To find the $x$ value at which the function $g(x) = x e^{-3x}$ attains a local maximum, we need to follow these steps:

1. Find the derivative $g'(x)$ and set it equal to 0 to find the critical points.
2. Check the second derivative or use the first derivative test to determine if these points are maxima, minima, or neither.

Step 1: First Derivative

Given $g(x) = x e^{-3x}$, we use the product rule to find the derivative: [ g'(x) = \frac{d}{dx} [x] \cdot e^{-3x} + x \cdot \frac{d}{dx} [e^{-3x}] ] $g'(x) = e^{-3x} + x \cdot (-3e^{-3x})$ $g'(x) = e^{-3x} (1 - 3x)$

Step 2: Set the First Derivative Equal to Zero

$e^{-3x} (1 - 3x) = 0$

Since $e^{-3x} \neq 0$ for any real $x$, we set: $1 - 3x = 0$

Solving for $x$: $3x = 1 \quad \Rightarrow \quad x = \frac{1}{3}$

Step 3: Confirm if it is a Maximum

To determine if $x = \frac{1}{3}$ is a maximum, we can use the second derivative test or the sign of the first derivative.

Conclusion

The local maximum occurs at: $x = \frac{1}{3}$

Would you like more details or have any questions about this solution? Here are some related questions for further exploration:

1. What is the second derivative $g''(x)$ and how can we use it to confirm a local maximum?
2. How does changing the coefficient in the exponent affect the location of the maximum?
3. Can this method be extended to find local extrema for more complex exponential functions?
4. What happens to the local maximum if the function is multiplied by a constant?
5. How does the behavior of $g(x)$ change as $x \to \infty$ and $x \to -\infty$?

Tip: When working with exponential functions, remember that they never equal zero, simplifying the process of finding critical points.