To find the expression $\frac{f(x) - f(1)}{x - 1}$, we first need to calculate the function $f(x)$ at $x = 1$ and then substitute the values into the expression.

Step 1: Evaluate $f(1)$

Given the function:

$f(x) = \frac{x + 7}{x + 3}$

Let's compute $f(1)$:

$f(1) = \frac{1 + 7}{1 + 3} = \frac{8}{4} = 2$

Step 2: Substitute into the Expression

Now, the expression we need to evaluate is:

$\frac{f(x) - f(1)}{x - 1} = \frac{\frac{x + 7}{x + 3} - 2}{x - 1}$

Step 3: Simplify the Expression

Let's simplify the numerator:

$f(x) - f(1) = \frac{x + 7}{x + 3} - 2 = \frac{x + 7 - 2(x + 3)}{x + 3}$

Simplify further:

$f(x) - f(1) = \frac{x + 7 - 2x - 6}{x + 3} = \frac{-x + 1}{x + 3} = \frac{-(x - 1)}{x + 3}$

Thus, our expression becomes:

$\frac{f(x) - f(1)}{x - 1} = \frac{\frac{-(x - 1)}{x + 3}}{x - 1}$

Step 4: Further Simplify

We can simplify this as:

$= \frac{-(x - 1)}{(x + 3)(x - 1)}$

Since $x \neq 1$, we can cancel out the $(x - 1)$:

$= \frac{-1}{x + 3}$

Final Answer

The simplified form of the expression is:

$\frac{f(x) - f(1)}{x - 1} = -\frac{1}{x + 3}$

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

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Here are 5 related questions for further exploration:

1. How do you find the derivative of a rational function using the limit definition?
2. What happens to the expression as $x \to -3$?
3. How can you find horizontal and vertical asymptotes for the function $f(x) = \frac{x + 7}{x + 3}$?
4. What are the critical points of $f(x) = \frac{x + 7}{x + 3}$?
5. How do you perform polynomial long division to simplify rational expressions?

Tip: Always check for values where the function is undefined, especially with rational expressions!