Let's tackle the problem shown in your image step by step. You are given the function $f(x) = \frac{x + 6}{x - 7}$, and asked to find:

(a) The average rate of change of $f(x)$ on the interval $[x, x + h]$

The average rate of change of a function $f(x)$ between two points $x$ and $x + h$ is given by:

$\text{Average rate of change} = \frac{f(x+h) - f(x)}{h}$

So, we'll first find $f(x+h)$:

$f(x+h) = \frac{(x+h) + 6}{(x+h) - 7} = \frac{x+h+6}{x+h-7}$

Now, compute the difference between $f(x+h)$ and $f(x)$:

$f(x+h) - f(x) = \frac{x+h+6}{x+h-7} - \frac{x+6}{x-7}$

The next step is to simplify this expression by getting a common denominator:

$f(x+h) - f(x) = \frac{(x+h+6)(x-7) - (x+6)(x+h-7)}{(x+h-7)(x-7)}$

This will give the numerator in terms of $h$. Afterward, divide by $h$ to complete the calculation for the average rate of change.

(b) The derivative of $f(x)$

The derivative of $f(x)$, denoted $f'(x)$, can be found using the quotient rule, since $f(x) = \frac{u(x)}{v(x)}$, where:

• $u(x) = x + 6$
• $v(x) = x - 7$

The quotient rule states:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

We first find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = 1, \quad v'(x) = 1$

Now apply the quotient rule:

$f'(x) = \frac{(1)(x - 7) - (x + 6)(1)}{(x - 7)^2} = \frac{x - 7 - x - 6}{(x - 7)^2} = \frac{-13}{(x - 7)^2}$

Thus, the derivative is:

$f'(x) = \frac{-13}{(x - 7)^2}$

Let me know if you'd like further details or explanations! Here are 5 related questions you could explore next:

1. What is the significance of the average rate of change in real-world applications?
2. How is the derivative interpreted in terms of rate of change?
3. Can you extend this method to find the average rate of change of other functions?
4. How does the quotient rule work in other complex rational functions?
5. What are the conditions for a function to have a well-defined derivative?

Tip: When applying the quotient rule, carefully handle the algebraic steps to avoid common sign errors, particularly when subtracting terms in the numerator.