Let's go through each of these problems step-by-step.

Problem 2

The question asks for the slope of the tangent line to the graph of $f(x) = x^2$ at the point $A(2, 4)$.

1. First, find the derivative of $f(x) = x^2$:
   $f'(x) = 2x$

2. Substitute $x = 2$ into $f'(x)$ to find the slope at that point:
   $f'(2) = 2 \times 2 = 4$

Answer for Problem 2: A) 4

Problem 3

We need to match each function with its derivative.

Given functions:

1. $y = x^4$
2. $y = \sin x$
3. $y = a^x$

List of derivatives:

• $A. \, y = -\cos x$
• $B. \, y = \frac{a^x}{\ln a}$
• $C. \, y = \frac{x^5}{5}$
• $D. \, y = \cos x$
• $E. \, y = a^x \ln a$
• $F. \, y = 4x^3$

Matching:

1. For $y = x^4$, the derivative is $y' = 4x^3$, which corresponds to F.
2. For $y = \sin x$, the derivative is $y' = \cos x$, which corresponds to D.
3. For $y = a^x$, the derivative is $y' = a^x \ln a$, which corresponds to E.

Answer for Problem 3: D) I-F, II-D, III-E

Problem 4

The question asks to compute the following limit: $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3}$

1. Factor the numerator: $x^2 - 5x + 6 = (x - 3)(x - 2)$

2. Substitute back into the limit: $\lim_{x \to 3} \frac{(x - 3)(x - 2)}{x - 3}$

3. Simplify by canceling $x - 3$: $= \lim_{x \to 3} (x - 2)$

4. Now substitute $x = 3$: $= 3 - 2 = 1$

Answer for Problem 4: C) 1

Summary of Answers:

• Problem 2: A) 4
• Problem 3: D) I-F, II-D, III-E
• Problem 4: C) 1

Would you like further details on any of these steps, or do you have additional questions?

Related Questions:

1. What is the derivative of $f(x) = x^3$ at any point $x$?
2. How do you find the tangent line to a curve at a specific point?
3. What is the chain rule in differentiation?
4. How can you simplify limits involving indeterminate forms?
5. What is the derivative of an exponential function with base $a$?

Tip:

When calculating a limit, always check if the function can be factored or simplified first to avoid indeterminate forms like $\frac{0}{0}$.