https://file.aiquickdraw.com/mathbot/file/1727881664141ft5j7rb0.jpg

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.

Find detailed solutions to derivative problems using the limit definition. Topics include linear and square root functions, and the application of the limit process to compute the derivative. Suitable for Grades 10-12.

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