To find the absolute maximum and minimum values of the function $f(x) = \frac{1}{x^2 + 6x}$ on the interval $(-6, 0)$, follow these steps:

Step 1: Check the Domain

The function $f(x)$ is undefined where $x^2 + 6x = 0$, which factors as: $x(x + 6) = 0 \quad \implies \quad x = 0 \text{ or } x = -6.$ Thus, the function is undefined at both endpoints of the interval $(-6, 0)$.

Step 2: Take the Derivative

The derivative of $f(x)$ is found using the quotient rule: $f'(x) = \frac{(0)(x^2 + 6x) - 1(2x + 6)}{(x^2 + 6x)^2} = \frac{-(2x + 6)}{(x^2 + 6x)^2}.$

Step 3: Solve $f'(x) = 0$ for Critical Points

The numerator of $f'(x)$ is $-(2x + 6)$, so setting it to zero gives: $-(2x + 6) = 0 \quad \implies \quad x = -3.$ Thus, $x = -3$ is a critical point.

Step 4: Evaluate $f(x)$ at the Critical Point

The value of $f(x)$ at $x = -3$ is: $f(-3) = \frac{1}{(-3)^2 + 6(-3)} = \frac{1}{9 - 18} = \frac{1}{-9} = -\frac{1}{9}.$

Step 5: Analyze Behavior Near the Endpoints

The function $f(x)$ approaches infinity as $x \to -6^+$ or $x \to 0^-$ because the denominator $x^2 + 6x \to 0^+$.

Step 6: Conclusion

The absolute maximum does not exist as $f(x) \to \infty$ near the endpoints. The absolute minimum is: $\text{Absolute Minimum: } -\frac{1}{9} \text{ at } x = -3.$

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Do you need a graph or further clarification? Here are 5 related questions to explore:

1. How does the behavior of $f(x)$ change outside the interval $(-6, 0)$?
2. What happens if the interval includes one of the undefined endpoints, like $[-6, 0)$?
3. How would the solution change if the denominator had a higher degree polynomial?
4. Can we approximate the location of maxima and minima for more complex rational functions?
5. How does the behavior of vertical asymptotes affect the extremum analysis?

Tip: Always check the domain of the function before evaluating for critical points or