The problem asks to show that:

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

where the function $f(x) = x^2 - 3x + 7$.

Step-by-Step Solution:

1. Start by calculating $f(x+h)$: $f(x+h) = (x+h)^2 - 3(x+h) + 7$ Expanding this expression: $f(x+h) = x^2 + 2xh + h^2 - 3x - 3h + 7$

2. Now calculate $f(x+h) - f(x)$: Using $f(x) = x^2 - 3x + 7$, we subtract: $f(x+h) - f(x) = \left( x^2 + 2xh + h^2 - 3x - 3h + 7 \right) - \left( x^2 - 3x + 7 \right)$ Simplifying this: $f(x+h) - f(x) = 2xh + h^2 - 3h$

3. Divide by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - 3h}{h}$ Simplify each term: $\frac{f(x+h) - f(x)}{h} = 2x + h - 3$

4. Conclusion: $\frac{f(x+h) - f(x)}{h} = 2x - 3 + h$ This confirms the given statement.

Let me know if you need further explanation or details!

Here are 5 related questions to deepen understanding:

1. What is the significance of finding $\frac{f(x+h) - f(x)}{h}$ in calculus?
2. How does this relate to the concept of the derivative?
3. What is the derivative of the given function $f(x) = x^2 - 3x + 7$?
4. What happens as $h \to 0$ in this expression?
5. How would you approach the problem if $f(x)$ were a different type of function, like a cubic function?

Tip: The expression $\frac{f(x+h) - f(x)}{h}$ is known as the difference quotient, which leads to the derivative of a function as $h \to 0$.