sparrow2
  • sparrow2
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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sparrow2
  • sparrow2
@ganeshie8 i think is does because of definition
freckles
  • freckles
Well a function doesn't increase/decrease at a point, it increases/decreases on an interval. I think it is better to leave the ( ).
sparrow2
  • sparrow2
i understant that but if you include a,b nothing is bad i thing,, it's more correct

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ganeshie8
  • ganeshie8
|dw:1436725268121:dw|
sparrow2
  • sparrow2
oh i see now,but sometimes you can include
ganeshie8
  • ganeshie8
I think you can include if the function is continuous in the closed interval
sparrow2
  • sparrow2
yeah it's true
freckles
  • freckles
|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)
ganeshie8
  • ganeshie8
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...
ganeshie8
  • ganeshie8
@sparrow2
freckles
  • freckles
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)
freckles
  • freckles
we base whether the function is increasing or decreasing base on what's derivative does
freckles
  • freckles
the function's derivative does not exist at x=1 or x=3
freckles
  • freckles
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 +
freckles
  • freckles
instead of "does not exist"
anonymous
  • anonymous
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
anonymous
  • anonymous
so basing your understanding of increasing on the existence of derivatives is a bad idea

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