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

To find the absolute maximum and minimum of \( f(x) = x + \frac{1}{x} \), we first calculate its derivative: \[ f'(x) = 1 - \frac{1}{x^2} \] Setting the derivative equal to zero to find critical points: \[ 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = 1 \text{ (since } x \text{ must be positive for } f(x) \text{ to be defined)} \] Next, we evaluate \( f \) at the critical point and the endpoints. Since \( f(x) \) is defined for \( x > 0 \), we examine the limit as \( x \to 0^+ \): \[ f(1) = 1 + \frac{1}{1} = 2 \] As \( x \to 0^+ \): \[ f(x) \to \infty \] And as \( x \to \infty \): \[ f(x) \to \infty \] So, the absolute minimum is \( 2 \) at \( x = 1 \) and there is no absolute maximum. Now, let's find the second derivative of \( f \): \[ f''(x) = \frac{2}{x^3} \] This is always positive for \( x > 0 \), indicating that the function is concave up in that domain. Now, to fill in the boxes: A. The absolute maximum is \( \text{none} \) at \( x=\text{none} \). There is no absolute maximum. B. The absolute minimum is \( 2 \) at \( x=1 \).