Let's analyze the problem step by step:

The function is $f(x) = 6 + 2x^2$. You're asked to calculate the following:

1. $f(a)$
2. $f(a + h)$
3. $\frac{f(a + h) - f(a)}{h}$

Step 1: Find $f(a)$:

Substitute $x = a$ into $f(x) = 6 + 2x^2$: $f(a) = 6 + 2a^2$

Step 2: Find $f(a + h)$:

Substitute $x = a + h$ into $f(x) = 6 + 2x^2$: $f(a + h) = 6 + 2(a + h)^2$ Expand $(a + h)^2$: $(a + h)^2 = a^2 + 2ah + h^2$ So: $f(a + h) = 6 + 2(a^2 + 2ah + h^2)$ $f(a + h) = 6 + 2a^2 + 4ah + 2h^2$

Step 3: Calculate $\frac{f(a + h) - f(a)}{h}$:

Using $f(a + h) = 6 + 2a^2 + 4ah + 2h^2$ and $f(a) = 6 + 2a^2$, subtract $f(a)$ from $f(a + h)$: $f(a + h) - f(a) = (6 + 2a^2 + 4ah + 2h^2) - (6 + 2a^2)$ Simplify: $f(a + h) - f(a) = 4ah + 2h^2$ Divide by $h$: $\frac{f(a + h) - f(a)}{h} = \frac{4ah + 2h^2}{h}$ Factor out $h$ from the numerator: $\frac{f(a + h) - f(a)}{h} = 4a + 2h$

Final Answers:

1. $f(a) = 6 + 2a^2$
2. $f(a + h) = 6 + 2a^2 + 4ah + 2h^2$
3. $\frac{f(a + h) - f(a)}{h} = 4a + 2h$

Would you like a deeper explanation of any step or help with similar problems? 😊

Related Questions:

1. What is the significance of the expression $\frac{f(a + h) - f(a)}{h}$?
2. How does this relate to the concept of derivatives in calculus?
3. Can you find $f'(x)$ directly for $f(x) = 6 + 2x^2$?
4. What happens to $\frac{f(a + h) - f(a)}{h}$ as $h \to 0$?
5. What are other examples of functions where similar steps are applied?

Tip:

When simplifying expressions like $\frac{f(a + h) - f(a)}{h}$, always fully expand and simplify terms before dividing by $h$ to avoid errors.