Problem: Determine which of the following compositions are convex or concave for $x \ge 0$: $f(x)=x^3$, $g(x)=x^2$, $h(x)=x^{1/3}$. 1. Compute $f(g(x))$. $f(g(x)) = (x^2)^3 = x^6$. The second derivative is $\frac{d^2}{dx^2}x^6 = 30x^4 \ge 0$ for $x \ge 0$. Hence A is convex. 2. Compute $g(f(x))$. $g(f(x)) = (x^3)^2 = x^6$. The second derivative is the same, $30x^4 \ge 0$ for $x \ge 0$. Hence B is convex. 3. Compute $h(f(x))$. $h(f(x)) = (x^3)^{1/3} = x$ for $x \ge 0$. A linear function is both convex and concave, so C is convex (true). 4. Compute $f(h(x))$. $f(h(x)) = (x^{1/3})^3 = x$ for $x \ge 0$. Again this is linear and therefore concave as well, so D is true. 5. Compute $h(g(x))$. $h(g(x)) = (x^2)^{1/3} = x^{2/3}$ for $x \ge 0$. The first derivative is $\frac{d}{dx}x^{2/3} = \frac{2}{3}x^{-1/3}$. The second derivative is $\frac{d^2}{dx^2}x^{2/3} = -\frac{2}{9}x^{-4/3} \le 0$ for $x>0$. Therefore $x^{2/3}$ is concave on $(0,\infty)$ and concave on $[0,\infty)$ in the usual sense, so E is true. Conclusion: All statements A, B, C, D, E are correct.