A rational function f(x)f(x) is defined for all real numbers except x=4 and x=3x=4 and x=3. We are given that x=4 and x=3x=4 and x=3 are vertical asymptotes for f(x)f(x). The first derivative of this function is given below:

f′(x)=(x^2−6x)/((x−4)^3(x−3)^3)


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.

A rational function f(x) is defined for all real numbers except x = 4 and x = 3. We are given that x = 4 and x = 3 are vertical asymptotes for f(x). The first derivative of this function is given below: f′(x)=(x^2−6x)/((x−4)^3(x−3)^3)

Explore the behavior of a rational function f(x) with vertical asymptotes at x = 4 and x = 3, using its first derivative f'(x) = (x^2 - 6x) / ((x - 4)^3 (x - 3)^3). Learn how to find critical points and understand the function's behavior near asymptotes, ideal for advanced high school or early college students.

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