To solve for the x-coordinates of the relative extrema of the function $g(x) = \frac{x^6}{e^{9x}}$, we need to find the critical points by setting the derivative equal to zero and analyzing the sign of the derivative to determine whether each critical point corresponds to a maximum or minimum.

Step-by-Step Solution:

1. Differentiate $g(x)$: The function $g(x) = \frac{x^6}{e^{9x}}$ is a quotient, so we apply the quotient rule: $g'(x) = \frac{(e^{9x})(6x^5) - (x^6)(9e^{9x})}{(e^{9x})^2} = \frac{6x^5 e^{9x} - 9x^6 e^{9x}}{e^{18x}}.$ This simplifies to: $g'(x) = \frac{e^{9x}(6x^5 - 9x^6)}{e^{18x}} = \frac{6x^5 - 9x^6}{e^{9x}}.$

2. Set $g'(x) = 0$: For $g'(x) = 0$, we only need the numerator to be zero, since $e^{9x} \neq 0$: $6x^5 - 9x^6 = 0.$ Factor out $3x^5$: $3x^5(2 - 3x) = 0.$ This gives the solutions: $x = 0 \quad \text{and} \quad x = \frac{2}{3}.$

3. Determine the nature of the critical points: To determine whether these critical points correspond to relative maxima or minima, we can use the second derivative test or analyze the sign changes of $g'(x)$ around these points.

   • For $x = 0$: Check values just to the left and right of 0 in $g'(x)$ to see if the derivative changes sign.
   • For $x = \frac{2}{3}$: Similarly, analyze the sign of $g'(x)$ around $x = \frac{2}{3}$.

4. Conclusions: By analyzing the behavior of $g(x)$ around these points, we conclude:

   • $x = 0$ is a relative maximum.
   • $x = \frac{2}{3}$ is a relative minimum.

Final Answer:

• x-coordinates of the relative maxima: $0$
• x-coordinates of the relative minima: $\frac{2}{3}$

Would you like a detailed explanation of the derivative sign analysis or further clarification on any part of this solution?

Here are five questions that expand on this solution:

1. What is the importance of the second derivative test in finding relative extrema?
2. How can the sign of the first derivative help determine the nature of critical points?
3. What would happen if there were no sign changes around the critical points?
4. Can the exponential function in the denominator affect the relative extrema of the function?
5. How does the presence of $x^6$ in the numerator influence the behavior of $g(x)$ at large values of $x$?

Tip: When analyzing critical points, checking the sign of the first derivative around each point can be faster than using the second derivative, especially for complex functions.