is greater than zero. Therefore, by the second derivative test, the function has a local minimum at \(x=\sqrt[3]{80}\).

Step 11 ：Since the function is defined over the interval \((0, \infty)\) and we have only one critical point, this is also the absolute minimum of the function.

Step 12 ：The function does not have an absolute maximum on the interval \((0, \infty)\) because as \(x\) approaches infinity, the function also approaches infinity.

Step 13 ：Therefore, the absolute minimum value of the function is \(f(\sqrt[3]{80})=(\sqrt[3]{80})^{2}+\frac{160}{\sqrt[3]{80}}=80+2\sqrt[3]{80}\) at \(x=\sqrt[3]{80}\), and there is no absolute maximum.