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The problem asks to find the derivative of \( \frac{f(x)}{x + 6} \) at \( x = 3 \), using the table provided. We can use the **quotient rule** for derivatives to solve this.

### Quotient Rule Formula:
For two functions \( u(x) \) and \( v(x) \), the derivative of their quotient is:

\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
\]

In this case, \( u(x) = f(x) \) and \( v(x) = x + 6 \).

### Step-by-Step Solution:

1. **Identify the functions:**
   - \( u(x) = f(x) \)
   - \( v(x) = x + 6 \)

2. **Find the derivatives:**
   - \( u'(x) = f'(x) \)
   - \( v'(x) = \frac{d}{dx}(x + 6) = 1 \)

3. **Apply the Quotient Rule:**

\[
\frac{d}{dx} \left( \frac{f(x)}{x + 6} \right) = \frac{f'(x) \cdot (x + 6) - f(x) \cdot 1}{(x + 6)^2}
\]

4. **Substitute \( x = 3 \):**
   From the table:
   - \( f(3) = 2 \)
   - \( f'(3) = 4 \)

Substitute these values into the equation:

\[
\frac{d}{dx} \left( \frac{f(x)}{x + 6} \right) \bigg|_{x=3} = \frac{4 \cdot (3 + 6) - 2 \cdot 1}{(3 + 6)^2}
\]

5. **Simplify:**
   - \( 4 \cdot 9 = 36 \)
   - The denominator is \( (3 + 6)^2 = 9^2 = 81 \)

Thus, the derivative at \( x = 3 \) is:

\[
\frac{36 - 2}{81} = \frac{34}{81}
\]

### Final Answer:
\[
\frac{34}{81}
\]

Would you like a further breakdown of any steps? 

Here are 5 related questions to deepen understanding:

1. How does the quotient rule differ from the product rule in calculus?
2. Can you explain why we need to square the denominator in the quotient rule?
3. What would happen if the denominator \( v(x) \) was a more complex function?
4. How do we handle higher-order derivatives using the quotient rule?
5. In what scenarios is the quotient rule most useful in calculus problems?

**Tip:** Always verify that both the numerator and denominator functions are differentiable before applying the quotient rule.

Use the table to the right to find the derivative of f(x)/(x + 6) at x = 3.

Find the derivative of f(x)/(x + 6) at x = 3 using the quotient rule and a table of values. This calculus problem is solved using the quotient rule for derivatives, with f(x) and f'(x) values provided for multiple x-values. Suitable for college-level Calculus I students.

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