Let's go through the odd-numbered problems one by one from the provided image.

1. $f(x) = \frac{1}{x}, \, a = 2$

We use the definition of the derivative: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

For $f(x) = \frac{1}{x}$ and $a = 2$: $f'(2) = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h}$

Simplify the expression: $f'(2) = \lim_{h \to 0} \frac{\frac{2 - (2+h)}{(2+h) \cdot 2}}{h} = \lim_{h \to 0} \frac{-h}{2(2+h)h} = \lim_{h \to 0} \frac{-1}{2(2+h)} = \frac{-1}{4}$

Thus, $f'(2) = -\frac{1}{4}$.

3. $f(x) = 3 - x^2, \, a = -1$

Using the same definition: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

For $f(x) = 3 - x^2$ and $a = -1$: $f'(-1) = \lim_{h \to 0} \frac{(3 - (-1+h)^2) - (3 - (-1)^2)}{h}$

Simplify: $f'(-1) = \lim_{h \to 0} \frac{(3 - (1 - 2h + h^2)) - (3 - 1)}{h} = \lim_{h \to 0} \frac{(2 + 2h - h^2) - 2}{h} = \lim_{h \to 0} \frac{2h - h^2}{h} = \lim_{h \to 0} (2 - h) = 2$

Thus, $f'(-1) = 2$.

5. $f(x) = \frac{1}{x}, \, a = 2$ (similar to problem 1)

We already computed this derivative in problem 1: $f'(2) = -\frac{1}{4}$

7. $f(x) = \sqrt{x + 1}, \, a = 3$

Again, using the definition of the derivative: $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

For $f(x) = \sqrt{x + 1}$ and $a = 3$: $f'(3) = \lim_{x \to 3} \frac{\sqrt{x+1} - \sqrt{4}}{x - 3} = \lim_{x \to 3} \frac{\sqrt{x+1} - 2}{x - 3}$

Multiply by the conjugate: $f'(3) = \lim_{x \to 3} \frac{(\sqrt{x+1} - 2)(\sqrt{x+1} + 2)}{(x - 3)(\sqrt{x+1} + 2)} = \lim_{x \to 3} \frac{x + 1 - 4}{(x - 3)(\sqrt{x+1} + 2)} = \lim_{x \to 3} \frac{x - 3}{(x - 3)(\sqrt{x+1} + 2)}$

Simplify: $f'(3) = \lim_{x \to 3} \frac{1}{\sqrt{x+1} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$

Thus, $f'(3) = \frac{1}{4}$.

9. Find $f'(x)$ if $f(x) = 3x - 12$

For a linear function $f(x) = 3x - 12$, the derivative is simply the coefficient of $x$, which is 3. Therefore: $f'(x) = 3$

Would you like more details on any of these solutions? Here are five follow-up questions:

1. How does multiplying by the conjugate help in simplifying limits?
2. What is the geometric interpretation of the derivative for $f(x) = 3x - 12$?
3. Why is the limit definition crucial for understanding the concept of a derivative?
4. How does the derivative of a square root function differ from a polynomial function?
5. Can the limit definition of a derivative be applied to all types of functions?

Tip: When dealing with square roots in limits, rationalizing using conjugates often simplifies the calculation.