sign and strictly decreasing/increasing

[extcar]

when you diff on cas, sometimes they give you sign(x)
so when you write out the equation, is it sign(x), x>0  and -sign(x) x<0 or is it x≥0
sorry if i'm not clear enough, i'll try to explain again if no one understands me

and for strictly decreasing/increasing functions,
the maximum/minimum value is included?

[Vincezor]

Have you had a read of this?

http://www.vcaa.vic.edu.au/correspondence/bulletins/2011/April/2011AprilSup2.pdf

Has examples with explanations...

Cheers to luffy for showing me this yesterday

[extcar]

awesome! thanks a lot!!!!!!!!!!!!!!!!!

[nacho]

Have you had a read of this?

http://www.vcaa.vic.edu.au/correspondence/bulletins/2011/April/2011AprilSup2.pdf

Has examples with explanations...

Cheers to luffy for showing me this yesterday

lol he showed me yesterday night too!

[paulsterio]

lol he showed me yesterday night too!

LOL! Do you and "Luffy" meet up at night?

[luffy]

lol he showed me yesterday night too!

LOL! Do you and "Luffy" meet up at night?

We actually do... occassionally I'm sure you do the same with some of your friends Paul xD (Ignoring any odd implications here.)

Also, just thought I would point out that I think strictly increasing/decreasing will be on the methods exam this year - though its mere speculation. I mean, VCAA must have introduced it into the current study design for a reason, otherwise it would be pointless. So, make sure you guys understand it by the definition just in case it does come up. Good luck everyone!

[paulsterio]

We actually do... occassionally I'm sure you do the same with some of your friends Paul xD (Ignoring any odd implications here.)

Also, just thought I would point out that I think strictly increasing/decreasing will be on the methods exam this year - though its mere speculation. I mean, VCAA must have introduced it into the current study design for a reason, otherwise it would be pointless. So, make sure you guys understand it by the definition just in case it does come up. Good luck everyone!

Hahaha, I'm glad you said "ignoring any odd implications"

What about the issue of Endpoints then, are endpoints included or excluded
So say we have a Parabola, y=x^2 for the domain [-1,3]
Is it strictly increasing for [0,3] or [0,3)

That's something to think about :S

[luffy]

We actually do... occassionally I'm sure you do the same with some of your friends Paul xD (Ignoring any odd implications here.)

Also, just thought I would point out that I think strictly increasing/decreasing will be on the methods exam this year - though its mere speculation. I mean, VCAA must have introduced it into the current study design for a reason, otherwise it would be pointless. So, make sure you guys understand it by the definition just in case it does come up. Good luck everyone!

Hahaha, I'm glad you said "ignoring any odd implications"

What about the issue of Endpoints then, are endpoints included or excluded
So say we have a Parabola, y=x^2 for the domain [-1,3]
Is it strictly increasing for [0,3] or [0,3)

That's something to think about :S

This is why I said to go by the definition. Remember that the terms strictly increasing/decreasing have to do with intervals - NOT gradient. Hence, the endpoints would logically be included (unless there is some exception that I am unaware of currently). So, its strictly increasing for [0,3].

Hope I helped.

[abeybaby]

I think its [0,3)... because you cant say that the function is increasing at x=3, when you dont know the gradient...

[paulsterio]

Well the definition of strictly increasing is that if f(b)>f(a) for b>a, then it's strictly increasing for [b,a]

so technically can "b" be an endpoint? :S

Well Luffy says [0,3] and Abes says [0,3)

And I say they both make sense =.="

[luffy]

I think its [0,3)... because you cant say that the function is increasing at x=3, when you dont know the gradient...

I really shouldn't be disagreeing with a legend like yourself, but I can't resist in this situation.

We are not referring to 'increasing', but a new concept called 'Strictly Increasing,' which VCAA introduced in the 2010 sample questions last year. While similar, the two terms differ completely in terms of definition and actually describe different things. Increasing, which is the term we have used in methods up til now, refers to a positive gradient and if that is the case, then yes, it would be (0,3).
However, going by the definition of Strictly Increasing, we are referring only to intervals. Gradient has absolutely no influence over whether something is 'Strictly Increasing.' Therefore, despite being an endpoint, the parabola would be Strictly Increasing over the interval [0,3]. Hope I explained it well enough.

Well the definition of strictly increasing is that if f(b)>f(a) for b>a, then it's strictly increasing for [b,a]

so technically can "b" be an endpoint? :S

Well Luffy says [0,3] and Abes says [0,3)

And I say they both make sense =.="

Abes is getting it confused with increasing. I just wanted to reiterate to the ATARnotes methods community this year that the two terms "increasing" and "strictly increasing" are different in definition, despite appearing very similar. The answer should be [0,3]. The only way I could be wrong is if there is an extra part of the definition that I have not seen.

[paulsterio]

Abes is getting it confused with increasing. I just wanted to reiterate to the ATARnotes methods community this year that the two terms "increasing" and "strictly increasing" are different in definition, despite appearing very similar. The answer should be [0,3]. The only way I could be wrong is if there is an extra part of the definition that I have not seen.

But if it's increasing then shouldn't it be (0,3) - cause increasing is where dy/dx>0

This gets confusing :S

I think its [0,3)... because you cant say that the function is increasing at x=3, when you dont know the gradient...

I really shouldn't be disagreeing with a legend like yourself, but I can't resist in this situation.

Let's go to his Spesh lecture!

Sighs, you are a legend yourself Luffy, equal top rank in spesh in a school as amazing as ours, only special people get that privelage ;P
Although the other top ranked kid is pretty awesome too

[luffy]

Abes is getting it confused with increasing. I just wanted to reiterate to the ATARnotes methods community this year that the two terms "increasing" and "strictly increasing" are different in definition, despite appearing very similar. The answer should be [0,3]. The only way I could be wrong is if there is an extra part of the definition that I have not seen.

But if it's increasing then shouldn't it be (0,3) - cause increasing is where dy/dx>0

This gets confusing :S

I think its [0,3)... because you cant say that the function is increasing at x=3, when you dont know the gradient...

I really shouldn't be disagreeing with a legend like yourself, but I can't resist in this situation.

Let's go to his Spesh lecture!

Sighs, you are a legend yourself Luffy, equal top rank in spesh in a school as amazing as ours, only special people get that privelage ;P
Although the other top ranked kid is pretty awesome too

I think I was editing my post, while you were writing yours. Yeah - should be increasing for the interval (0,3).

We still got one SAC left buddy. Maybe this last one will finally separate us It'd be quite funny if we both flunked it and a new rank 1 rose up xD

[paulsterio]

I think I was editing my post, while you were writing yours. Yeah - should be increasing for the interval (0,3).

We still got one SAC left buddy. Maybe this last one will finally separate us It'd be quite funny if we both flunked it and a new rank 1 rose up xD

Yepp, I think so (:
What's annoying is that around the internet, I haven't been able to find much on STRICTLY increasing, just increasing (which is dy/dx>0) - I think this is a VCAA made up concept =.=

Haha, well yeah, we have the 1 last SAC, but it's skills =.=" but the fact that I was owned on the last Skills is worrying! D:
It would be funny if a new rank 1 rises up for sure
But I still wonder what would happen if that last SAC doesn't separate us, which might happen :\

Btw, thanks for looking out for us Methods Kids, some of whom, like me, narrowly escaped the Cone of Death, by not doing Methods last year

[luffy]

I think I was editing my post, while you were writing yours. Yeah - should be increasing for the interval (0,3).

We still got one SAC left buddy. Maybe this last one will finally separate us It'd be quite funny if we both flunked it and a new rank 1 rose up xD

Yepp, I think so (:
What's annoying is that around the internet, I haven't been able to find much on STRICTLY increasing, just increasing (which is dy/dx>0) - I think this is a VCAA made up concept =.=

Haha, well yeah, we have the 1 last SAC, but it's skills =.=" but the fact that I was owned on the last Skills is worrying! D:
It would be funny if a new rank 1 rises up for sure
But I still wonder what would happen if that last SAC doesn't separate us, which might happen :\

Btw, thanks for looking out for us Methods Kids, some of whom, like me, narrowly escaped the Cone of Death, by not doing Methods last year

http://en.wikipedia.org/wiki/Monotonic_function

Talks about strictly increasing/decreasing a bit there, so clearly the concept was not made up by VCAA.

As for the cone of death question, I think your intelligent enough to have gotten it if you did the exam last year I am probably one of a select few, silly enough to use two triangles which aren't even similar. =.= Oh well, life goes on.