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.