TITLE: Is it possible to find inflection points by setting the first derivative to 0?
QUESTION [5 upvotes]: I have the following
$$y = \frac{x^2}{2}-\ln x$$
$$y'= x - \frac1x$$
I learned that inflection points were found by setting the $2^{nd}$ derivative equal to $0$, however, if I do that in this case I would get $i$, and I already checked and such is not possible in this case. However when I sent the $1^{st}$ derivative equal to $0$ I get,
$$x = \pm1$$
as possible inflection points which makes more sense.
Do I have a misconception as to how to find inflection points ? Or am I missing something ?

REPLY [9 votes]: An inflection point is where the sign of curvature changes, from concave up to down or vice versa, hence the necessity of the vanishing second derivative.
But $f(x) = \frac{x^2}{2} - \ln x$ is concave up (convex) before and after the stationary point $x = 1$. This can be seen from the graph, or by noting that for all $x \in \mathbb{R}$ $$f''(x) = 1 + \frac{1}{x^2} > 0$$ 
So there can't be any inflection point.