To find the absolute maximum of the function $f(x) = 8x - 4x \ln(x)$ on $(0, \infty)$, follow these steps:

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Step 1: Compute the derivative of $f(x)$

The derivative is: $f'(x) = \frac{d}{dx}(8x) - \frac{d}{dx}(4x \ln(x))$ Using derivative rules: $f'(x) = 8 - \left(4 \ln(x) + 4 \right)$ $f'(x) = 8 - 4\ln(x) - 4$ $f'(x) = 4 - 4\ln(x)$

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Step 2: Set $f'(x) = 0$ to find critical points

$4 - 4\ln(x) = 0$ $\ln(x) = 1$ x = e \quad \text{(since the natural logarithm of e is 1)}

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Step 3: Verify the nature of the critical point

To determine whether $x = e$ is a maximum, check the second derivative: $f''(x) = \frac{d}{dx}(4 - 4\ln(x)) = -\frac{4}{x}$ At $x = e$, $f''(e) = -\frac{4}{e}$, which is negative. Thus, $f(x)$ has a local maximum at $x = e$.

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Step 4: Compute the value of $f(x)$ at $x = e$

$f(e) = 8e - 4e\ln(e)$ $f(e) = 8e - 4e(1)$ $f(e) = 8e - 4e = 4e$

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Step 5: Conclusion

The absolute maximum value of $f(x)$ on $(0, \infty)$ is: $4e \quad \text{at} \quad x = e$

Final Answer:

$\text{Absolute maximum: } 4e \quad \text{(approximately 10.87)}$ $\text{at } x = e \quad \text{(approximately 2.72)}$

Let me know if you'd like detailed calculations or further clarifications.

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Related Questions:

1. How does the logarithmic function impact the behavior of $f(x)$ as $x \to \infty$?
2. What happens to $f(x)$ as $x \to 0^+$?
3. How can you verify $x = e$ is the global maximum over $(0, \infty)$?
4. Why is the second derivative useful in classifying critical points?
5. How would the solution change if the interval were different?

Tip:

Always check the domain of the logarithmic function when working with optimization problems to ensure critical points lie within valid intervals.