Find the absolute maximum and minimuin, if eilher exists, for \( f(x)=x+\frac{4}{x} \). Find the second derivative of \( f \). \( f^{\prime \prime}(x)=\frac{8}{x^{3}} \) 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.

To find the absolute maximum and minimum of the function \( f(x) = x + \frac{4}{x} \), we start by identifying the critical points. First, we need to find the derivative: \[ f'(x) = 1 - \frac{4}{x^2} \] Setting the derivative to zero for critical points: \[ 1 - \frac{4}{x^2} = 0 \implies \frac{4}{x^2} = 1 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2 \] Examining the endpoints of the function and the critical points: Since \( f(x) \) is undefined at \( x = 0 \), we'll consider the intervals \( (-\infty, -2] \) and \( [2, \infty) \). Calculating \( f(-2) \) and \( f(2) \): \[ f(-2) = -2 + \frac{4}{-2} = -2 - 2 = -4 \] \[ f(2) = 2 + \frac{4}{2} = 2 + 2 = 4 \] As \( x \to -\infty \), \( f(x) \to -\infty \) and as \( x \to \infty \), \( f(x) \to \infty \). Therefore, within the relevant intervals: - The absolute minimum occurs at \( x = -2 \) with \( f(-2) = -4 \). - The absolute maximum occurs at \( x = 2 \) with \( f(2) = 4 \). So, the values we fill in are: A. The absolute maximum is \( 4 \) at \( x = 2 \). The absolute minimum is \( -4 \) at \( x = -2 \). Now, let's consider the second derivative, which you provided as \( f^{\prime \prime}(x) = \frac{8}{x^3} \). This indicates the concavity of the function: - For \( x > 0 \), \( f''(x) > 0 \) meaning the function is concave up. - For \( x < 0 \), \( f''(x) < 0 \) meaning the function is concave down. Thus, we establish the behavior of \( f(x) \) across its critical points, confirming that the maximum and minimum calculated earlier are indeed correct.