According to the definition of a local maximum, I have to take an open interval around 'a' so it seems like 'a' is a local maximum. I don't understand why it isn't.
It is a local maximum, just as you suspect.
The definition you're using that claims that it isn't a local maximum might have been misworded, or worded correctly but subtly in a way that you've overlooked. The important thing is that you've obviously understood what's actually meant by a local maximum.
(That you've detected this discrepancy means that you're engaged - thinking, questioning, and reflecting rather than just passively absorbing the material. I'd wager that you're a very good student.)
At the boundary of a domain, you consider the open interval intersect the domain. (Technically, that's what you're doing everywhere, except that distinction between open interval and open interval intersect the domain doesn't have any tangible consequences at most points in the domain.)
Local maximum (minimum) means that, when resticted to domain values that are some fixed distance from the domain value of the maximum (minimum), that the function's value is greater than or equal to (less than or equal to) all other values of the function in that region of the domain. It means that "nearby, the function is never greater (lesser)."
I was very confused because the textbook states that it is not a maxima, but yet the right endpoint is a maxima. I guess this is an error in the textbook, but I just wanted to double check since my professor's office hours are not until the day after tomorrow.
Anyone (any book) is free to define anything they want in any way they want. But 100 out of 100 mathematicians will say that x=a corresponds to a local maximum for that function you sketched.
That's assuming, of course, that a is actually in the domain! - meaning that the domain is like [a, b) and not like (a, b). Could *that* be the issue here?
f(x) = 3x on [4, 10) has a minimum of 12 at x=4, but no maximum. It "wants" to have a maximum of 30 at x=10, but x=10 isn't in the domain.
Well, that was the graph they gave, without an interval. It would be ridiculous of the author to expect readers to assume that the interval is (a,b] just based on this graph, in my opinion at least. I'm just going to go with the textbook having an error.
alane - that's what his or her book says. That's what you say. I'd say otherwise, and Phizkid thinks otherwise. This is just a question of definitions. So I'm curious, is your statement here because you remember that that's what you were taught, or are you just accepting and reinforcing the book definition as Phizkid has stated it?
Wikipedia also says that only interior points are allowed to count as local maximums. So, maybe that is now the normative definition. Those authors are entitled to make whatever definitions they want - and I'm entitled to declare that their definitions are stupid.
1) You're presuming that I don't have a PhD.
2) "it is what about every math professor that I have talked to has said." Now, seriously, how many math professors have you actually asked about this? I've got big money saying the answer is zero.
I have over thirty different calculus text books.
On this topic they are all over the place.
I like how Einar Hille solves the question. He simply says that $\displaystyle a$ is a endpoint maximum.