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.

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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. The absolute minimum of $$ f(x) = x + \frac{4}{x} $$ is 4 at $$ x = 2 $$. There is no absolute maximum.

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