1. The problem asks us to analyze the function $f$ based on the graph of its second derivative $f''(x)$. 2. Given that $f''(0) = 4 > 0$, the second derivative is positive at $x=0$. 3. The sign of $f''(x)$ determines the concavity of $f$: if $f''(x) > 0$, $f$ is convex upward (concave up); if $f''(x) < 0$, $f$ is concave downward. 4. Since $f''(0) = 4 > 0$ and the vector points left at $y=4$ on the positive $y$-axis, it suggests $f''(x)$ is positive near $x=0$. 5. For $x < 0$, the problem does not provide explicit values of $f''(x)$, but the vector at $x=0$ is positive, so $f''(x)$ is likely positive near zero from the left side as well. 6. Because $f''(x)$ does not change sign at $x=0$, there is no inflection point at $x=0$. 7. Therefore: - (a) $f$ is convex upward in $]-,0[$ is true if $f''(x) > 0$ there. - (b) $f$ has an inflection point at $x=0$ is false because $f''(x)$ does not change sign. - (c) $f$ is convex upward in $]0,[$ is true since $f''(0) > 0$ and likely positive nearby. - (d) $f$ has no local maximum value cannot be concluded from $f''(x)$ alone. Final conclusion: (a) and (c) are true, (b) and (d) are false. Answer: The curve of the function $f$ is convex upward in both $]-,0[$ and $]0,[$, so statements (a) and (c) are true, (b) and (d) are false.