Determine the inflection points of $f(x)=x^4-10x^3$.

We have seen this function a couple of times before when we found its maxima and minima and determined its concavity. This time we are going to find it's inflection points by first finding the points where $f\,''(x)=0$ and then looking at each of those points to see if $f\,''(x)$ changes on each side.

We have used this function several times and know that $f\,'(x)=4x^3-30x^2$ and $f\,''(x)=12x^2-60x$. Now let's find the values of $x$ for which $f\,''(x)=0$. Let's do some quick algebra:
$$\begin{eqnarray} 12x^2-60x&=&0\\ x(12x-60)&=&0. \end{eqnarray}$$
Now we can easily see that $f\,''(x)=0$ when $x=0$ and $x=5$. (Recall $12\cdot 5=60$.) So we now just need to examine the second derivative when $x$ is negative (i.e. left of $x=0$), when $0\lt x\lt 5$ (i.e. to the right of zero and left of $5$) and when $x\gt 5$ (i.e. to the right of $5$).

Our arguments are now identical to when we determined its concavity earlier. We are just doing the same things again.

The first thing to note is that when $x\lt 0$ then $f\,''(x)\gt 0$. This is because $12x-60$ is negative and a negative number times a negative number ($x$) is always positive. So when $x$ is negative $f(x)$ is concave up.

Now let's look at $0\lt x\lt 5$. In this situation $x$ is (obviously) positive and $12x-60$ is negative. This means that $f\,''(x)\lt 0$ and is concave down. So to the left of $x=0$ we are concave up and to the right we are concave down. This means that at $x=0$ we have a change in concavity and $x=0$ is an inflection point.

When $x\gt 5$, then $12x-60\gt 0$ (positive) making $f\,''(x)\gt 0$. Therefore, when $x\gt 5$, $f(x)$ is concave up. Again, at $x=5$ the concavity has changed so $x=5$ is an inflection point.