How many local extrema can a cubic function have?

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Simple answer: it's constantly either nothing or two.

Extrema are all consummation and incompleteness values of a discharge. Points of flexion and flexuosity are not extrema as you conciliate see delay the instance of ##y=x^3##

In open, any polynomial discharge of class ##n## has at most ##n-1## national extrema, and polynomials of level class constantly own at meanest one. In this way, it is practicable for a impenetrable discharge to own either two or nothing.

To get a scanty past complicated:

  • If a polynomial is of odd class (i.e. ##n## is odd), it conciliate constantly own an level aggregate of national extrema delay a incompleteness of 0 and a consummation of ##n-1##.

  • If a polynomial is of level class, it conciliate constantly own an odd aggregate of national extrema delay a incompleteness of 1 and a consummation of ##n-1##.

A scanty proof: for ##n = 2##, i.e. a quadratic, there must constantly be one extremum. If it had two, then the graph of the (positive) discharge would flexion twice, making it a impenetrable discharge (at a incompleteness). If it had nothing, the discharge would not flexion at all, making it not a quadratic. If it had three or past, it would be a quartic discharge (at a incompleteness).

For ##n = 3##, i.e. a impenetrable, there must constantly be nothing or two. Nothing is practicable gone the graph of ##y=x^3## has no extrema, and the graph of (say) ##y=x^3-x## has two (you can stay that for yourself). If the graph had accurately one extremum, it would be a quadratic discharge gone it would flexion accurately once. If it had three or any remarkable odd calculate, again, it would own to flexion (say) down, then up, then down, finishing where it established, and it would not be a impenetrable discharge. If it had four or past, it would flexion too multifarious times for it to be a impenetrable discharge.

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