1. The problem states that the graph of the second derivative $f''$ of a continuous function $f$ is given on the interval $[0,9]$, and $f$ has only one critical point at $x=6$. 2. Recall that a critical point of $f$ occurs where the first derivative $f'$ is zero or undefined. Here, $x=6$ is the only critical point. 3. To classify the critical point, we use the second derivative test: $$\text{If } f''(x) > 0, \text{ then } f \text{ is concave up and } f \text{ has a relative minimum at } x.$$ $$\text{If } f''(x) < 0, \text{ then } f \text{ is concave down and } f \text{ has a relative maximum at } x.$$ 4. From the graph of $f''$, at $x=6$, $f''(6) > 0$ (the graph is above the $x$-axis), so $f$ is concave up at $x=6$. 5. Therefore, $f$ has a relative minimum at $x=6$. 6. To determine if this is an absolute minimum, consider the behavior of $f''$ and $f'$ on the interval: - Since $x=6$ is the only critical point, $f'$ changes sign only there. - The concavity changes and the shape of $f''$ suggest $f$ decreases before $6$ and increases after $6$, consistent with a relative minimum. - However, without information about $f$ at the endpoints, we cannot confirm if this minimum is absolute. 7. Hence, the correct statement is: (A) $f$ has a relative minimum at $x=6$ but not an absolute minimum. Final answer: (A)