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.