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To solve this problem, we need to determine when the function \( f(x) = x^3 \ln(x) \) is concave down.

### Step 1: Find the second derivative
1. **First derivative:** Use the product rule for differentiation:
   \[
   f'(x) = \frac{d}{dx}(x^3 \ln(x)) = 3x^2 \ln(x) + x^2.
   \]

2. **Second derivative:** Differentiate \( f'(x) \) using the product rule again:
   \[
   f''(x) = \frac{d}{dx}(3x^2 \ln(x) + x^2) = 6x \ln(x) + 5x.
   \]

### Step 2: Determine concavity
- A function is concave down where its second derivative is less than zero:
  \[
  f''(x) = 6x \ln(x) + 5x < 0.
  \]
- Factor out \( x \):
  \[
  x(6 \ln(x) + 5) < 0.
  \]

### Step 3: Find critical points
- For \( f''(x) \) to be negative, the term inside the parentheses should be negative:
  \[
  6 \ln(x) + 5 < 0 \implies \ln(x) < -\frac{5}{6}.
  \]
- Solve for \( x \):
  \[
  x < e^{-5/6}.
  \]

### Step 4: Determine the interval
Since \( x > 0 \) due to the domain of \( \ln(x) \), the interval where the function is concave down is:
\[
(0, e^{-5/6}).
\]

### Final Answer:
The correct choice is **(G) \((0, e^{-5/6})\)**.

Would you like a more detailed explanation of any of the steps? Here are some related questions:

1. What is the role of the second derivative in determining concavity?
2. How does the domain of the logarithmic function affect the solution?
3. What happens to the concavity of \( f(x) \) outside the interval \( (0, e^{-5/6}) \)?
4. How would the solution change if the function were \( x^2 \ln(x) \) instead?
5. Can you explain the product rule in differentiation?

**Tip:** Always remember to check the domain of logarithmic functions, as they are only defined for positive values of \( x \).

Let f(x) = x^3 ln x. On what interval is f concave down?

Explore how to find the interval where the function f(x) = x^3 ln x is concave down. This problem demonstrates key calculus concepts like differentiation and concavity, including step-by-step solutions and explanations. Suitable for high school students in Grades 11-12.

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