The given problem is asking to find and simplify the following expression:

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

where the function $f(x) = 2x^2 - 1$.

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

We substitute $x = a+h$ into the function $f(x) = 2x^2 - 1$:

$f(a+h) = 2(a+h)^2 - 1$

Now expand $(a+h)^2$:

$f(a+h) = 2(a^2 + 2ah + h^2) - 1 = 2a^2 + 4ah + 2h^2 - 1$

Step 2: Find $f(a)$

Substitute $x = a$ into the function $f(x) = 2x^2 - 1$:

$f(a) = 2a^2 - 1$

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

Now subtract $f(a)$ from $f(a+h)$:

$f(a+h) - f(a) = (2a^2 + 4ah + 2h^2 - 1) - (2a^2 - 1)$

Simplify the expression:

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

Step 4: Divide by $h$

Now divide the result by $h$:

$\frac{f(a+h) - f(a)}{h} = \frac{4ah + 2h^2}{h}$

Simplify the expression:

$\frac{f(a+h) - f(a)}{h} = 4a + 2h$

Final Answer:

$\frac{f(a+h) - f(a)}{h} = 4a + 2h$

Would you like more details on any part of the solution?

Here are 5 related questions to expand on this topic:

1. How do you interpret the expression $\frac{f(a+h) - f(a)}{h}$ in the context of calculus?
2. Can you explain how to apply the limit as $h \to 0$ to this expression?
3. What is the significance of the difference quotient in terms of derivatives?
4. How would this problem change if the function $f(x)$ had a higher degree, say $f(x) = 3x^3 - 1$?
5. How does expanding $(a+h)^2$ apply to other quadratic expressions?

Tip: The expression $\frac{f(a+h) - f(a)}{h}$ is the foundation of finding derivatives, which measure the rate of change of a function at a point.