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Proposition 3. (Sufficient conditions for an extremum in terms of the first derivative). Let $f:U(x_0)\to \mathbb{R}$ be a function defined on neighborhood $U(x_0)$ of the point $x_0$, which is continuous at the point itself and differentiable in a deleted neighborhood $\mathring{U}(x_0)$. Let $\mathring{U}^-(x_0)=\{x\in U(x_0): x<x_0\}$ and $\mathring{U}^+(x_0)=\{x\in U(x_0): x>x_0\}$. Then the following conclusions are valid:

a) $(\forall x\in\mathring{U}^-(x_0) (f'(x)<0))\ \land \ (\forall x\in\mathring{U}^+(x_0) (f'(x)<0)) \Rightarrow (f \ \text{has no extremum at} \ x_0);$

b) $(\forall x\in\mathring{U}^-(x_0) (f'(x)<0))\ \land \ (\forall x\in\mathring{U}^+(x_0) (f'(x)>0)) \Rightarrow (x_0 \ \text{is a strict local minimum of} \ f);$

c) $(\forall x\in\mathring{U}^-(x_0) (f'(x)>0))\ \land \ (\forall x\in\mathring{U}^+(x_0) (f'(x)<0)) \Rightarrow (x_0 \ \text{is a strict local maximum of} \ f);$

d) $(\forall x\in\mathring{U}^-(x_0) (f'(x)>0))\ \land \ (\forall x\in\mathring{U}^+(x_0) (f'(x)>0)) \Rightarrow (f \ \text{has no extremum at} \ x_0);$

Briefly, but less precisely, one can say that if the derivative changes sign in passing through the point, then the point is an extremum, while if the derivative does not change sign, the point is not extremum.

We remark immediately, however, that these sufficient conditions are not necessary for an extremum, as one can verify the following example.

Example: Let $$f(x) = \begin{cases} 2x^2+x^2\sin \frac{1}{x}, & \text{if }x\neq 0 \\ 0, & \text{if }x=0 \end{cases}$$

Since $x^2\leq f(x)\leq 3x^2$, it is clear that the function has a strict local minimum at $x_0=0$, but the derivative $f'(x)=4x+2x\sin \frac{1}{x}-\cos \frac{1}{x}$ is not of constant sign in any deleted one-sided neighborhood of this point. This same example shows the misunderstandings that can arise in connection with the abbreviated statement of Proposition 3 just given.

This is an excerpt from Zorich's book (Volume 1, page 238). Most of it I understood but some moments are not crystal clear. So let me ask you questions please:

1) This example shows us that the converse of b) in Proposition 3 is not valid, right? What is an example of a function which shows that the converse of a) is not valid?

2) I am a bit confused with the phrase "This same example shows the misunderstandings that can arise in connection with the abbreviated statement of Proposition 3 just given." I have spent some time trying to understand what is wrong but gave up.

I'd be very grateful for answers to my question.

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(1) For a counterexample to the converse of (a), just take $f(x)=x^3$.

(2) The example given shows that it doesn’t make sense to speak of $f’$ “changing sign” at the local minimum. The cases where it makes sense to speak of a sign change are those where $f’$ has constant sign on left and right sided deleted neighborhoods.

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  • $\begingroup$ Thanks for your reply! But your answer to part b) is not so clear. $\endgroup$
    – ZFR
    Apr 24, 2021 at 22:11
  • $\begingroup$ @ZFR: What isn’t clear? If there is no one-sided deleted neighborhood of a point on which the sign of $f’$ is constant, it makes no sense to speak of $f’$ “changing sign” at that point. Read that sentence several times until you get it. (Such language only makes sense if you can say “the sign on the left is ..., while the sign on the right is ...” But “the sign on the left” literally makes no sense if there is no left-side deleted neighborhood on which $f’$ has constant sign.) It is not possible to be clearer. $\endgroup$ Apr 24, 2021 at 22:16

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