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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Rosie on April 25, 2008, 05:53:19 pm

Title: Chapter 1 - 5 HELP!
Post by: Rosie on April 25, 2008, 05:53:19 pm: I don't understand the purpose of a maximal domain. Isn't it the same thing as a domain just that the maximal domain is the maximum number that an x-value can take?
Title: Re: Chapter 1 - 5 HELP!
Post by: droodles on April 25, 2008, 06:00:46 pm: yes it is, implied domain is also the same thing as maximal domain
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 25, 2008, 06:04:38 pm: Oh ok, thanks
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 25, 2008, 06:07:25 pm: Quote from: Rosie on April 25, 2008, 05:53:19 pm
I don't understand the purpose of a maximal domain. Isn't it the same thing as a domain just that the maximal domain is the maximum number that an x-value can take?
as well as the minimum number an x-value can take, but discluding anything in the middle that it cannot include

in more pedantic terms, it is the "largest subset of R that an x-value can take".

e.g. for $\frac{1}{x}$ , the maximal/implied domain is $x\neq 0$ or $x\in R \backslash \{ 0 \}$ or $x\in (-\infty,0) \cup (0,\infty)$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 25, 2008, 06:23:06 pm: With modulus functions, the answer always has to be positive. Is this true?
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 25, 2008, 06:27:44 pm: Quote from: Rosie on April 25, 2008, 06:23:06 pm
With modulus functions, the answer always has to be positive. Is this true?
positive or zero
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 25, 2008, 06:31:28 pm: thanks
Title: Re: Chapter 1 - 5 HELP!
Post by: plato on April 25, 2008, 08:38:18 pm: Quote from: Rosie on April 25, 2008, 06:23:06 pm
With modulus functions, the answer always has to be positive. Is this true?

Let's take a step back for a minute. This thread began with a question re maximal domain and so the "answer" will not always be positive if the question is "what is the maximal domain?" The maximal domain, or largest set of allowable x-values, can include negative values such as in $x \in R \backslash \{0\}$

On the other hand, your question about the "answer" with modulus functions always being positive suggests you could be asking about the range rather than the domain. The range of a modulus function is always positive if the functions are such as $f(x)= |x^3-4x^2+2| or g(x)=|cos x| etc$

But, if such moduls functions are translated downward or have other functions added or subtracted, then the range may include negative values.
Examples are $f(x)=|x^3+5|-8 or g(x)=3x+|x-2|$
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 25, 2008, 09:19:11 pm: plato: i believe they were two separate questions :P
but nice examples

modulus function by definition:

$\begin{array}{ccc}<br /> & x & x\ge 0\\<br />|x|=\{ & & \\<br />& -x & x<0\\<br />\end{array}$

hence, it is always positive
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 05:35:14 pm: Here's a question for everyone.
The maximal domain of the function f with rule f(x) = log_e(5x-2) is:
Title: Re: Chapter 1 - 5 HELP!
Post by: /0 on April 26, 2008, 05:42:55 pm: In any logarithm function $f(n) = \log_a n$ , with positive base $a$ , $n$ must be greater than zero.

So for $f(x) = \log_{e}(5x-2)$ ,

$5x-2 > 0$

$5x > 2$

$x > \frac{2}{5}$

Or $x \in (\frac{2}{5}, \infty)$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 05:47:49 pm: Wow. Thanks. So you're saying that whenever an implied domain is asked for a log function and tha base is positive, the steps required to answer this problem will be almost the same. n will be > 0
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 05:55:49 pm: I also have a similar question.
For the function with rule f(x) = (8-4x)^1/2, what is the maximal domain?
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 05:59:01 pm: I have solved this question using an example from my text book but I don't understand why you solve it like that.
Title: Re: Chapter 1 - 5 HELP!
Post by: /0 on April 26, 2008, 06:23:19 pm: Yep, in the logarithm all you need to do is make sure that bit is greater than zero (also, I don't think they would ever give you a negative base)

$f(x)=\sqrt{8-4x}$

We require that $8-4x \geq 0$ , since the domain is a subset of $R$ .

Solving gives $x \leq 2 \Rightarrow x \in (-\infty, 2]$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 06:37:58 pm: Ok. I think I get it.
For the functions f: (-infinity, 2] for all real numbers, f(x) = 2-x^1/2 and g with rule g(x) = IxI teh composition f o g is defined i fthe domain is:

(sorry about the full writing)
Title: Re: Chapter 1 - 5 HELP!
Post by: dcc on April 26, 2008, 06:49:35 pm: The thing about implied domain is that you are restricting the domain of a function to only values which make sense

for example:

$f(x) = \sqrt{x}$

What is the implied domain of this function? well, think about the square root function.

$f(0) = \sqrt{0} = 0$

$f(4) = \sqrt{4} = 2$

$f(-1) = \sqrt{-1} = \mbox{undefined}$

As you should know, you cannot square root negative numbers, so in this case, you have to ensure that the domain of f(x) only allows for zero and positive values.

Another example:

$g(x) = \dfrac{1}{\sqrt{x}}$

So we have another square root graph, except now its the reciprocal.

trying some values again:

$g(1) = \dfrac{1}{1} = 1$

$g(4) = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}$

$g(0) = \dfrac{1}{0} = \mbox{undefined}$

$g(-1) = \dfrac{1}{\sqrt{-1}} = \mbox{undefined}$

So now in this example, the domain has been restricted to $x > 0$ to ensure that you do not get:
a. divide by zero
or
b. square root of negative numbers.
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 26, 2008, 06:58:41 pm: Thanks dcc. I liked that explanation. Things are clearer now.
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 09:13:24 am: What about that question?
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 09:18:15 am: Just to speed things up, I have another question.

The range of the function f:R -> R, f(x) = Ix^2 - 4I + 3 is
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 09:23:05 am: Also, for this question
P(x) = x³ + ax² + bx - 9 has zeros at x=1 and x=-3. The values of a and b are:

Can this be solved using two simultaneous equations by subing the values 1 and -3 in and then finding a and b.
Title: Re: Chapter 1 - 5 HELP!
Post by: Odette on April 27, 2008, 09:32:27 am: Quote from: Rosie on April 27, 2008, 09:23:05 am
Also, for this question
P(x) = x³ + ax² + bx - 9 has zeros at x=1 and x=-3. The values of a and b are:

Can this be solved using two simultaneous equations by subing the values 1 and -3 in and then finding a and b.

Yes. :)
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 09:58:04 am: Yep I got the answer now thanks. I just need help with those other questions I posted above.
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 27, 2008, 11:29:02 am: Quote from: Rosie on April 27, 2008, 09:18:15 am
Just to speed things up, I have another question.

The range of the function f:R -> R, f(x) = Ix^2 - 4I + 3 is

firstly, it'll look better if you used the key "|" located above Enter
and it'll be even better if you could use LaTeX: http://vcenotes.com/forum/index.php/topic,3137.0.html
that is how many of us can type fancy-looking maths

as for the question:
1) looking at the function inside the modulus, the range of that is $[-4,\infty)$ (we know this as it is a quadratic)

2) that is, after applying the modulus, the range of the modulus will be $[0,\infty)$

3) the modulus then has 3 added to it, hence the range of f is $[3,\infty)$

you need to be careful with step 1, a modulus only makes negative numbers positive, so its range is not always $[0,\infty)$ If you had $f:R\to R,\mbox{ where }f(x)=|x^2+4|+3$ , the range of the modulus will actually be $[4,\infty)$

hope that made sense

PS Rosie:
can you please avoid double/triple (quintuple) posting =)
if you need to add things to your latest post (before someone else responds to it), there are modify functions available.
as much as I would like to see someone finally achieving sextuple posting, it can get annoying.
thanks
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 11:39:28 am: Thanks for that. I didn't know how to get that key
Say, is the range of a modulus function always [0, infinity)
So, step 1 is not really needed as we only really need to know the range of the modulus function and whether the y-value has moved up or down. Is this correct?
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 27, 2008, 11:57:04 am: Quote from: Rosie on April 27, 2008, 11:39:28 am
Say, is the range of a modulus function always [0, infinity)
So, step 1 is not really needed as we only really need to know the range of the modulus function and whether the y-value has moved up or down. Is this correct?

no, I made the point that the range of a modulus function is not necessarily $[0,\infty)$ , it can be $[4,\infty)$ if the inner function's minimum value is 4, etc
step 1 is important, always check the range of the inner function first.
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 27, 2008, 10:02:10 pm: ok. thanx
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 28, 2008, 06:29:20 pm: Hey, this was a question on my sac today and I was abit confused with how to solve it: (multiple-choice question on the sac)

$t = x \div x - a$ , solve for x

I put my answer as $x = 1 \div t-a$ just because I couldn't think of a way to solve it.

Anyone help?
Title: Re: Chapter 1 - 5 HELP!
Post by: plato on April 28, 2008, 06:36:09 pm: Quote from: Rosie on April 28, 2008, 06:29:20 pm
Hey, this was a question on my sac today and I was abit confused with how to solve it: (multiple-choice question on the sac)

$t = x \div x - a$ , solve for x

I put my answer as $x = 1 \div t-a$ just because I couldn't think of a way to solve it.

Anyone help?

I assume the question is $t = \frac{x}{x-a}$

Then $t(x-a) = x$
   $\therefore tx-ta=x$
   $\therefore tx-x=ta$
   $\therefore x(t-1)=ta$
   $\therefore x=\frac{ta}{t-1}$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 28, 2008, 06:49:53 pm: Whoops. I got that question wrong. Looked like a tricky question but I guess I was wrong. That was a great explanation. Thanks
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 29, 2008, 04:02:16 pm: Can someone please solve this question for me.
If ax³ + x² + 6 is exactly divisible by x + 1, the value of a is:
Title: Re: Chapter 1 - 5 HELP!
Post by: ed_saifa on April 29, 2008, 04:15:12 pm: Since you know that $x +1$ is a factor of $ax^3 +x^2 +6$ then we know that $P(-1)=0$
Therefore

$P(-1)= -a + 1 +6=0$

$-a= -7$

$a=7$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on April 30, 2008, 06:45:15 pm: Does anyone know how to change your topic name. Can you do that?
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on April 30, 2008, 07:40:35 pm: go to the first post (by you) and click "modify"
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on June 17, 2008, 03:36:16 pm: Passage in my textbook:
The definition of a function tells us that for each x in the domain of f there is a unique element,y, in the range such that (x,y) $\in$ f. The element y is called the image of x under f or the value of f at x and is denoted by f(x). If (x,y) $\in$ f, then x is called a pre-image of y.

I'm preparing my reference book for methods and this paragraph makes no sense to me. Can anyone explain it to me?

What does image and pre-image mean. (is image the x values and pre-image the y values)

I also wanted to ask what the codomain is?
Title: Re: Chapter 1 - 5 HELP!
Post by: Mao on June 17, 2008, 05:08:30 pm: mmm nice question:

1. i have never heard of "image" or "pre-image" used in the sense of the cartesian plane.

2. don't worry about the codomain. For the purpose of methods, just accept that for the functions you will be dealing with, the codomain will be R.
but if you do want a definition, it is the plane at which the values of the function exist on.
Title: Re: Chapter 1 - 5 HELP!
Post by: kj_ on June 17, 2008, 05:13:49 pm: the words pre- image and image are just fancy terms for subbing a value in as f(x) and subbing a value in as x. i can't remember which one is which, though - but i've never seen them in any question for 3/4 methods, so unless you were referring to 1/2 methods, don't worry about it at all :)

edit: just went through my 1/2 book, and to clarify,

preimage, say, f(x) = x +1, the preimage of 7, let 7 = f(x) and solve for x
image, say, f(x) = x + 1, the image of 1, let 1 = x, and solve for f(x)

you should be able to try and put a literal meaning to the words "pre-image" and "image" now :)
Title: Re: Chapter 1 - 5 HELP!
Post by: /0 on June 18, 2008, 01:15:05 am: Put simply:
$Range \subseteq Codomain$

E.g Consider $f: [1, \infty) \rightarrow R, \ f(x) = x^2-2x$

The $[1, \infty)$ bit will be the domain, and $R$ will be the codomain. As you can see from observing the graph, the range is actually $[-1, \infty)$
Title: Re: Chapter 1 - 5 HELP!
Post by: Rosie on June 18, 2008, 08:01:10 am: I understand now but it is in my 3/4 methods book. There's actually one or two questions about finding the pre-image. Thanks for that

Q. For {(2,y):y $\in$ Z} Is this a function and state the domain and range.
Q. For y $\ge$ 3x+2 Is this a function and state the domain and range.
How would you go about this. Thanks
Title: Re: Chapter 1 - 5 HELP!
Post by: /0 on June 18, 2008, 06:40:21 pm: Q.1

You could rephrase it as: the points (2,y) such that y is an integer, ...(2, -1), (2, 0), (2, 1), (2, 2)...

So the domain will be {2} and the range will be Z.

Q.2

$y \geq 3x+2$ isn't a function because for each x coordinate there is more than one y coordinate. The domain and range will both be R because if u draw the graph the required region will go from $-\infty$ to $\infty$ in both the x and y directions.