Schrodinger
  • Schrodinger
This isn't a numerical question but moreso conceptual. So, my Calculus textbook has the following paragraph regarding Local Extrema:
Mathematics
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chestercat
  • chestercat
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Schrodinger
  • Schrodinger
"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.
Schrodinger
  • Schrodinger
Whoops, missed a portion. Right after "in some half-open interval" for an endpoint maxima for a:\[[a, a+\delta), \delta > 0\]
Schrodinger
  • Schrodinger
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?

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anonymous
  • anonymous
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
amistre64
  • amistre64
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

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