First Derivative Test for inflection points

Question

Mathworld, "First Derivative Test" states:

Suppose $f(x)$ is continuous at a stationary point $x_0$. ...

If $f'(x)$ has the same sign on an open interval extending left from $x_0$ and on an open interval extending right from $x_0$, then $f(x)$ has an inflection point at $x_0$.

I am however having trouble proving the above claim and I suspect that without more conditions, it is false.

My questions: If the above claim is true, how do I prove it? If false, how do I strengthen the assumptions so that it becomes true?

Answer 1

Mathworld's claim is false.

This issue is actually discussed by Kouba (1995), who gives this counterexample:

Define $f:\mathbb R \rightarrow \mathbb R$ by:

$$f\left(x\right)=\begin{cases} x^{3}+x^{4}\sin\frac{1}{x} & \text{ for }x\neq0,\\ 0 & \text{ for }x=0. \end{cases}$$

Kouba shows that $f$ is continuous at the stationary point $0$. Also, $f'(x)>0$ for all $x\in(-0.5,0.5)$. However, $0$ is not an inflection point of $f$.

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Kouba then shows that we can fix this by requiring also that the second derivative has only a finite number of zeros in the neighborhood of $c$:

Theorem. Let function $f$ be twice continuously differentiable on the interval $[a, b]$, and suppose the second derivative $f"$ has only a finite number of zeros on this interval. Suppose $f'(x)$ is strictly positive at all points $x$ in $(a,b)$ (or strictly negative at all such points), except that $f'(c) = 0$, $a < c < b$. Then $f$ has an inflection point at $x = c$.