## 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 ?

1. sparrow2

@ganeshie8 i think is does because of definition

2. 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 ( ).

3. sparrow2

i understant that but if you include a,b nothing is bad i thing,, it's more correct

4. ganeshie8

|dw:1436725268121:dw|

5. sparrow2

oh i see now,but sometimes you can include

6. ganeshie8

I think you can include if the function is continuous in the closed interval

7. sparrow2

yeah it's true

8. 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)

9. 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...

10. ganeshie8

@sparrow2

11. 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)

12. freckles

we base whether the function is increasing or decreasing base on what's derivative does

13. freckles

the function's derivative does not exist at x=1 or x=3

14. 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 +

15. freckles

16. anonymous

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

so basing your understanding of increasing on the existence of derivatives is a bad idea