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The problem provided involves analyzing the difference quotient for the function \( f(x) = \frac{1}{x + 8} \). We need to simplify the expression

\[
\frac{f(6 + h) - f(6)}{h}
\]

and then express it in the form \( \frac{-1}{ah + b} \).

### Step-by-Step Solution:

1. **Substitute \( f(x) \) into the difference quotient:**

   Since \( f(x) = \frac{1}{x + 8} \), we can compute \( f(6 + h) \) and \( f(6) \) as follows:
   
   - \( f(6 + h) = \frac{1}{(6 + h) + 8} = \frac{1}{h + 14} \)
   - \( f(6) = \frac{1}{6 + 8} = \frac{1}{14} \)

   So, the difference quotient becomes:

   \[
   \frac{f(6 + h) - f(6)}{h} = \frac{\frac{1}{h + 14} - \frac{1}{14}}{h}
   \]

2. **Simplify the numerator:**

   Find a common denominator for \( \frac{1}{h + 14} - \frac{1}{14} \):

   \[
   \frac{1}{h + 14} - \frac{1}{14} = \frac{14 - (h + 14)}{14(h + 14)} = \frac{14 - h - 14}{14(h + 14)} = \frac{-h}{14(h + 14)}
   \]

   So now we have:

   \[
   \frac{f(6 + h) - f(6)}{h} = \frac{\frac{-h}{14(h + 14)}}{h} = \frac{-h}{14(h + 14) \cdot h}
   \]

3. **Cancel \( h \) from the numerator and denominator:**

   \[
   \frac{-h}{14(h + 14) \cdot h} = \frac{-1}{14(h + 14)}
   \]

4. **Final Form:**

   Now we have simplified the expression to:

   \[
   \frac{f(6 + h) - f(6)}{h} = \frac{-1}{14(h + 14)}
   \]

### Answer:

The quotient can be simplified to:

\[
\frac{-1}{ah + b} \quad \text{where } a = 14 \text{ and } b = 196.
\]

Would you like further details on this solution? 

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### Related Questions:

1. What is the difference quotient, and why is it used in calculus?
2. How would the solution change if the function were \( f(x) = \frac{1}{x + c} \) with a different constant \( c \)?
3. What is the significance of finding the limit as \( h \to 0 \) in the difference quotient?
4. How does this result relate to finding the derivative of \( f(x) = \frac{1}{x + 8} \)?
5. What other techniques could be used to simplify complex rational expressions?

### Tip:

In difference quotient problems, always simplify the numerator carefully by finding a common denominator to avoid errors in handling complex fractions.

Let f(x) be the function 1/(x + 8). Then the quotient (f(6 + h) - f(6)) / h can be simplified to -1 / (ah + b) for:

Explore how to simplify the difference quotient for the function f(x) = 1 / (x + 8). This solution includes function evaluation, algebraic simplification, and the use of the difference quotient formula, suitable for students in Grades 11-12.

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