Find the absolute maximum and minimum, if either exists, for \( f(x)=x+\frac{4}{x} \) Find the second derivative of \( f \). \( f^{\prime \prime}(x)=\square \)

To find the absolute maximum and minimum of the function \( f(x) = x + \frac{4}{x} \), we start by finding the derivative: \[ f'(x) = 1 - \frac{4}{x^2} \] Setting \( f'(x) = 0 \) to find critical points gives: \[ 1 - \frac{4}{x^2} = 0 \implies \frac{4}{x^2} = 1 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2 \] Next, we check the second derivative to determine the concavity and find the maximum or minimum: \[ f''(x) = \frac{8}{x^3} \] Now we can evaluate \( f \) at the critical points \( x = 2 \) and \( x = -2 \): 1. For \( x = 2 \): \[ f(2) = 2 + \frac{4}{2} = 2 + 2 = 4 \] 2. For \( x = -2 \): \[ f(-2) = -2 + \frac{4}{-2} = -2 - 2 = -4 \] Now we can check the behavior as \( x \) approaches the boundaries: As \( x \to 0^+ \), \( f(x) \to +\infty \). As \( x \to 0^- \), \( f(x) \to -\infty \). As \( x \to +\infty \), \( f(x) \to +\infty \). As \( x \to -\infty \), \( f(x) \to -\infty \). Thus, the absolute maximum is \( 4 \) at \( x = 2 \) and the absolute minimum is \( -4 \) at \( x = -2 \). In summary, the answer is: - Absolute maximum is \( 4 \) at \( x = 2 \) - Absolute minimum is \( -4 \) at \( x = -2 \) The second derivative is: \[ f''(x) = \frac{8}{x^3} \]