Let's work through the given AP Calculus problem step by step.

Problem Overview:

The graph of a function $f$ is given, alongside a table of selected values of $f(x)$. The graph and the table offer information about the behavior of the function around specific points.

We are asked to address the following parts: (a) Approximate the value of $f'(4.5)$ using data from the table. (b) Determine if there is a value of $x$, for $0 < x < 5$, such that $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 0$. (c) Evaluate three different limit expressions.

Part (a): Approximate $f'(4.5)$

To approximate $f'(4.5)$, we'll use the definition of the derivative: $f'(4.5) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}$ We can use values of $f(x)$ near $x = 4.5$ from the table.

From the table, we have: $f(4) = 2.533 \quad \text{and} \quad f(5) = 1$

Now, calculate the average rate of change between $x = 4$ and $x = 5$: $f'(4.5) \approx \frac{f(5) - f(4)}{5 - 4} = \frac{1 - 2.533}{1} = -1.533$ So, $f'(4.5) \approx -1.533$.

Part (b): Is there a value $x$ such that $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 0$?

The expression given is the definition of the derivative, $f'(x)$. The question asks if there is a value of $x$ for which the derivative is equal to zero.

From the graph, a horizontal tangent (which indicates $f'(x) = 0$) occurs at $x = 2$, as stated in the problem description.

Therefore, the value of $x$ where $f'(x) = 0$ is $x = 2$.

Part (c): Evaluate the given limits

1. $\lim_{h \to 0^+} \frac{f(5+h) - f(5)}{h}$

   This limit represents the right-hand derivative at $x = 5$. Using the table, we have:

   • $f(5) = 1$
   • $f(5.5) = 1.125$

   Approximate the derivative: $\frac{f(5.5) - f(5)}{5.5 - 5} = \frac{1.125 - 1}{0.5} = 0.25$

   So, $\lim_{h \to 0^+} \frac{f(5+h) - f(5)}{h} = 0.25$.

2. $\lim_{h \to 0^-} \frac{f(5+h) - f(5)}{h}$

   This limit represents the left-hand derivative at $x = 5$. Using the table, we have:

   • $f(5) = 1$
   • $f(4.5) = 3.193$

   Approximate the derivative: $\frac{f(4.5) - f(5)}{4.5 - 5} = \frac{3.193 - 1}{-0.5} = \frac{2.193}{-0.5} = -4.386$

   So, $\lim_{h \to 0^-} \frac{f(5+h) - f(5)}{h} = -4.386$.

3. $\lim_{h \to 0} \frac{f(5+h) - f(5)}{h}$

   Since the right-hand and left-hand derivatives at $x = 5$ do not match, the derivative does not exist at $x = 5$. Hence, this limit does not exist.

Conclusion:

• (a) $f'(4.5) \approx -1.533$
• (b) $f'(2) = 0$, so the value of $x$ is 2.
• (c)(i) $\lim_{h \to 0^+} \frac{f(5+h) - f(5)}{h} = 0.25$
• (c)(ii) $\lim_{h \to 0^-} \frac{f(5+h) - f(5)}{h} = -4.386$
• (c)(iii) The limit does not exist.

Do you need more details on any part of this solution? Here are five related questions to explore this concept further:

1. How do we know when to use a difference quotient to approximate a derivative?
2. Why is the limit of the derivative different from the left and right at $x = 5$?
3. What is the significance of concavity and how does it affect derivative signs?
4. How can we identify points where the derivative of a function is zero on a graph?
5. What conditions cause the derivative of a function to not exist at a point?

Tip: When approximating derivatives from data points, make sure to select values of $x$ that are as close as possible to the point of interest to get a better approximation.