To find the absolute maximum of the function $f(x) = \ln(x + 72) + \frac{1}{x}$ on the interval $[-71, -1]$, we'll proceed by evaluating the endpoints and critical points.

Step-by-Step Solution

1. Find the derivative of $f(x)$: $f(x) = \ln(x + 72) + \frac{1}{x}$ Differentiating with respect to $x$: $f'(x) = \frac{1}{x + 72} - \frac{1}{x^2}$

2. Set $f'(x) = 0$ to find critical points: $\frac{1}{x + 72} - \frac{1}{x^2} = 0$ This simplifies to: $\frac{x^2 - (x + 72)}{x^2(x + 72)} = 0$ which further simplifies to: $x^2 - x - 72 = 0$ Solving this quadratic equation using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-72)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 288}}{2} = \frac{1 \pm \sqrt{289}}{2}$ $x = \frac{1 \pm 17}{2}$ Thus, $x = 9$ or $x = -8$.

   Since we are restricted to the interval $[-71, -1]$, only $x = -8$ lies within this interval.

3. Evaluate $f(x)$ at the endpoints and at $x = -8$:

   • At $x = -71$: $f(-71) = \ln(-71 + 72) + \frac{1}{-71} = \ln(1) - \frac{1}{71} = 0 - \frac{1}{71} = -\frac{1}{71}$

   • At $x = -1$: $f(-1) = \ln(-1 + 72) + \frac{1}{-1} = \ln(71) - 1$

   • At $x = -8$: $f(-8) = \ln(-8 + 72) + \frac{1}{-8} = \ln(64) - \frac{1}{8} = \ln(64) - \frac{1}{8}$

4. Compare values:

   • $f(-71) = -\frac{1}{71}$
   • $f(-1) = \ln(71) - 1$
   • $f(-8) = \ln(64) - \frac{1}{8}$

   Since $\ln(64) = 4 \ln(2)$, and $\ln(71) > 4 \ln(2)$, the largest value is $f(-1) = \ln(71) - 1$.

Conclusion

The absolute maximum of $f(x)$ on $[-71, -1]$ is: $\boxed{\ln(71) - 1}$

Do you have any questions or want more details?

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Related Questions

1. How do you find the absolute minimum of a function over a closed interval?
2. What are the conditions for a function to achieve absolute extrema over a closed interval?
3. How does the natural logarithm function $\ln(x)$ behave on the interval $(0, \infty)$?
4. Why is it necessary to check endpoints when finding absolute extrema on a closed interval?
5. How can the quadratic formula be used to find critical points for more complex functions?

Tip: Always verify if critical points lie within the interval before evaluating the function at these points.