Consider the function $y(x)=x^4-x^3+2$. It holds that
- $y'(x)=4x^3-3x^2$;
- $y''(x)=12x^2-6x$.
To find the stationary points of $y(x)$ we solve $y'(x)=0$.
$$y'(x)=0\Leftrightarrow 4x^3-3x^2=0\Leftrightarrow x^2(4x-3)=0\Leftrightarrow x=0\mbox{ or } x=\frac{3}{4}.$$
Plugging these two points into $y''(x)$ gives the following results.
- $y''(0)=0$. A sign chart of $y'(x)$ shows that the function is decreasing for all $x\leq \frac{3}{4}$. Therefore, $x=0$ is an inflection point and not an extremum location;
- $y''(\frac{3}{4})>0$, hence $x=\frac{3}{4}$ is a minimum location.
From this it follows that $y(\frac{3}{4})=1\frac{229}{256}$ is a minimum of the function $y(x)$.