ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: kdgamz on July 12, 2009, 05:19:38 pm

Title: heinemann textbook
Post by: kdgamz on July 12, 2009, 05:19:38 pm
hey does anyone have heinemann specialist worked solutions?  could you please post them up cos im dead stuck on loads of questions.
I'd really appreciate it.
Thanks :)
Title: Re: heinemann textbook
Post by: TrueTears on July 12, 2009, 05:20:59 pm
Maybe post up some questions you can't do and it would benefit everybody.
Title: Re: heinemann textbook
Post by: TonyHem on July 12, 2009, 05:24:21 pm
Yeah post up the questions :), I haven't seen stuff from the Heinemann book before.
Title: Re: heinemann textbook
Post by: kdgamz on July 13, 2009, 10:00:12 am
the question is:

find the domain of the derivative of:

y= cos-1(2√(x))

the derivative is -1/(√(x-4x2) but i keep getting the domain wrong!

Thanks

btw..if any1 has the teacher CD or anything like that...could you post it up???

thanks again :)
Title: Re: heinemann textbook
Post by: kdgamz on July 13, 2009, 03:56:07 pm
i take it no one knows the answer????
Title: Re: heinemann textbook
Post by: NE2000 on July 13, 2009, 04:13:52 pm
i take it no one knows the answer????

Is it (0, 1/4) ? by any chance
Title: Re: heinemann textbook
Post by: hyperblade01 on July 13, 2009, 04:17:19 pm
It depends on the domain of the function and since its inverse trig graph, it has to be one-to-one

Find out the domain of the inverse cosine graph that makes it one-to-one and there you go






Title: Re: heinemann textbook
Post by: kdgamz on July 14, 2009, 09:08:35 am
yeh thats the answer, but isnt the domain of a inverse cosine graph [-1,1]???

Title: Re: heinemann textbook
Post by: GerrySly on July 14, 2009, 09:50:55 am
yeh thats the answer, but isnt the domain of a inverse cosine graph [-1,1]???
But you can't take the square root of a negative number over R
Title: Re: heinemann textbook
Post by: kdgamz on July 14, 2009, 04:43:18 pm
oh yeh...thanks