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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: xce on February 26, 2008, 10:07:53 pm

Title: Families of functions question
Post by: xce on February 26, 2008, 10:07:53 pm: Hi all,

We have been given an assignment on Chapter 3: Families of functions (Essential Math Methods 3&4 textbook). I have been able to answer all questions on the assignment easily (the assignment is quite easy), apart from one of them. Any help is appreciated.

Express $\frac{2x+1}{5x+3}$ in the form $\frac{a}{x+b}+c$

Thanks!
Title: Re: Families of functions question
Post by: bucket on February 26, 2008, 10:10:41 pm: i know this one!..i think?

editsorry, it turns out i only know how to work it out without the coefficients too =\ lol. how embarrasing

$\frac{2x+1}{5x+3}$

$\frac{2x+3-2}{5x+3}$

$\frac{2x+3}{5x+3}-\frac{2}{5x+3}$

thats as far as I can go =\ sorry.
Title: Re: Families of functions question
Post by: xce on February 26, 2008, 10:13:12 pm: Heard of the question? It's from the assignment they ship on the Teacher's CD-ROM. I'd imagine quite a few schools use it.

I should also add that the textbook gives an example on how to do it, but only for questions where there is no coefficient in front of the x term. Therefore, the example was unhelpful.
Title: Re: Families of functions question
Post by: AppleXY on February 26, 2008, 10:22:04 pm: --> Disregard. Miscalc.
Title: Re: Families of functions question
Post by: bucket on February 26, 2008, 10:40:08 pm: apple are you sure that is right? because when I put your equation into the calculator with x=2, it doesn't get the same answer as the original equation.
Title: Re: Families of functions question
Post by: AppleXY on February 26, 2008, 10:49:50 pm: Disregard. Something went terribly wrong. Sorry, I have to check over it again, but now I gotta do my Identity & Belonging crap for english. :P
Title: Re: Families of functions question
Post by: dcc on February 26, 2008, 10:53:00 pm: I'd do it using long division, as such:

$\dfrac{2}{5}$

$5x + 3 | 2x + 1$

$-(2x + \dfrac{6}{5})$

Remainder: $-\dfrac{1}{5}$

Thus remembering our form for dividing:

$f(x) = q(x) + \dfrac{r(x)}{d(x)}$

$= \dfrac{2}{5} - \dfrac{\dfrac{1}{5}}{5x + 3}$

$= \dfrac{2}{5} - \dfrac{1}{25(x + \dfrac{3}{5})}$

$\mbox{a = -\dfrac{1}{25}, b = \dfrac{3}{5}, c = \dfrac{2}{5}}$
Title: Re: Families of functions question
Post by: AppleXY on February 26, 2008, 11:05:56 pm: lol, bloody hell why use my "alegbraic equality way". Urgh, it was an alternative.

Synthetic Division is always a winnar in these sort of questions.

EDIT: Found way. I'll type the alternative way soon. I SHOULD GET MY ENGLISH DONE :(
Title: Re: Families of functions question
Post by: Toothpaste on February 26, 2008, 11:21:39 pm: $\frac{2x+1}{5x+3}$ = $\frac{A}{5x+3} + \frac{B}{1}$
times through by 5x+3

$2x+1 = A + (5x+3)B$

To find A eliminate B:
let x = $\frac{-3}{5}$

$A = 2(\frac{-3}{5})+1$

$\therefore \implies A = \frac{-1}{5}$

now for B:
$2x+1 = A + (5x+3)B$

let x = 1
$3 = A +6B$ ... (1)

let x = 0
$1 = A+B$
rearrange
$A = 1-B$ ...(2)

sub (2) into (1):

$3 = (1-B) +6B$
$\implies 2 = 5B$
$\implies B = \frac{2}{5}$

$\frac{2x+1}{5x+3}$ = $\frac{A}{5x+3} + \frac{B}{1}$
put A and B back

$\therefore \frac{2x+1}{5x+3}$ = $\frac{-1}{5(5x+3)} + \frac{2}{5}$

$= \frac{-1}{(25x+15)} + \frac{2}{5}$

$=\frac{-1}{25(x+\frac{3}{5})} + \frac{2}{5}$

... same as dcc's :D ... with partial fractions
Title: Re: Families of functions question
Post by: humph on February 27, 2008, 11:30:49 am: Quote from: xce on February 26, 2008, 10:07:53 pm
Hi all,

We have been given an assignment on Chapter 3: Families of functions (Essential Math Methods 3&4 textbook). I have been able to answer all questions on the assignment easily (the assignment is quite easy), apart from one of them. Any help is appreciated.

Express $\frac{2x+1}{5x+3}$ in the form $\frac{a}{x+b}+c$

Thanks!
wow, i'd just do that kinda in my head...

$\frac{2x+1}{5x+3}= \frac{2x+\frac{6}{5}-\frac{1}{5}}{5x+3}= \frac{2x+\frac{6}{5}}{5x+3} - \frac{\frac{1}{5}}{5x+3} =\frac{(-1/25)}{x+\frac{3}{5}} + \frac{2}{5}$

i always just find it easier to manipulate the top line to match the bottom than to do long division or anything like that...
Title: Re: Families of functions question
Post by: lanvins on February 27, 2008, 04:00:29 pm: for the way dcc did it, how did u get q(X) r(x)?
Title: Re: Families of functions question
Post by: dcc on February 27, 2008, 08:15:03 pm: q(x) is like the quotient, if you divide 12 by 4, 3 is the quotient.
r(x) is the remainder
Title: Re: Families of functions question
Post by: AppleXY on February 27, 2008, 08:23:41 pm: Mmm. In synthetic division (aka Long Div) if you divide q(x) with r(x) this results in partial fractions if the remainder is not 0 :) [if zero, this will be a factor of q(x) ].

oh btw, when you obtain a remainder r(x) will be on the denominator.

i.e. $Q(x)$ + $\frac{Y}{r(x)}$ where Y is the constant.
Title: Re: Families of functions question
Post by: Mao on February 27, 2008, 09:10:42 pm: let me have a crack at solving it without long division:

$\frac{2x+1}{5x+3}$ in the form $\frac{a}{x+b}+c$

therefore:

$\frac{2x+1}{5x+3}=\frac{a}{x+b}+c$

$\frac{2x+1}{5x+3}=\frac{a+c(x+b)}{x+b}$

cross multiplying:

$(2x+1)(x+b)=(5x+3)(a+c(x+b))$

$2x^2+(2b+1)x+b=(5x+3)(cx+(a+bc))$

$2x^2+(2b+1)x+b=5cx^2+(3c+5(a+bc))x+3(a+bc)$

therefore:

[1]: $2=5c$ , [2]: $2b+1=3c+5(a+bc)$ , [3]: $b=3(a+bc)$

from [1]: $c=\frac{2}{5}$ ,

substituting into [2]:

$2b+1=\frac{6}{5}+5a+2b$

$-\frac{1}{5}=5a$

$a=-\frac{1}{25}$

and hence, substituting a and c gives:

$b=\frac{3}{5}$

yay i got the same as everyone else

$\frac{(-1/25)}{x+\frac{3}{5}} + \frac{2}{5}$

.. :D
Title: Re: Families of functions question
Post by: AppleXY on February 27, 2008, 09:38:49 pm: lol, theres many ways to do. But I'd do the synthetic or equalities using >1 coffecients (toothpick's posts).
Title: Re: Families of functions question
Post by: Mao on February 27, 2008, 09:48:55 pm: mmm on a closer inspection, toothpick's method was the easiest, and probably will become my preference now

but partial fractions is not part of the methods course, and is only used in spec. for differentiation (or antidifferentiation? i forget), and for someone doing methods, i'm guessing that simple algebra (with a few more steps) and equality of polynomials should suffice :P

but, PARTIAL FRACTIONS FTW!
Title: Re: Families of functions question
Post by: dcc on February 27, 2008, 10:20:06 pm: $\mbox{Let f(x) }= \dfrac{2x + 1}{5x + 3}\mbox{ and let F(x) be the antiderivative of f(x)}$

$\mbox{Let }F(x) = \int \dfrac{2x + 1}{5x + 3} \mbox{ } dx$

$\mbox{Let }u = 5x + 3$

$\dfrac{du}{dx} = 5$

$\dfrac{1}{5} \int \dfrac{2x + 1}{u}\mbox{ } du$

Now obviously we have to get rid of the $2x + 1$ from the numerator, so let us figure out what 2x + 1 equals:

$u = 5x + 3\mbox{ therefore: } 5x = u - 3$

$2x = \dfrac{2}{5}(u - 3)$

$2x = \dfrac{2}{5}u - \dfrac{6}{5}$

adding 1 to find 2x + 1, we get $2x + 1 = \dfrac{2}{5}u - \dfrac{1}{5}$

So we can rewrite our integral:

$\dfrac{1}{5} \int \dfrac{\dfrac{2}{5}u - \dfrac{1}{5}}{u}\mbox{ }du$

Which when expanded out is:

$\dfrac{1}{5} \int \dfrac{2}{5} - \dfrac{1}{5u}\mbox{ }du$

Anti differentiating this, we get:

$F(x) = \dfrac{1}{5} ( \dfrac{2}{5}u - \dfrac{1}{5} \log_e{u} ) + c$

Which when getting rid of the 1/5's, becomes:

$F(x) = \dfrac{1}{25} (2u - log_e{u}) + c$

Substituing in for u, we get:

$F(x) = \dfrac{1}{25} (2(5x + 3) - log_e(5x + 3)) + c$

$F(x) = \dfrac{1}{25} (10x + 6 - log_e(5x + 3)) + c$

Now, differentiating F(x), we will arrive back at f(x):

$f(x) = \dfrac{1}{25}(10 - \dfrac{5}{5x + 3})$

When expanded, this becomes:

$f(x) = \dfrac{2}{5} - \dfrac{1}{25x + 15}$

$f(x) = \dfrac{2}{5} - \dfrac{1}{25}(\dfrac{1}{x + \dfrac{3}{5}})$

Therefore: $a = -\dfrac{1}{25}\mbox{, }b = \dfrac{3}{5}\mbox{ and last but not least, }c = \dfrac{2}{5}$
Title: Re: Families of functions question
Post by: Mao on February 27, 2008, 10:23:46 pm: OMG I LOVE CALCULUS!!!!!!!!!!!!!!!!!!!!!

that was so smart xD
Title: Re: Families of functions question
Post by: AppleXY on February 27, 2008, 10:34:31 pm: Quote from: Mao on February 27, 2008, 10:23:46 pm
OMG I LOVE CALCULUS!!!!!!!!!!!!!!!!!!!!!

that was so smart xD

Amen.
Title: Re: Families of functions question
Post by: Glockmeister on February 27, 2008, 10:35:57 pm: 2nd Amen here.

Still prefer long division though.
Title: Re: Families of functions question
Post by: Collin Li on February 27, 2008, 10:40:49 pm: The best way is just to match the denominator:

$\frac{2x+1}{5x+3} = \frac{\frac{2}{5}\left(5x + \frac{5}{2}\right)}{5x+3} = \frac{\frac{2}{5}\left(5x + 3 - \frac{1}{2}\right)}{5x+3} = \frac{\frac{2}{5}\left(5x + 3\right) - \frac{1}{5}}{5x+3}$

$= \frac{2}{5} + \frac{-\frac{1}{5}}{5x+3} = \frac{2}{5} + \frac{-\frac{1}{25}}{x+\frac{3}{5}}$

Always use this trick when you're dividing a linear function by another one. It's so easy! Here's a simple example:

$\frac{x+2}{x+1} = \frac{(x+1)+1}{x+1} = 1 + \frac{1}{x+1}$

The hassle of long division is avoided!
Title: Re: Families of functions question
Post by: humph on February 28, 2008, 12:13:53 am: precisely! only i generally do most of that working out in my head ;D
Title: Re: Families of functions question
Post by: Neobeo on February 28, 2008, 12:46:23 am: The main problem with every method discussed here is that if you make one wrong step, you get everything wrong. In all the above solutions, a, b and c are very closely dependent on each other. Therefore a wrong answer in a will result in a wrong answer for b and c. This is of course very bad for marks. So let's see how we can solve each variable one at a time.

$\text{Let }f(x)=\frac{2x+1}{5x+3}\text{ and }g(x)=\frac{a}{x+b}+c$

By equating $f(x)$ with $g(x)$ we can obtain the solution, but again this will be prone to "one mistake leads to everything wrong" kind of situation as mentioned.

We can find c by itself, using limits.

$\lim\limits_{x \to \infty }f(x)=\frac{2}{5}\mbox{, }\lim\limits_{x \to \infty }g(x)=c$

$\therefore c = \frac{2}{5}$

We can also find b by itself, also using limits.

$\lim\limits_{x \to -b}f(x)=\frac{1-2b}{3-5b}\mbox{, }\lim\limits_{x \to -b}g(x)=\infty$

$f(x)=g(x)\mbox{ so }3-5b=0$

$\therefore b = \frac{3}{5}$

Finally we now we find a. This should be less taxing since it uses no limits at all. Only derivatives.

$f'(x)=\frac{1}{(3+5x)^2}\mbox{, }g'(x)=-\frac{a}{(b+x)^2}$

$\frac{1}{f'(x)}=(3+5x)^2\mbox{, }\frac{1}{g'(x)}=-\frac{(b+x)^2}{a}$

Keeping in mind that both sides are equal, we differentiate them two more times

$\frac{d^2}{dx^2}\frac{1}{f'(x)}=50\mbox{, }\frac{d^2}{dx^2}\frac{1}{g'(x)}=-\frac{2}{a}$

$\therefore a = -\frac{2}{50} = -\frac{1}{25}$

As you can see, the solutions for each variable are independent from each other. Now you do not have to worry about making a mistake from the first step. Even if one of them is wrong, the other two can still be right!

In conclusion: $c = \dfrac{2}{5}\mbox{, }b = \dfrac{3}{5}\mbox{ and last but not least, }a = -\dfrac{1}{25}$
Title: Re: Families of functions question
Post by: xce on February 28, 2008, 07:02:41 am: Thanks so much for all your help, everyone! I'll read and digest it all tonight :P Out of curiosity, how did you know about synthetic long division? We have only covered polynomial long division in class?
Title: Re: Families of functions question
Post by: AppleXY on February 28, 2008, 07:44:31 am: http://en.wikipedia.org/wiki/Synthetic_division
Title: Re: Families of functions question
Post by: xce on February 28, 2008, 07:46:39 pm: Heh, I read up when I saw dcc's post. So you just read mathematics-related Wikipedia pages? You haven't been taught this in class? Is it part of the course?
Title: Re: Families of functions question
Post by: Glockmeister on February 28, 2008, 07:48:29 pm: I don't think this is specifically in the curriculum, but I think the MHS people do get taught this ('cause my brother knows it)
Title: Re: Families of functions question
Post by: Collin Li on February 28, 2008, 07:57:43 pm: Synthetic division is nothing more than just doing the long division in your head. It's simple enough: just write it down term by term, storing the next number in your head (or sometimes the next two).
Title: Re: Families of functions question
Post by: Ahmad on February 28, 2008, 09:06:10 pm: Convoluted solution (upon request) with absolutely no benefit, using generating functions:

LHS:
$\frac{2x + 1}{5x + 3} = \frac{2x+1}{3(\frac{5}{3}x + 1)}$

$= \frac{2x+1}{3} \cdot \displaystyle \sum_{k=0}^\infty (-\frac{5}{3}x)^k$

$=\frac{1}{3}\left( 1 + \displaystyle \sum_{k=1}^\infty \left[ 2(\frac{-5}{3})^{k-1} x^k + (\frac{-5}{3}x)^k \right] \right)$

$= \frac{1}{3} + \frac{-1}{15} \displaystyle \sum_{k=1}^\infty (-\frac{5}{3}x)^k$

RHS:
$\frac{a}{x + b} + c = \displaystyle (c+\frac{a}{b}) + \frac{a}{b} \sum_{k=1}^\infty (-\frac{x}{b})^k$

$\frac{-5}{3} = \frac{-1}{b} \implies b = \frac{3}{5}$

$\frac{a}{b} = \frac{-1}{15} \implies a = \frac{-1}{25}$

$c + \frac{a}{b} = \frac{1}{3} \implies c = \frac{2}{5}$
Title: Re: Families of functions question
Post by: AppleXY on March 01, 2008, 01:38:17 pm: Wow. Well there you go, generating functions for Partial fractions. Amazing. But again, SUPER impractical :P
Title: Re: Families of functions question
Post by: Ahmad on March 02, 2008, 11:58:43 am: It's not meant to be practical. We were doing a 'most convoluted solution' thing. :P