Evaluate the difference quotient for the given function. Simplify your answer.
f(x) =
x + 8/x + 3,    f(x) − f(2)/x − 2

To solve the problem of evaluating the difference quotient for the function \( f(x) = \frac{x + 8}{x + 3} \) at the point \( x = 2 \), we need to calculate the following expression:

\[
\frac{f(x) - f(2)}{x - 2}
\]

### Step 1: Calculate \( f(2) \)
Substitute \( x = 2 \) into the function \( f(x) \):

\[
f(2) = \frac{2 + 8}{2 + 3} = \frac{10}{5} = 2
\]

### Step 2: Substitute \( f(x) \) and \( f(2) \) into the difference quotient
The difference quotient becomes:

\[
\frac{f(x) - 2}{x - 2}
\]

Where \( f(x) = \frac{x + 8}{x + 3} \).

So,

\[
\frac{\frac{x + 8}{x + 3} - 2}{x - 2}
\]

### Step 3: Simplify the expression
First, simplify the numerator \( \frac{x + 8}{x + 3} - 2 \):

\[
\frac{x + 8}{x + 3} - 2 = \frac{x + 8 - 2(x + 3)}{x + 3} = \frac{x + 8 - 2x - 6}{x + 3} = \frac{-x + 2}{x + 3}
\]

Now, the difference quotient is:

\[
\frac{\frac{-x + 2}{x + 3}}{x - 2}
\]

This can be rewritten as:

\[
\frac{-x + 2}{(x + 3)(x - 2)}
\]

### Final Answer
The simplified form of the difference quotient is:

\[
\frac{2 - x}{(x + 3)(x - 2)}
\]

Would you like more details, or do you have any questions about this process?

### Related Questions:
1. How do you find the difference quotient for a polynomial function?
2. What are the applications of the difference quotient in calculus?
3. How does the difference quotient relate to the derivative of a function?
4. What happens to the difference quotient as \( x \) approaches 2 in this case?
5. Can the difference quotient be used to determine the rate of change in real-world scenarios?

**Tip:** Understanding how to simplify complex fractions is crucial for solving difference quotients efficiently.

Learn how to evaluate the difference quotient for the function f(x) = (x + 8) / (x + 3). Step-by-step solution provided, including calculating f(2), simplifying the expression, and understanding its application in calculus. Suitable for college-level students and calculus enthusiasts.

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