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Schrodinger

This isn't a numerical question but moreso conceptual. So, my Calculus textbook has the following paragraph regarding Local Extrema:

  • one year ago
  • one year ago

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  1. Schrodinger
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    "If the domain of ƒ is the closed interval [a, b], then ƒ has a local maximum at the endpoint \[x=a\] if \[f(x) \le f(a)\] for all x in some half-open interval Likewise, f has a local maximum at an interior point \[x = c\] if \[f(x) \le f(c)\]for all x in some open interval \[(c -\delta, c+\delta), \delta > 0\] and a local maximum at the endpoint \[x = b\]if \[f(x) \le f(b)\]for all x in some half-open interval \[(b -\delta,b],\delta > 0.\] The inequalities are reversed for local minimum values. In Figure 4.5, the function ƒ has local maxima at c and d and local minima at a, e, and b. Local extrema are also called relative extrema. Some functions can have infinitely many local extrema, even over a finite interval." I get why [a, b] has to be a closed interval, but I don't understand what the consequences would be for the above statements regarding local extrema if they had closed intervals. i.e., why are some of them half-open? I don't understand how them being closed would be an issue or challenge the logic of the statements.

    • one year ago
  2. Schrodinger
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    Whoops, missed a portion. Right after "in some half-open interval" for an endpoint maxima for a:\[[a, a+\delta), \delta > 0\]

    • one year ago
  3. Schrodinger
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    Is it just because delta can't be actually quantified/defined and is essentially an arbitrary number, while [a,b] are definite fixed end points? So basically the open-ended statement is just a reinforcement that the half-open interval can be anything?

    • one year ago
  4. Xavier
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    Having the point in question is necessary. In order to have a, for example, in the interval you construct it needs to be closed on one side at least since it it at the end of your [a,b]. For c to be within, since it is in (a,b) it is possible to have an open interval around in on both sides. Why intervals are kept open when involving the delta I'm not exactly sure

    • one year ago
  5. amistre64
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    it might have to do with the infinitesimal property of the a+g, you would not use a closed interval to express infinity, so it might be improper to use it on an infinitesimal as well

    • one year ago
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