sign and strictly decreasing/increasing

[nacho]

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.

I need to clear something up here.
Are consequential marks awarded in methods?
Because if you got part A) wrong for using the wrong similar triangles, then shouldnt you have at least been awarded marks for the other parts of the question?
I don't get how you lost all 5 marks?

also luf, what is your opinion on the itute exam? how did you score on them? I dont see how you found them easier than kilrofl

[vea]

I didn't get the cone question but I got consequential marks for it. Don't have a SoM to prove it or anything but i had an A+ for exam 1 and the cutoff was like 36.5/40 or something!

[thushan]

My two cents on the difference between increasing and strictly increasing:

Firstly, forget gradients. These are merely accessories in this case. The gradient of the function is not in the definition of the terms. Think intervals, think mapping between two sets (which is what a function is).

A function is INCREASING over an interval [a,b] IF a > b <=> f(a) >= f(b) (i meant greater or equal to). So by definition, f(x) = 1 is an INCREASING function. Weird, but correct.

A function is STRICTLY INCREASING over an interval [a,b] IF a>b <=> f(a) > f(b). So something like y = x^2 is STRICTLY increasing over the interval [0, infinity), but f(x) = 1 is NOT a strictly increasing function.

So, IF a function is STRICTLY increasing over an interval, THEN it is also INCREASING over that interval as well. However, the converse is not always true, as shown by the f(x) = 1 example.

So basically, in almost all cases when you get a question that asks for an interval over which a function is INCREASING or STRICTLY INCREASING you include the endpoint - you use the [ ] brackets.

[paulsterio]

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.

Alright then, I didn't see that before, but I think whether or not the end point is included is a matter of definition, not theory, you can't "prove" whether an endpoint is included or not?

Oh thanks! you know, I just realised I have never had a conversation with you that has not involved that question =.=" But you're right, life goes on

I need to clear something up here.
Are consequential marks awarded in methods?
Because if you got part A) wrong for using the wrong similar triangles, then shouldnt you have at least been awarded marks for the other parts of the question?
I don't get how you lost all 5 marks?

Consequentials should be awarded but I don't know why they weren't awarded in his case, unless he did something drastically wrong - but I'd be careful with consequentials, they're hotly debated and there are times where they aren't awarded properly

I didn't get the cone question but I got consequential marks for it. Don't have a SoM to prove it or anything but i had an A+ for exam 1 and the cutoff was like 36.5/40 or something!

Luffy lost all 5 marks on the Cone question - 35/40 according to his SoM but he got an A+ too? :S

My two cents on the difference between increasing and strictly increasing:

Firstly, forget gradients. These are merely accessories in this case. The gradient of the function is not in the definition of the terms. Think intervals, think mapping between two sets (which is what a function is).

A function is INCREASING over an interval [a,b] IF a > b <=> f(a) >= f(b) (i meant greater or equal to). So by definition, f(x) = 1 is an INCREASING function. Weird, but correct.

A function is STRICTLY INCREASING over an interval [a,b] IF a>b <=> f(a) > f(b). So something like y = x^2 is STRICTLY increasing over the interval [0, infinity), but f(x) = 1 is NOT a strictly increasing function.

So, IF a function is STRICTLY increasing over an interval, THEN it is also INCREASING over that interval as well. However, the converse is not always true, as shown by the f(x) = 1 example.

So basically, in almost all cases when you get a question that asks for an interval over which a function is INCREASING or STRICTLY INCREASING you include the endpoint - you use the [ ] brackets.

Hmm, yeah, that makes some sense, thanks Thushan


One more thing to think about guys:
Say y = (x-1)^3/(x-1)
This is a parabola with a discontinuity at x = 1
If we were to find the Minimum Value of y, would it be 0? :S because technically that point's undefined
So what would the Minimum Value be? :S

[luffy]

My two cents on the difference between increasing and strictly increasing:

Firstly, forget gradients. These are merely accessories in this case. The gradient of the function is not in the definition of the terms. Think intervals, think mapping between two sets (which is what a function is).

A function is INCREASING over an interval [a,b] IF a > b <=> f(a) >= f(b) (i meant greater or equal to). So by definition, f(x) = 1 is an INCREASING function. Weird, but correct.

A function is STRICTLY INCREASING over an interval [a,b] IF a>b <=> f(a) > f(b). So something like y = x^2 is STRICTLY increasing over the interval [0, infinity), but f(x) = 1 is NOT a strictly increasing function.

So, IF a function is STRICTLY increasing over an interval, THEN it is also INCREASING over that interval as well. However, the converse is not always true, as shown by the f(x) = 1 example.

So basically, in almost all cases when you get a question that asks for an interval over which a function is INCREASING or STRICTLY INCREASING you include the endpoint - you use the [ ] brackets.

Wow - even I was wrong there. I never knew the definition of increasing functions - I was always taught that it was the gradient is positive. Thanks a heap there Thushan! Hence, my point stands - learn everything by definition!

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.

I need to clear something up here.
Are consequential marks awarded in methods?
Because if you got part A) wrong for using the wrong similar triangles, then shouldnt you have at least been awarded marks for the other parts of the question?
I don't get how you lost all 5 marks?

also luf, what is your opinion on the itute exam? how did you score on them? I dont see how you found them easier than kilrofl

I've had this conversation with you at least 10 times (and like 100+ times with paul since he loves the topic). Hence, I can't really be bothered answering it xD. I never ordered a statement of exam script, and hence couldn't debate anything anyway.

As for the marking scheme, from what I have been told, I think that if you use the correct method, you will get a "method" mark for consequentials, but you don't get the answer mark. Though, a variety of sources have told me differently, so confirmation is probably required.

Lastly, I always supported kilbaha over iTute. Theres something about iTute exams that just seemed dodgy to me. I also felt that they divulged away from the study design at times. Hence, my love and passion for kilbaha stands

[brightsky]

in response to the above question..that's just a normal parabola. there's no discontinuity. hence the minimum of y is 0.

[luffy]

in response to the above question..that's just a normal parabola. there's no discontinuity. hence the minimum of y is 0.

O.o Why isn't there a discontinuity? The graph is undefined for x = 1?

I don't think you can state a minimum value for a function like that. It would just approach 0. Its sorta like stating a minimum value for y= 1/x where the domain is (0, infinity).

[paulsterio]

Lastly, I always supported kilbaha over iTute. Theres something about iTute exams that just seemed dodgy to me. I also felt that they divulged away from the study design at times. Hence, my love and passion for kilbaha stands

I find it funny how when you were writing up your table for practice exams in Specialist Maths, I had to remind you to put Kilbaha in, you so forgot about it

in response to the above question..that's just a normal parabola. there's no discontinuity. hence the minimum of y is 0.

There is a discontinuity at x = 1 (:

I don't think you can state a minimum value for a function like that. It would just approach 0. Its sorta like stating a minimum value for y= 1/x where the domain is (0, infinity).

Well here's the issue, a function such as y = 1/x is a strictly decreasing function for (0,infinity) so it doesn't have a minimum, but the graph that I was talking about does have a "minimum" cause one side is increasing and the other is decreasing? :S

LOL! these are quite fun to think about!

[brightsky]

hmm..upon some afterthought, yeah i think you're right. definition given by wikipedia for well definition.

In common mathematical usage, an arithmetic expressionE(x,y,...) is called undefined for certain values x0, y0,... of the variables x, y, ... in two different cases:

1. A meaning for E(x0,y0,...) has not been specified in the definition of E.
2. The same, and in addition there is no "sensible" way to extend the existing definition to cover the case x0,...

although the second case i don't quite understand. but in the case that it's undefined, there wouldn't be any minimum since:

A function has a global (or absolute) minimum point at x∗ if f(x∗) ≤ f(x) for all x, but if we have x* = 0 as the minimum, f(x*) doesn't exist.

[thushan]

best way to express the minimum value would be

lim (x approaches 1) of f(x)

where f(x) = (x-1)^3/(x-1)

[luffy]

best way to express the minimum value would be

lim (x approaches 1) of f(x)

where f(x) = (x-1)^3/(x-1)

But wouldn't expressing it as a limit of the function be wrong? Because, you can then evaluate that limit to equal 0. I think I'm thinking too much into this and self-confusing myself xD.

[thushan]

Actually, you're right. My mistake.

Maybe this expression - the smallest possible number greater than 0

[paulsterio]

Actually, you're right. My mistake.

Maybe this expression - the smallest possible number greater than 0

0 + 1/infinity LOL!

but hmm, i'm still stumped

[abeybaby]

whats this graph everyone's talking about?

[b^3]

One more thing to think about guys:
Say y = (x-1)^3/(x-1)
This is a parabola with a discontinuity at x = 1
If we were to find the Minimum Value of y, would it be 0? :S because technically that point's undefined
So what would the Minimum Value be? :S