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. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute minimum is \( \square \) at \( x=\square \). B. There is no absolute minimum.

To find the absolute maximum and minimum of \( f(x)=x+\frac{1}{x} \), we first look at its critical points by taking the derivative: \[ f'(x) = 1 - \frac{1}{x^2} \] Setting \( f'(x) = 0 \) gives: \[ 1 - \frac{1}{x^2} = 0 \] \[ \frac{1}{x^2} = 1 \] \[ x^2 = 1 \] \[ x = 1 \text{ or } x = -1 \] However, since \( f(x) \) is undefined at \( x = 0 \), we consider the intervals \( (-\infty, -1) \), \( (-1, 0) \), and \( (0, 1) \), \( (1, \infty) \). Evaluating the endpoints and critical points: 1. As \( x \to 0^- \), \( f(x) \to -\infty \) 2. As \( x \to 0^+ \), \( f(x) \to +\infty \) 3. At \( x = -1 \), \( f(-1) = -1 + \frac{1}{-1} = -2 \) 4. At \( x = 1 \), \( f(1) = 1 + \frac{1}{1} = 2 \) Thus, there is no absolute maximum as \( f(x) \to +\infty \) when approaching \( 0 \) from the right, and the absolute minimum occurs at \( x=-1 \) with \( f(-1)=-2 \). Now let's find the second derivative: Calculating the first derivative again: \[ f'(x) = 1 - \frac{1}{x^2} \] Taking the second derivative: \[ f''(x) = \frac{2}{x^3} \] So, \( f^{\prime \prime}(x)=\frac{2}{x^3} \). Now, let's summarize the results: A. The absolute maximum is there is no absolute maximum. B. The absolute minimum is \( -2 \) at \( x=-1 \).