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To determine the inflection points of the function \( f(x) = \frac{x}{e^{6x}} \), we need to identify the values of \( x \) where the second derivative \( f''(x) \) changes sign. This involves the following steps:

1. **Compute \( f'(x) \)**, the first derivative of \( f(x) \).
2. **Compute \( f''(x) \)**, the second derivative of \( f(x) \).
3. **Find the points where \( f''(x) = 0 \) or is undefined**.
4. **Check for sign changes in \( f''(x) \) around these points** to confirm they are inflection points.

Let’s go through these steps.

### Step 1: First Derivative \( f'(x) \)
Using the quotient rule for \( f(x) = \frac{x}{e^{6x}} \):
\[
f'(x) = \frac{e^{6x} \cdot 1 - x \cdot 6e^{6x}}{(e^{6x})^2} = \frac{e^{6x} - 6x e^{6x}}{e^{12x}} = \frac{1 - 6x}{e^{6x}}
\]

### Step 2: Second Derivative \( f''(x) \)
Now we differentiate \( f'(x) = \frac{1 - 6x}{e^{6x}} \) again, applying the quotient rule:
\[
f''(x) = \frac{e^{6x} \cdot (-6) - (1 - 6x) \cdot 6e^{6x}}{(e^{6x})^2} = \frac{-6e^{6x} - 6(1 - 6x)e^{6x}}{e^{12x}}
\]
Simplify this expression:
\[
f''(x) = \frac{-6 - 6 + 36x}{e^{6x}} = \frac{36x - 12}{e^{6x}} = \frac{12(3x - 1)}{e^{6x}}
\]

### Step 3: Find Points Where \( f''(x) = 0 \)
Set \( f''(x) = 0 \):
\[
\frac{12(3x - 1)}{e^{6x}} = 0
\]
Since \( e^{6x} \neq 0 \) for all \( x \), we only need to solve:
\[
3x - 1 = 0 \Rightarrow x = \frac{1}{3}
\]

### Step 4: Verify the Inflection Point
To confirm \( x = \frac{1}{3} \) is an inflection point, we check the sign of \( f''(x) \) on either side of \( x = \frac{1}{3} \):

- For \( x < \frac{1}{3} \), say \( x = 0 \): \( f''(0) = \frac{12(3 \cdot 0 - 1)}{e^{6 \cdot 0}} = -12 \), which is negative.
- For \( x > \frac{1}{3} \), say \( x = 1 \): \( f''(1) = \frac{12(3 \cdot 1 - 1)}{e^{6 \cdot 1}} = \frac{24}{e^6} \), which is positive.

Since \( f''(x) \) changes sign at \( x = \frac{1}{3} \), this point is indeed an inflection point.

### Final Answer
The inflection point occurs at:
\[
x = \frac{1}{3}
\]

Let me know if you’d like more details on any of these steps or have further questions. 

Here are some related questions for deeper understanding:

1. How does the quotient rule work in derivative calculations?
2. What is the significance of an inflection point in a function’s graph?
3. Why is the exponential function \( e^{6x} \) never zero?
4. How do we interpret the sign change of \( f''(x) \) in the context of concavity?
5. Can inflection points exist for functions without sign changes in the second derivative?

**Tip:** Always verify sign changes in the second derivative to confirm an inflection point, as setting \( f''(x) = 0 \) alone isn’t sufficient.

Determine the exact values at which inflection points occur, if any, of the function f(x) = x / e^(6x).

Find the inflection points for the function f(x) = x / e^(6x) by calculating the first and second derivatives. This involves using the quotient rule and verifying sign changes in the second derivative. Suitable for undergraduate calculus students.

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