If $f$ continuous such that $f'(0) = 0$ and $f'(x) > 0 $ on $(-\infty, 0) \cup (0, \infty),$ then is $x = 0$ an inflection point? If $f$ is a continuous function such that $f'(0) = 0$ and $f'(x) > 0 $ on $(-\infty, 0) \cup (0, \infty),$ then is there always an inflection point at $x = 0?$ I know that with the First Derivative Test, there is neither a max or min at $x = 0,$ which the original question was asking about, but I was then thinking about whether there would have to be an inflection point there, and I can't seem to think of a counterexample where there would not be one. Can anyone confirm or deny that there is always an inflection point at $x = 0?$
The definition of an inflection point that I'm using is a point in the domain of $f$ where the concavity changes from concave up to concave down or concave down to concave up.
 A: By hypothesis, $f' > 0$ except at $0$ and $f'(0) = 0$, so the derivative $f'$ has an absolute minimum at $0$.
If we assume $f''$ is continuous (which may be part of your definition of inflection point), the first derivative test applied to $f'$ guarantees there exists a real $\delta > 0$ such that $f''$ is negative in $(-\delta, 0)$ and positive in $(0, \delta)$. Thus, $f''$ changes sign at $0$, so $f$ has an inflection point at $(0, f(0))$.

Without the hypothesis of continuity of $f''$, there exists a differentiable function $f$ whose derivative is
$$
f'(x) = \begin{cases}
  x^{2}(1 + \tfrac{1}{2}\sin(1/x)) & x \neq 0, \\
  0 & x = 0.
\end{cases}
$$
This function is differentiable everywhere, even at $0$, and positive everywhere except $0$. The chain rule shows that for $x \neq 0$ we have
$$
f''(x) = 2x(1 + \tfrac{1}{2}\sin(1/x)) - \tfrac{1}{2}\cos(1/x),
$$
while the limit definition gives $f''(0) = 0$. Consequently, the function $f''$ is not negative in any interval $(-\delta, 0)$, nor it is positive in any interval $(0, \delta)$, so $f''$ does not "change sign" in the sense usually meant when discussing inflection points.
