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So I know the first derivative of a function $f$ must be $0$ (or undefined) at a point $x_0$ for there to be a critical point at $f(x_0)$. But does the converse hold true? i.e. if the first derivative is $0$ is it definitely a critical point?

Thanks

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  • $\begingroup$ What is your definition of a critical point? en.wikipedia.org/wiki/Critical_point_(mathematics) $\endgroup$ – saulspatz Jan 7 at 16:07
  • $\begingroup$ I believe it is defined as a point $x_0$ in the domain of a function $f$ such that $f'(x_0)$ is $0$ or undefined? $\endgroup$ – Achuan Chen Jan 7 at 16:16
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    $\begingroup$ Right, it's the definition. All definitions are if and only if. I don't understand your question. $\endgroup$ – saulspatz Jan 7 at 16:18
  • $\begingroup$ I knew that given a critical point exists the first derivative at that point must be $0$ or undefined, but I wasn't sure if the converse holds true - whether the first derivative being 0 always implies there's a critical point. $\endgroup$ – Achuan Chen Jan 7 at 16:25
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The critical points are defined to be the points where the derivative is zero or undefined. Whether those points are turning points or not is up in the air. For instance, the derivative of $f(x)=x^3$ is zero when $x=0$, but $(0,0)$ is neither a local maximum nor a local minimum point of $f$.

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    $\begingroup$ Great explanation! THank you $\endgroup$ – Achuan Chen Jan 7 at 16:13

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