# Critical Numbers of the Second Derivative

Given $f(x)=x^5+5x^4-40x^2$ I have to find the concavity of the function using the second derivative, and it's critical numbers. But a problem arrived when I graphed the second derivative I saw that at one of the critical numbers ($x=-2$) the function doesn't change concavity. What is $x=-2$ if its not a point of inflection in the graph?

• It's just a point. Nothing too special. Commented Dec 25, 2016 at 19:59

I assume you found that $f''(x)=(x+2)^2(x-1)$. Since the multiplicity of the zero $x=-2$ is even, the sign of $f''(x)$ does not change at $x=-2$, which is why it's not an inflection point.
Think about it this way: For $x < -2$, $f''(x) < 0$ and the function is concave down. However, for $x > -2$, we also have $f''(x) < 0$ and the function is still concave down, so nothing actually changed. Therefore, $x=-2$ isn't really important.
However, for $x < 1$, $f''(x) < 0$ and the function is concave down and then for $x > 1$, $f''(x) > 0$ and the function is concave up. This means the concavity changed, so this is something important because it means $x=1$ is an inflection point.
In short, not all zeroes of $f''(x)$ are important because if the sign does not change, then the concavity does not change and nothing really happens. You should look at all of the zeroes of $f''(x)$ because they could be inflection points, but if the sign of $f''(x)$ does not change at that zero like what happened with $x=-2$ here, then it's not actually an inflection point.