Although asked several times..

[kenhung123]

I still don't understand 'strictly' decreasing and 'strictly' increasing as apparently the 'strictly' means something different to just simply decreasing or increasing.

Is there any particular way to find the strictly decreasing/increasing interval? I don't really get the definition..

[m@tty]

Without the strictly it means that as you move in the positive x-direction, the y values either increase (for increasing) or decrease (for decreasing) OR they stay the same.

However, when the word 'strictly' is introduced, it cannot stay constant - this means that as you move right the y value is either always increasing (for strictly increasing) or always decreasing (for strictly decreasing).


EDIT:
For a function where the interval is a subset of the domain;

Strictly increasing:

Increasing:

Strictly decreasing:

Decreasing:

[kenhung123]

Ok...so how would we determine the strictly increasing/decreasing interval for a question?

[/0]

strictly increasing =>
strictly decreasing =>

[m@tty]

Without the strictly, you need to include the points where .

Also, if the end-point of the strictly increasing/decreasing interval is a stationary point of inflection, then you will have to extend it accordingly. (Because after the point of inflection it continues either increasing or decreasing as before..)

But VCAA questions probably won't be like this.. but you never know..

[kenhung123]

how about when f'(x)=0, which is the hard part :S
Someone told me when f'(x)=0, its both strictly increasing and decreasing so you include it all the time. True?

[LachyMc]

yeah i am pretty sure that when f'(x) = 0 you include it when it is "strictly" increasing or decreasing as i just did a practice exam with this question

[kenhung123]

Err did anyone realise, neap 2010 took strictly increasing as just increasing by saying f'(x)>0 for x>(x value for turning point)

[ks04]

according to school, (our head of math writes the spec exam) you can use either square or round brackets (including or exclusing the turning point) for strictly decreasing as it is defined differently in different text books.

[kenhung123]

but the 2010 sample paper does explicitly state that students need to be aware that strictly decreasing for the turning point of that graph needs to be included...

[Blakhitman]

Contact VCAA (E-mail) and ask them whether they will accept both or not.

[/0]

Increasing is a bit of a vague term since it usually means the same as non-decreasing. i.e. . But the 'strict' part refers to a strict inequality, i.e. instead of , so I would have thought strictly increasing means , without including stationary points.

[Blakhitman]

Apparently it's the other way though. The "strict" means you do include the stationary point (from the sample questions).

Found here http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/cas/publications/mmcas1-sampfin-w.pdf

Question 4

[rubiks]

Are there any worked solutions to the vcaa sample questions?

[_avO]

Exam 1 soln: http://www.itute.com/mathline/mm_vcaa_34_sample_exam_1_sol.pdf

Exam 2 soln: http://www.itute.com/mathline/mm_vcaa_34_sample_exam_2_sol.pdf

My bad these are for non-cas