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Let's go through each part of this problem step-by-step.

### Given Information
We are given:
- A function \( f(x) \) defined for \( x \geq 0 \).
- A table with values of \( f(x) \), \( f'(x) \), and \( f''(x) \) for various intervals of \( x \):
  - At \( x = 0 \): \( f(x) = 2 \)
  - For \( 0 \leq x < 3 \): \( f'(x) < 0 \) and \( f''(x) > 0 \)
  - At \( x = 3 \): \( f(x) = -2 \), \( f'(x) = 0 \), and \( f''(x) = 0 \)
  - For \( x > 3 \): \( f'(x) < 0 \) and \( f''(x) < 0 \)

Using this, we can answer the questions.

### Part (a)
The problem states that the graph of \( y = f(x) \) has a stationary point of inflection.

A stationary point of inflection is a point where:
1. The derivative \( f'(x) = 0 \).
2. The second derivative \( f''(x) \) changes sign (indicating a change in concavity).

From the table, we see that at \( x = 3 \):
- \( f'(3) = 0 \), so it's a stationary point.
- \( f''(3) = 0 \), and the concavity changes from \( f''(x) > 0 \) for \( x < 3 \) to \( f''(x) < 0 \) for \( x > 3 \).

Thus, \( x = 3 \) is a stationary point of inflection.

**Answer for Part (a):** The coordinates of the stationary point of inflection are \( (3, -2) \).

### Part (b)
For \( 0 \leq x < 3 \), we are asked about the concavity of \( y = f(x) \).

From the table, we know:
- For \( 0 \leq x < 3 \): \( f''(x) > 0 \), which means the curve is concave upwards (or convex) on this interval.

**Answer for Part (b):** The correct answer is **concave upwards (convex)**.

### Part (c)
We need to sketch a continuous smooth curve for \( y = f(x) \) for \( x \geq 0 \), using the information from parts (a) and (b).

Here's the process:
1. **At \( x = 0 \):** The value of \( f(x) \) is \( 2 \).
2. **From \( 0 \leq x < 3 \):** \( f(x) \) is decreasing (since \( f'(x) < 0 \)) and concave upwards (since \( f''(x) > 0 \)).
3. **At \( x = 3 \):** The function reaches \( f(x) = -2 \), and there is a stationary point of inflection here, where the graph changes from concave upwards to concave downwards.
4. **For \( x > 3 \):** \( f(x) \) continues to decrease (since \( f'(x) < 0 \)) and is concave downwards (since \( f''(x) < 0 \)).

In your sketch:
- Start at \( (0, 2) \).
- Draw a decreasing, concave-up curve toward \( (3, -2) \).
- At \( x = 3 \), have the curve inflect and switch to concave downwards, continuing to decrease for \( x > 3 \).

Would you like more details on any part of the solution? Here are five additional questions for practice:

1. What defines a stationary point in calculus?
2. How do we identify an inflection point on a graph?
3. Why does \( f(x) \) have to be concave upwards for \( 0 \leq x < 3 \)?
4. What is the significance of the concavity change at \( x = 3 \)?
5. How does the behavior of \( f(x) \) for \( x > 3 \) influence the overall shape of the graph?

**Tip:** Remember, a stationary point of inflection requires both the first and second derivatives to be zero at that point, and a change in concavity around the point.

Consider the function f(x), defined for x ≥ 0. The following information is given: f(0) = 2, f(x) has negative first derivatives and positive second derivatives for 0 ≤ x < 3, f(3) = -2 with both first and second derivatives at zero, and f(x) has negative first and second derivatives for x > 3. Answer the following:

a. The graph of y = f(x) has a stationary point of inflection. What are the coordinates of this stationary point of inflection?

b. Select the correct answer for the following: For 0 ≤ x < 3, the shape of the curve of y = f(x) is concave upwards (convex) or concave downwards.

c. Sketch a continuous smooth curve of the graph of y = f(x) for x ≥ 0, using the given information and your answers to parts a and b.

Explore the function f(x) defined for x ≥ 0, with a focus on finding the stationary point of inflection and analyzing the concavity for intervals 0 ≤ x < 3 and x > 3. This calculus problem includes differentiation concepts and a sketch of the function’s behavior.

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