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.