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In this article https://www.jstor.org/stable/2310782 i found this proposition:

Let $f$ be a differentiable function defined on an open interval $(a, b)$ containing the point $x_0$.

Let:

(B) There exists an open interval $I\subset (a, b)$, $x_0\in I$, such that on $I$ the function $f'$ attains a (local) maximum or minimum at $x_0$.

(C) There exists an open interval $I\subset (a, b)$, $x_0\in I$ such that on I we have $T\ge f$ on one side of $x_0$ and $T\le f$ on the other side. Here $T$ is the tangent to the graph of $f$ at the point $x_0$.

Then (B) implies (C).

I ask help for proof thats if $x_0$ is local minimum of $f'$ (i.e. $\exists \delta>0$ : $f'(x)\ge f'(x_0)$ for $x_0-\delta<x<x_0+\delta$), then $f\le T$ in $(x_0-\delta,x_0)$ and $f\ge T$ in $(x_0,x_0+\delta)$. $[T(x)=f(x_0)+f'(x_0)(x-x_0)]$.

Ps. On Wikipedia (https://en.wikipedia.org/wiki/Inflection_point) i found that ' If all extrema of $f'$ are isolated (that is, in some neighborhood, x is the one and only point at which f' has a (local) minimum or maximum), then an inflection point is a point on the graph of f at which the tangent crosses the curve.'

user791759
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  • If the derivative is decreasing on $(x_0-\delta,x_0)$, then $f$ is concave down on that interval, which means that the graph lies under the tangent. – Arturo Magidin Mar 20 '24 at 18:53
  • You inserted "(local)" into condition (B), but there's no need to do so. If $f'$ has a local minimum on $I$ at $x_0$, the interval $I$ can be chosen to be small enough so that $f'$ has an absolute minimum on $I$ as $x_0$. – Matthew Leingang Mar 20 '24 at 19:49

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By $T$, the author is referring to the linearization of $f$ about $x=x_0$: $$ T(x) = f(x_0) + f'(x_0)(x-x_0) $$ By the Mean Value Theorem, for all $x \in (a,b)$ there exists a point $\xi$ between $x$ and $x_0$, such that $$ f(x) = f(x_0) + f'(\xi)(x-x_0) $$ Now suppose that $f'(x_0)$ is the minimum value of $f'$ on $I$, and $x$ is a point in $I$ less than $x_0$. Then $$\begin{aligned} f'(\xi) &\geq f'(x_0) \\\implies f'(\xi)(x-x_0) &\leq f'(x_0)(x-x_0) \\\implies f(x_0) + f'(\xi)(x-x_0) &\leq f(x_0) + f'(x_0)(x-x_0) \\\implies f(x) &\leq T(x) \end{aligned}$$

Matthew Leingang
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