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Let's solve the problem in the image: 

We are given the function \( f(x) = x^3 \ln(x) \) and asked to find on what interval(s) the function is concave down.

### Step 1: Find the second derivative of \( f(x) \).

The function is \( f(x) = x^3 \ln(x) \).

#### First Derivative:
To find concavity, we first need the first derivative, \( f'(x) \). We use the product rule since \( f(x) = x^3 \cdot \ln(x) \).

\[
f'(x) = 3x^2 \ln(x) + x^2
\]
(This is applying the product rule and derivative of \( \ln(x) \)).

#### Second Derivative:
Next, we differentiate \( f'(x) = 3x^2 \ln(x) + x^2 \).

Using the product rule again for \( 3x^2 \ln(x) \), we get:

\[
f''(x) = 6x \ln(x) + 3x + 2x
\]
\[
f''(x) = 6x \ln(x) + 5x
\]

### Step 2: Set the second derivative equal to zero to find inflection points.
We want to find when \( f''(x) = 0 \), which will help us determine when the concavity changes.

\[
6x \ln(x) + 5x = 0
\]
Divide both sides by \( x \) (assuming \( x > 0 \)):

\[
6 \ln(x) + 5 = 0
\]
Solve for \( \ln(x) \):

\[
\ln(x) = -\frac{5}{6}
\]

Exponentiate both sides to solve for \( x \):

\[
x = e^{-5/6}
\]

### Step 3: Analyze the concavity.
To determine where the function is concave down, we analyze the sign of \( f''(x) \) around the critical point \( x = e^{-5/6} \).

- For \( x < e^{-5/6} \), \( \ln(x) \) is negative, making \( f''(x) \) positive, so the function is concave up.
- For \( x > e^{-5/6} \), \( \ln(x) \) is positive, making \( f''(x) \) negative, so the function is concave down.

### Step 4: Conclusion
The function is concave down for \( x > e^{-5/6} \).

Thus, the correct answer is **(B) \( (e^{-5/6}, \infty) \)**.

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Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. What is the interval of concavity for \( f(x) = x^2 \ln(x) \)?
2. How do you find the points of inflection for \( f(x) = x^3 \ln(x) \)?
3. What does it mean for a function to be concave up or concave down?
4. How would the solution change if the function were \( f(x) = x^3 \ln(2x) \)?
5. How do you determine whether a function is increasing or decreasing?

**Tip**: Always check the behavior of the second derivative around critical points to determine intervals of concavity.

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

Explore how to determine the interval of concavity for the function f(x) = x^3 ln(x). This problem includes finding the second derivative, analyzing inflection points, and identifying when the function is concave down.

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