1. **State the problem:** We are given a piecewise function $$f(x) = \begin{cases} x^3 + 2, & x > 0 \\ 3 - 2x^2, & x < 0 \end{cases}$$ We want to find the intervals where the curve of $f$ is convex upwards. Convex upwards means the second derivative $f''(x) > 0$. 2. **Recall the derivatives:** Given: $$f'(x) = \begin{cases} 3x^2, & x > 0 \\ -4x, & x < 0 \end{cases}$$ $$f''(x) = \begin{cases} 6x, & x > 0 \\ -4, & x < 0 \end{cases}$$ 3. **Analyze convexity for $x > 0$:** For $x > 0$, $$f''(x) = 6x$$ Since $x > 0$, $6x > 0$, so $f$ is convex upwards on the interval $]0, \infty[$. 4. **Analyze convexity for $x < 0$:** For $x < 0$, $$f''(x) = -4$$ Since $-4 < 0$, $f$ is concave (not convex upwards) on $]-\infty, 0[$. 5. **Conclusion:** The function is convex upwards only on the interval $]0, \infty[$. **Final answer:** The curve is convex upwards when $x \in ]0, \infty[$. This corresponds to option (a).