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sparrow2
 one year ago
when a function increase/decrease on interval (a,b) does it also increase/decrease on interval [a,b] if a,b are in domain ?
sparrow2
 one year ago
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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sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 i think is does because of definition

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Well a function doesn't increase/decrease at a point, it increases/decreases on an interval. I think it is better to leave the ( ).

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0i understant that but if you include a,b nothing is bad i thing,, it's more correct

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1436725268121:dw

sparrow2
 one year ago
Best ResponseYou've already chosen the best response.0oh i see now,but sometimes you can include

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think you can include if the function is continuous in the closed interval

freckles
 one year ago
Best ResponseYou've already chosen the best response.3dw: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)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4But (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...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so looking at that function I drew it's derivative would be something like this: dw:1436726620507:dw this is above the xaxis so this tells us the function is increasing on (1,3) since the function's derivative exists on (1,3)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we base whether the function is increasing or decreasing base on what's derivative does

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the function's derivative does not exist at x=1 or x=3

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so 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 +

freckles
 one year ago
Best ResponseYou've already chosen the best response.3instead of "does not exist"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0increasing/decreasing is not defined in terms of derivatives; \(f:(a,b)\to\mathbb{R}\) is increasing on \((a,b)\) iff \(x<y\implies f(x)\le f(y)\) for all \(x,y\in(a,b)\). it just so happens, however, if \(f\) is differentiable and we have \(f'\ge0\) then by the mean value theorem \(f(y)f(x)=f'(c)(yx)\) that \(f'\ge 0\) suggests that \(yx>0\implies f(y)f(x)\ge 0\) or in other words \(x<y\implies f(x)\le f(y)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so basing your understanding of increasing on the existence of derivatives is a bad idea
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