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.

We are given the function \[ f(x)=x+\frac{4}{x}, \] with \(x\ne0\). **Step 1. Find the critical points.** Differentiate \(f\) with respect to \(x\): \[ f'(x)=1-\frac{4}{x^2}. \] Set \(f'(x)=0\): \[ 1-\frac{4}{x^2}=0 \quad\Longrightarrow\quad \frac{4}{x^2}=1 \quad\Longrightarrow\quad x^2=4. \] Thus, the critical points are \[ x=2 \quad \text{and} \quad x=-2. \] **Step 2. Evaluate \(f\) at the critical points.** At \(x=2\): \[ f(2)=2+\frac{4}{2}=2+2=4. \] At \(x=-2\): \[ f(-2)=-2+\frac{4}{-2}=-2-2=-4. \] **Step 3. Analyze the behavior on the entire domain \(\mathbb{R}\setminus\{0\}\).** Notice that the domain of \(f\) is not a closed bounded interval (it is \(\mathbb{R}\setminus\{0\}\)). Therefore, we must consider what happens as \(x\) approaches the endpoints (in this case, \(0\) and \(\pm\infty\)). - As \(x\to0^+\): \[ f(x)=x+\frac{4}{x}\to +\infty. \] - As \(x\to\infty\): \[ f(x)\to x\to +\infty. \] - As \(x\to0^-\): \[ f(x)=x+\frac{4}{x}\to -\infty. \] - As \(x\to-\infty\): \[ f(x)=x+\frac{4}{x}\to x\to -\infty. \] Thus: - On the positive side (\(x>0\)), \(f\) has a local (and absolute on this part of the domain) minimum at \(x=2\) with value \(f(2)=4\), but no absolute maximum since \(f(x)\to+\infty\) both as \(x\to0^+\) and \(x\to \infty\). - On the negative side (\(x<0\)), \(f\) has a local maximum at \(x=-2\) with value \(f(-2)=-4\), but no absolute minimum since \(f(x)\to-\infty\) both as \(x\to0^-\) and \(x\to -\infty\). Since the overall domain is not bounded or closed, the function does not have a global absolute maximum or an absolute minimum when considering both pieces together. If we interpret the question in the context where \(x>0\) (which is common for functions of the form \(x+\frac{4}{x}\)), then: - The absolute minimum is \(4\) at \(x=2\). - There is no absolute maximum. **Step 4. Confirm the second derivative.** We are given that \[ f''(x)=\frac{8}{x^{3}}. \] This confirms, for example, that at \(x=2\), \[ f''(2)=\frac{8}{2^3}=\frac{8}{8}=1>0, \] so \(x=2\) gives a local minimum. Similarly, at \(x=-2\), \[ f''(-2)=\frac{8}{(-2)^3}=\frac{8}{-8}=-1<0, \] so \(x=-2\) gives a local maximum. **Final Answer (Select the correct choice):** B. The absolute minimum is \(4\) at \(x=2\). There is no absolute maximum.