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Is a smooth, non-constant, continuous function with more than one critical point necessarily not convex?

I figure if you have two critical points, you have an inflection point between by Rolle's Theorem. If that inflection point isn't also a critical point, it implies a sign change of the second derivative, which means it switches from convex to concave or vice versa.

So id conclude convex shape implies at most one Critical point, and so a unique minimum or maximum ignoring boundaries of the domain.

TurlocTheRed
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  • I assume you exclude constant functions from consideration. – Parcly Taxel Oct 21 '18 at 15:49
  • Correct! I should mention that. – TurlocTheRed Oct 21 '18 at 15:49
  • please use convex/concave instead of convex up/down; you need that the function is not constant between the two critical points – LinAlg Oct 22 '18 at 01:01
  • Thanks for the suggestion, I've updated the concavity part. Does smooth and non-constant imply non-constant between the points? As I recall, smooth means differentiable to first order, A transition from non-constant over one interval to constant over another would introduce a cusp I think. – TurlocTheRed Oct 22 '18 at 14:23

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After some research I think I've found an answer I'm satisfied with.

Concavity can change at inflection points. The second derivative being zero and the next higher order, non-zero derivative being an odd power implies an inflection point.

Critical points and inflection points may not be such under Euclidean Transformations of the curve.

Curvature and its derivative remain unchanged under such rotations and give you information about concavity that holds after such transformations.

TurlocTheRed
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