Find the absolute maximum and minimum, if eilher exists, for \( f(x)=x+\frac{1}{x} \) Find the second derivative of t . \( \mathrm{f}^{\prime \prime}(\mathrm{x})=\square \) Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute maximum is \( \square \) at \( x=\square \). B. There is no absolute maximum. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute minimum is \( \square \) at \( x=\square \). O. There is no absolute minimurn.

To find the absolute maximum and minimum of the function \( f(x) = x + \frac{1}{x} \), we first determine its critical points. The derivative of \( f(x) \) is: \[ f'(x) = 1 - \frac{1}{x^2} \] Setting \( f'(x) = 0 \), we solve for \( x \): \[ 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1 \text{ (since } x > 0 \text{ is required for the domain of } f(x)\text{)} \] Next, we evaluate \( f(x) \) at the critical point and the endpoints of the domain \( (0, \infty) \). As \( x \to 0^+ \), \( f(x) \to \infty \), and as \( x \to \infty \), \( f(x) \to \infty \). Thus: 1. At \( x = 1 \): \( f(1) = 1 + 1 = 2 \) Since the function goes to infinity on both ends and only has one critical point, the absolute minimum is \( 2 \) at \( x = 1 \). The second derivative of \( f(x) \) is: \[ f''(x) = \frac{2}{x^3} \] This indicates the concavity of \( f \): since \( f''(x) > 0 \) for \( x > 0 \), the function is concave up at the critical point, confirming it is a minimum. A. The absolute maximum is ⧬ at \( x= ⧬ \). B. There is no absolute maximum. A. The absolute minimum is \( 2 \) at \( x=1 \). O. There is no absolute minimum. --- Exploring the real-world application of \( f(x) = x + \frac{1}{x} \) reveals its connection to optimization problems, such as minimizing the cost of materials in engineering. This function represents a balance between a linear cost and a diminishing returns factor, crucial in fields like construction and manufacturing. For problem-solving tips, ensure you always test the critical points and examine the behavior of the function at the boundaries of its domain. A common mistake is neglecting to consider the limits of the function, especially in rational expressions, as they can dramatically influence the existence of absolute maxima or minima.