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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 ?

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  1. sparrow2
    • one year ago
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    @ganeshie8 i think is does because of definition

  2. freckles
    • one year ago
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    Well a function doesn't increase/decrease at a point, it increases/decreases on an interval. I think it is better to leave the ( ).

  3. sparrow2
    • one year ago
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    i understant that but if you include a,b nothing is bad i thing,, it's more correct

  4. ganeshie8
    • one year ago
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    |dw:1436725268121:dw|

  5. sparrow2
    • one year ago
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    oh i see now,but sometimes you can include

  6. ganeshie8
    • one year ago
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    I think you can include if the function is continuous in the closed interval

  7. sparrow2
    • one year ago
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    yeah it's true

  8. freckles
    • one year ago
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    |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)

  9. ganeshie8
    • one year ago
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    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...

  10. ganeshie8
    • one year ago
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    @sparrow2

  11. freckles
    • one year ago
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    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)

  12. freckles
    • one year ago
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    we base whether the function is increasing or decreasing base on what's derivative does

  13. freckles
    • one year ago
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    the function's derivative does not exist at x=1 or x=3

  14. freckles
    • one year ago
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    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 +

  15. freckles
    • one year ago
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    instead of "does not exist"

  16. anonymous
    • one year ago
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    increasing/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 \(f(y)-f(x)=f'(c)(y-x)\) that \(f'\ge 0\) suggests that \(y-x>0\implies f(y)-f(x)\ge 0\) or in other words \(x<y\implies f(x)\le f(y)\)

  17. anonymous
    • one year ago
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    so basing your understanding of increasing on the existence of derivatives is a bad idea

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