We are tasked with constructing and simplifying the difference quotient for the function $f(x) = 8 - x^2$.

The difference quotient formula is:

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

Let's break this down step by step:

Step 1: Find $f(x+h)$

Substitute $x+h$ into the function $f(x)$: $f(x+h) = 8 - (x+h)^2$ Now expand $(x+h)^2$: $(x+h)^2 = x^2 + 2xh + h^2$ Thus: $f(x+h) = 8 - (x^2 + 2xh + h^2)$ Simplify: $f(x+h) = 8 - x^2 - 2xh - h^2$

Step 2: Find $f(x+h) - f(x)$

Now subtract $f(x)$ from $f(x+h)$: $f(x+h) - f(x) = (8 - x^2 - 2xh - h^2) - (8 - x^2)$ Simplify the expression: $f(x+h) - f(x) = - 2xh - h^2$

Step 3: Divide by $h$

Now divide by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2}{h}$ Simplify: $\frac{f(x+h) - f(x)}{h} = -2x - h$

Final Simplified Difference Quotient:

$\frac{f(x+h) - f(x)}{h} = -2x - h$

Would you like further details or have any questions?

Here are five related questions to explore further:

1. How would the difference quotient change if the function was $f(x) = 8 - x^3$?
2. What happens to the difference quotient as $h$ approaches 0?
3. How does the difference quotient relate to the derivative of the function?
4. Can the process for constructing the difference quotient be applied to trigonometric functions?
5. How would the result change if we were dealing with a function of multiple variables?

Tip: The difference quotient gives an average rate of change between two points, and as $h$ approaches 0, it approaches the derivative, which represents the instantaneous rate of change.