when a function increase/decrease on interval (a,b) does it also increase/decrease on interval [a,b] if a,b are in domain ?

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when a function increase/decrease on interval (a,b) does it also increase/decrease on interval [a,b] if a,b are in domain ?

Mathematics
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@ganeshie8 i think is does because of definition
Well a function doesn't increase/decrease at a point, it increases/decreases on an interval. I think it is better to leave the ( ).
i understant that but if you include a,b nothing is bad i thing,, it's more correct

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|dw:1436725268121:dw|
oh i see now,but sometimes you can include
I think you can include if the function is continuous in the closed interval
yeah it's true
|dw:1436726038544:dw| you could say this function is increasing on (1,2] union [2,3) but that is weird. when you can just say it is increasing on (1,3)
But (1, 3) is more general/loose compared to [1, 3] So if a theorem requires the function to be increasing/continuous in (a, b), then you must leave it like that. Asking for [a, b] is more restrictive as you can see...
so looking at that function I drew it's derivative would be something like this: |dw:1436726620507:dw| this is above the x-axis so this tells us the function is increasing on (1,3) since the function's derivative exists on (1,3)
we base whether the function is increasing or decreasing base on what's derivative does
the function's derivative does not exist at x=1 or x=3
so does that really mean we should include x=1 or/and x=3 so we say the function is increasing on [1,3] I just don't know if I like to say that because when I read y=f(x) is increasing on [1,3] should I not think that f'(1) is + and f'(3) is +
instead of "does not exist"
increasing/decreasing is not defined in terms of derivatives; \(f:(a,b)\to\mathbb{R}\) is increasing on \((a,b)\) iff \(x0\implies f(y)-f(x)\ge 0\) or in other words \(x
so basing your understanding of increasing on the existence of derivatives is a bad idea

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