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The problem states: suppose you have a function $f(x)$ and a line $g(x)$, $f(x)$ intersects $g(x)$ 3 times in: $x_1<x_2<x_3$. $f(x)$ also has 3 derivatives and its third doesn't vanish in $[x_1,x_3]$. Prove that $f(x)$ has at least one inflection point in $(x_1,x_3)$.

I do not know if my attempt is right:

(1) $\exists z\in (x_1,x_3)|~{}$ $f''(z)=0$: Suppose that it is not the case, by the fact that $f'''(x)\neq 0$ we know that $f''(x)$ constantly increases or decreases. In any case, if $f''(x)\neq 0$ in the interval then $f'(x)$ does in fact constantly decrease or increase, wich an absurd because it has to reach $g'(x)$ at least 2 times, one in the interval $(x_1,x_2)$ and the other on the interval $(x_2,x_3)$.

(2) Definition of inflection point: there are multiple definition for inflection point, if we define it as a point where $f''(x)$ vanish, we have finished. We can also define it as a point $z$ where for every $\epsilon>0 \exists z_1,z_2$ such that $|z-z_i|<\epsilon$ and the graph changes from being behind to being above the funcion. We are going to use this last one for the last step.

(3) Taylor's expansion: At any point in the interval we can take the Taylor's expansion with the Lagrange remainder as: $f(x)=f(z)+f'(z)(x-z)+\frac{f''(z)(x-z)^2}{2!}+\frac{f''(z)(x-z)^3}{3!}$ so, regrouping and taking the $z$ where $f''(z)=0$ we have: $f(x)- (f(z)+f'(z)(x-z))=\frac{f''(z)(x-z)^3}{3!}$ that is, the graph minus the tangent line changes sign from $z-\epsilon$ to $z+\epsilon$ for every $\epsilon>0$.

Paul Frost
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MiguelCG
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  • Your paragraph (2) is very hard to understand. Moreover, it is wrong to say that an inflection point is a point where $f''$ vanishes: Consider $f(x)=x^4$ at the origin. At an inflection point, concavity must change. You also have a typo in your equation for the Taylor expansion, but your argument is almost correct. Why does the fact that $f'''$ doesn't vanish tell you that it cannot change sign? – Ted Shifrin Apr 19 '25 at 19:08

1 Answers1

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In (1) you write

by the fact that $f'''(x)\neq 0$ we know that $f''(x)$ constantly increases or decreases.

This conclusion requires to know that $f'''(x) > 0$ or $f'''(x) < 0$, and you need an argument to show this. In fact, we must not assume that $f'''$ is continuous (in which case it is trivial). However, it is well-known that derivatives have the Intermediate Value Property. This implies that if $f'''(a) > 0$ and $f'''(b) < 0$ for some $a, b \in [x_1,x_3]$, then $f'''(\xi ) = 0$ for some $\xi$ between $a$ and $b$, and this is impossible.

An inflection point $\xi$ is not defined by the property $f''(\xi) = 0$. The latter is just a necessary condition. See here. We need to know that $f''$ changes sign at $\xi$. But this follows from (1) because $f''$ is strictly monotonic.

Paul Frost
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