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Is this true?

$f^{'} (x_{0}) = 0$ and f′ changes sign nowhere near $x_{0} \Rightarrow f$ is constant near $x_{0}$

No — it is false.

The counterexample

This one comes with the Notes pack.

85 counterexamples are free — including most of the ones Part C leans on hardest.

The kind of mistake this is

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

differentiation

Others that fail the same way

From Continuity and Differentiation › Differentiability, mean value theorems, Taylor, L'Hôpital

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