Consider the function $y(x)=x^4-x^3+2$. It holds that
- $y'(x)=4x^3-3x^2$;
- $y''(x)=12x^2-6x$.
Solving $y''(x)=0$ gives
$$12x^2-6x=0\Leftrightarrow 12x(x-\frac{1}{2})=0\Leftrightarrow x=0\mbox{ or } x=\frac{1}{2}.$$
We distinguish the following cases:
- $y''(x)\geq 0$ for every $x\leq 0$;
- $y''(x)=0$ for $x=0$;
- $y''(x)\leq 0$ for every $0\leq x\leq \frac{1}{2}$;
- $y''(x)=0$ for $x=\frac{1}{2}$;
- $y''(x)\geq 0$ for every $x\geq \frac{1}{2}$.
Conclusion: $y(x)$ is concave on the interval $[0,\frac{1}{2}]$, $y(x)$ is convex on the intervals $(-\infty,0]$ and $[\frac{1}{2},\infty)$, and $x=0$ and $x=\frac{1}{2}$ are inflection points.