Given the information about the rational function $f(x)$ and its first derivative:

$f'(x) = \frac{x^2 - 6x}{(x - 4)^3 (x - 3)^3}$

Key observations:

1. Vertical Asymptotes: The problem states that $x = 4$ and $x = 3$ are vertical asymptotes for the function $f(x)$. This aligns with the fact that the denominator of $f'(x)$ includes terms $(x - 4)^3$ and $(x - 3)^3$, which go to infinity as $x \to 4$ or $x \to 3$.

2. Critical Points: To find the critical points (where the derivative $f'(x) = 0$), set the numerator of the derivative equal to zero: $x^2 - 6x = 0$ Factor the quadratic: $x(x - 6) = 0$ So, the critical points are: $x = 0 \quad \text{and} \quad x = 6$ These are the values where the slope of the function is zero, meaning there could be local maxima, minima, or inflection points at these values.

3. Behavior Near Vertical Asymptotes:

   • Near $x = 4$, the derivative has a cubic term $(x - 4)^3$ in the denominator, indicating a steep slope as $x$ approaches 4.
   • Near $x = 3$, the derivative behaves similarly due to the $(x - 3)^3$ term. Thus, the function $f(x)$ has steep vertical asymptotes at these points.

4. Sign of the Derivative:

   • For $x < 0$, the numerator $x^2 - 6x$ is positive, and since both denominator terms are positive, $f'(x) > 0$, indicating increasing behavior.
   • For $0 < x < 6$, the sign of the numerator and denominator will determine increasing or decreasing behavior, with critical points at $x = 0$ and $x = 6$ influencing the slope.

Further analysis:

• To classify the critical points $x = 0$ and $x = 6$, you would need the second derivative $f''(x)$, or analyze the sign changes of $f'(x)$.
• The behavior at the asymptotes $x = 3$ and $x = 4$ needs to be considered as these are points where $f(x)$ has vertical asymptotes.

Would you like to delve deeper into the behavior of the function around critical points and asymptotes, or explore the second derivative for concavity analysis?

Here are 5 related questions for further understanding:

1. What happens to the function $f(x)$ as $x \to 4^+$ and $x \to 4^-$?
2. What is the behavior of $f(x)$ around $x = 0$ and $x = 6$?
3. How can we use the second derivative $f''(x)$ to classify the critical points?
4. What are the horizontal asymptotes of the function $f(x)$, if any?
5. What role do the powers of $(x-4)^3$ and $(x-3)^3$ in the denominator play in shaping the graph?

Tip: To fully understand the behavior of a rational function, consider analyzing the asymptotes, critical points, and intervals of increase and decrease.