# Can we consolidate the definition of “critical point” in one-variable calculus?

Instead of saying that a critical point is a point x such that $f’(x) = 0$ or $f$ is not differentiable at $x$, why not just say that it is a point $x$ such that $\dfrac{1}{f’(x)}$ fails to be defined?

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We want to direct attention to the derivative, which has a natural connection to the usual question (relative extrema). The reciprocal of the derivative has no such natural connection. – André Nicolas Sep 7 '11 at 22:59

We could do that, but would doing so gain us anything except increased obfuscation?

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I guess you're right, so I've up-voted and accepted your answer, but I was also curious to see if I was overlooking something technical, and I see that I wasn't. Anyway, thanks for the perspective. BTW, it seems like this could be additionally-mentioned to an advanced student as a sort of iceing-on-the-cake kind of thing. – Mike Jones Sep 8 '11 at 0:12
That reminds me of the 'quote': "eschew obfuscation, espouse elucidation". :-) – Srivatsan Sep 8 '11 at 0:17
-which in turn reminds me of the quote: "It's bad luck to be superstitious." – Mike Jones Sep 8 '11 at 0:25
BTW, the Wikipedia article on "stationary point" mentions that it is often, erroneously, taken to be synonymous with "critical point". My proposal would eliminate that error, but perhaps the obfuscation cure wold be worse than the confusion disease. – Mike Jones Sep 8 '11 at 0:29

While the definition you propose works, it by its nature fails to differentiate between the two separate types of critical points. While both Stationary Points (those with zero slope) and undifferentiable points are critical points, the have different properties and implicates for the nature and continuity of the function, and the presence of maxima/minima. Your definition fails to make the distinction (without looking at the derivative, which reduces to the standard definition.

Additionally, observing the derivative itself near critical values can tell us more about the nature of those points, while your definition fails to provide additional information with more calculation.

It's a good thought, but its seems that, in practice, such a definition would be inferior to the current standard.

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OK, you get an up-vote from me, but please fix the typos in your answer:) – Mike Jones Sep 8 '11 at 0:31
Whoops, that's an embarrassing mistake for my first answer here. What I get for typing quickly and trusting Lion to spell check I suppose. Thanks. – Drew Christianson Sep 8 '11 at 1:33