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Problem: Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f'''(x)<0$ for every $x\not=0$ and $f''(0)=0$. Prove that $M(0,f(0))$ is an inflection point of $f$.

I use the following definition of inflection points:

Dfn: Let $f:(\alpha,\beta)\to \mathbb{R}$ be a function that is differentiable in $(\alpha,\beta)-\{x_0\}$ that is continuous (but not necessarily differentiable) at $x_0$. If $f$ is concave up at $(\alpha,x_0)$ and concave down at $(x_0,\beta)$ or vice versa and the curve of $f$ admits a tangent line at $(x_0,f(x_0))$, then $(x_0,f(x_0))$ is an inflection point of $f$.

Attempt: the result follows immediately with the additional assumption that $f''$ is continous at $x=0$.

1123581321
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  • @Surb How can we write $f'''(0)$? $f$ is three times differentiable at $\mathbb{R}-{0}$ – 1123581321 Oct 04 '23 at 06:19
  • sorry, I thought it was written $f'''(0)<0$. But the argue is more or less the same. Hint Since $f'''(x)<0$ for all $x\neq 0$, $f''$ is decreasing for all $x\neq 0$. In particular, $f''(x)>0$ for all $x<0$ and $f''(x)<0$ for all $x>0$. Therefore $f'$ has a maximum at $x=0$.... – Surb Oct 04 '23 at 06:27
  • @Surb I can see that $f''$ is decreasing at $(\infty,0)$ and$(0,+\infty)$ but how can we show that it is decreasing for all $x\not=0$ ? – 1123581321 Oct 04 '23 at 06:29
  • acctually, saying that $f''$ is decreasing for all $x\neq 0$ is useless. Since $f''$ is continuous, you can directly say that $f''$ is decreasing on $\mathbb R$. – Surb Oct 04 '23 at 06:41
  • @Surb but $f''$ is not continous at $x=0$ ... at least I do not see that – 1123581321 Oct 04 '23 at 06:47
  • If the condition $f \in \mathcal C^2(\mathbb R)\cap\mathcal C^3(\mathbb R-{0})$ is not satisfy, the exercice is wrong... – Surb Oct 04 '23 at 07:46
  • @Surb is there any counterexample? – 1123581321 Oct 04 '23 at 07:50

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