Let's start by solving the first problem:

Problem #90:

For the function $f(x) = 4x - 2x^2$, find $f(a + h) - f(a)$.

1. Find $f(a + h)$: Substitute $x = a + h$ into the function $f(x) = 4x - 2x^2$:

   $f(a + h) = 4(a + h) - 2(a + h)^2$

   Now, expand the expression:

   $f(a + h) = 4a + 4h - 2[(a + h)(a + h)]$ $f(a + h) = 4a + 4h - 2(a^2 + 2ah + h^2)$ $f(a + h) = 4a + 4h - 2a^2 - 4ah - 2h^2$ $f(a + h) = 4a - 2a^2 + 4h - 4ah - 2h^2$

2. Find $f(a)$: Now, substitute $x = a$ into the original function:

   $f(a) = 4a - 2a^2$

3. Subtract $f(a)$ from $f(a + h)$:

   $f(a + h) - f(a) = (4a - 2a^2 + 4h - 4ah - 2h^2) - (4a - 2a^2)$

   Simplifying:

   $f(a + h) - f(a) = 4a - 2a^2 + 4h - 4ah - 2h^2 - 4a + 2a^2$ $f(a + h) - f(a) = 4h - 4ah - 2h^2$

Thus, the final answer is:

$f(a + h) - f(a) = 4h - 4ah - 2h^2$

Would you like me to solve the other problems as well? Let me know if you want more details or clarifications.

Related Questions:

1. What is the significance of the difference quotient in calculus?
2. How can the expression $f(a + h) - f(a)$ be used to find the derivative of a function?
3. What happens when $h \to 0$ in this problem?
4. How do we simplify expressions involving binomials like $(a + h)^2$?
5. How can this process be extended to more complex functions?

Tip: The difference $f(a + h) - f(a)$ is a crucial part of the definition of the derivative, which is the rate of change of the function at $x = a$.