Are local extremum possible at endpoints of a closed bounded interval?

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I thought local extremum did not exist at the endpoints of a closed bounded interval, however my textbook claims this.

Wikipedia:

"A continuous (real-valued) function on a compact set always takes maximum and minimum values on that set. An important example is a function whose domain is a closed (and bounded) interval of real numbers (see the graph above). The neighborhood requirement precludes a local maximum or minimum at an endpoint of an interval."
 
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If you consider a 'neighborhood' to be a 'neighborhood in the domain' your textbook is right. If you consider it to be a 'neighborhood in the reals' then Wikipedia is right. There are a lot of terms that are defined somewhat differently in different references. I think you'd better live by your textbooks definition. Until you change textbooks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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