We are tasked with analyzing the function $f(x) = \frac{3x}{\sqrt{x - 4}}$ over the interval $[6, 12]$.

Let's break down each part of the problem.

Part 1: Find the critical points of $f(x)$ on the interval $[6, 12]$.

A critical point occurs where the derivative $f'(x)$ is either zero or undefined.

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

The function is $f(x) = \frac{3x}{\sqrt{x - 4}}$.

To differentiate, we use the quotient rule:

$f'(x) = \frac{d}{dx} \left( \frac{3x}{\sqrt{x - 4}} \right)$

The quotient rule states that:

$f'(x) = \frac{(v \cdot u' - u \cdot v')}{v^2}$

where $u(x) = 3x$ and $v(x) = \sqrt{x - 4}$.

First, compute $u'(x) = 3$ and $v'(x)$. To differentiate $v(x) = (x - 4)^{1/2}$, we apply the chain rule:

$v'(x) = \frac{1}{2}(x - 4)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x - 4}}$

Now apply the quotient rule:

$f'(x) = \frac{\sqrt{x - 4} \cdot 3 - 3x \cdot \frac{1}{2\sqrt{x - 4}}}{(\sqrt{x - 4})^2}$

Simplify the numerator and denominator:

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

Now, to find the critical points, we set $f'(x) = 0$:

$3\sqrt{x - 4} - \frac{3x}{2\sqrt{x - 4}} = 0$

Multiply through by $2\sqrt{x - 4}$ to clear the fractions:

$6(x - 4) - 3x = 0$

Simplify:

$6x - 24 - 3x = 0$

$3x = 24$

$x = 8$

So, the only critical point is $x = 8$ within the interval $[6, 12]$.

Part 2: Identify the absolute minimum or minima of the function on the given interval.

We need to evaluate $f(x)$ at the critical point $x = 8$ and at the endpoints of the interval, $x = 6$ and $x = 12$.

• At $x = 6$:

$f(6) = \frac{3(6)}{\sqrt{6 - 4}} = \frac{18}{\sqrt{2}} = \frac{18\sqrt{2}}{2} = 9\sqrt{2}$

• At $x = 8$:

$f(8) = \frac{3(8)}{\sqrt{8 - 4}} = \frac{24}{\sqrt{4}} = \frac{24}{2} = 12$

• At $x = 12$:

$f(12) = \frac{3(12)}{\sqrt{12 - 4}} = \frac{36}{\sqrt{8}} = \frac{36}{2\sqrt{2}} = \frac{18}{\sqrt{2}} = 9\sqrt{2}$

Thus, the function value is $9\sqrt{2}$ at both $x = 6$ and $x = 12$, and $12$ at $x = 8$.

Since $f(8) = 12$ is the highest value and $f(6) = f(12) = 9\sqrt{2} \approx 12.727$, the absolute minimum value of $f(x)$ occurs at $x = 8$.

Part 3: Identify the absolute maximum or maxima of the function on the given interval.

Since the values at the endpoints are equal and smaller than the critical point, the absolute maximum occurs at $x = 6$ and $x = 12$, with a value of $9\sqrt{2} \approx 12.727$.

Summary of Answers:

1. Critical point(s): $x = 8$
2. Absolute minimum: $12$ at $x = 8$
3. Absolute maximum: $9\sqrt{2}$ at $x = 6$ and $x = 12$

Let me know if you'd like to explore these calculations further or need clarification!

Follow-up questions:

1. How do you differentiate more complex functions using the chain and quotient rules?
2. What is the significance of critical points in real-world applications?
3. How does the graph of the function reflect the critical points and extreme values?
4. What happens to the behavior of the function as $x$ approaches 4 from the right?
5. How would you find critical points for higher-order rational functions?

Tip: Always check whether the endpoints of an interval are included in your analysis when finding absolute extrema!