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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: vamsiaus on October 07, 2009, 05:27:01 pm

Title: Question
Post by: vamsiaus on October 07, 2009, 05:27:01 pm: Hey this is one of the multiple choice questions from the Neap06 exam 2. I am just not sure how to find the domain of the function.

If f(x) = sqrt(9-x^2) for x ---> [-3,3] and g(x) = sqrt(x) for x ---> [0,infinite), then f(g(x))= sqrt(9-x) for

A) x = R
B) x = [0,9]
C) x = [-3,3]
D) x = [0,infinite)
E) x= (-infinite, 9]

what I thought, was that the domain of the inside must be in the domain of the outside and worked it out as [0,3]. But none of the options had it.
If somebody could shed some light on how to work out the domain and ranges for these sort of questions, it'd be great. Thanks.
Title: Re: Question
Post by: TrueTears on October 07, 2009, 05:37:38 pm: $f(x) = \sqrt{9-x^2}$ and $g(x) = \sqrt{x}$

Now for $f(g(x))$ remember $\mbox{ran} g(x) \subset \mbox{dom} f(x)$

Lets check whether this is so.

$\mbox{ran} g(x) = [0, \infty)$

$\mbox{dom} f(x) = [-3,3]$

So no, $\mbox{ran} g(x) \subset \mbox{dom} f(x)$ is not satisfied.

We must now restrict $\mbox{ran} g(x)$ to $[0, 3]$

Now remember $\mbox{dom} f(g(x)) = \mbox{dom} g(x) = [0, 9]$

So, B.
Title: Re: Question
Post by: c23 on October 07, 2009, 06:05:54 pm: lol is this vamsi from hallam ?
Title: Re: Question
Post by: vamsiaus on October 07, 2009, 06:08:48 pm: yes, whos this C23 with an orange picture.
Title: Re: Question
Post by: c23 on October 07, 2009, 06:31:25 pm: ROFL! i knew it from ur name
its charlene
from ur methods class? ROFL
this is funny