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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.

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### 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**.

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### 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.

The graph of the function f and a table of selected values of f(x) are shown. Approximate f'(4.5), find if there’s a value of x for which the derivative is zero, and evaluate three different limit expressions at x = 5.

Learn how to approximate the derivative at f'(4.5) using given data, identify a point where the derivative is zero, and evaluate right-hand and left-hand limits for a function at x = 5. This problem involves differentiation, limits, and analyzing concavity.

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