To evaluate the expression $\frac{g(x + h) - g(x)}{h}$ for the function $g(x) = 6 - x^2$, follow these steps:

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

Substitute $x + h$ into the function $g(x)$: $g(x + h) = 6 - (x + h)^2$

Step 2: Expand $g(x + h)$

Expand the square in the expression: $g(x + h) = 6 - \left(x^2 + 2xh + h^2\right)$ Simplify this: $g(x + h) = 6 - x^2 - 2xh - h^2$

Step 3: Find $g(x + h) - g(x)$

Subtract $g(x)$ from $g(x + h)$: $g(x + h) - g(x) = \left(6 - x^2 - 2xh - h^2\right) - \left(6 - x^2\right)$ Simplify the expression by canceling out like terms: $g(x + h) - g(x) = -2xh - h^2$

Step 4: Divide by $h$

Now, divide the expression by $h$: $\frac{g(x + h) - g(x)}{h} = \frac{-2xh - h^2}{h}$ Simplify by dividing each term by $h$: $\frac{g(x + h) - g(x)}{h} = -2x - h$

So, the evaluated expression is: $\frac{g(x + h) - g(x)}{h} = -2x - h$

Would you like to explore more about this problem or have any questions?

Here are 5 related questions for further exploration:

1. What is the significance of the expression $\frac{g(x + h) - g(x)}{h}$ in calculus?
2. How would the answer change if $g(x)$ was a different quadratic function?
3. What happens to the expression as $h$ approaches 0?
4. How does this result relate to the derivative of $g(x)$ at $x$?
5. What if $g(x)$ was a cubic function, how would the process differ?

Tip: When evaluating expressions like $\frac{g(x + h) - g(x)}{h}$, you're essentially finding the average rate of change of the function over the interval $x$ to $x + h$.