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From: Ivan on 8 Mar 2005 05:37
Hi,

At least, it's a problem that *I* am finding difficult (read "impossible")!

p(x)=[abs(x-1)+x-2] / [abs(x-1)-3x]

The questions read:
(a) Sketch p
(b) Express p(x) as a piecewise defined function without absolute values.

By using the graphics calculator I have managed to sketch p! It seems that
there is a vertical asymptote at x=0.25, a right hand horizontal asymptote
at y=-0.5 and a left hand horizontal asymptote at y=0. Is there a better
way to determine what asymptotes exist than guessing and testing while
looking at the graph on the calculator?

I considered that calculating the limits as x->0-,0+,oo- and oo+ might be a
good idea but then I realised that I don't know how to do this with
absolute value functions!

The other idea that I came up with was that there must be a vertical
asymptote when the divisor = 0. That gave me x-1-3x=0 and 1-x-3x=0. Solving
for x gave me *two* values, x=-0.5 and x=0.25 but the graph definitely
shows only one vertical asymptote. Obviously my theory is half right but I
don't understand what half or why!

Even with the advantage of having the graph in front of me I have no idea
how to go about expressing this function in piecewise form. It's obviously
some form of reciprical function with domains x<0.25 and x>0.25 but beyond
that I am completely lost!

Any help at all will be appreciated - I've spent at least 4 hours (probably
more) playing with this.

Thanks in advance,

Ivan.
From: Michael J�rgensen on 8 Mar 2005 07:23

"Ivan" <ivan999(a)remove34this.iprimus.com.au> wrote in message
news:Xns9613BE6525929ivan999remove34thisi(a)203.134.67.67...
> Hi,
>
> At least, it's a problem that *I* am finding difficult (read
"impossible")!
>
> p(x)=[abs(x-1)+x-2] / [abs(x-1)-3x]
>
> The questions read:
> (a) Sketch p
> (b) Express p(x) as a piecewise defined function without absolute values.

As so often is the case, it looks more difficult than it really is :-)

There is really not very much to think about. The key is to get rid of the
abs() functions. In your problem, you have two instances of abs(x-1). So the
first step is to divide into two cases: x>1 and x<1. In each interval you
will have an expression for p(x), written without abs() functions. Note that
the two expressions for p(x) will not be the same.

Now that the abs() is gone, you're back on safe ground again.

Hope that helps.

-Michael.


From: Ivan on 8 Mar 2005 07:25
<SNIP original message>

Progress at last!

It finally dawned on me that if my proposition is that x>1 and I obtain x=-
0.5 when I solve the equation then it is as simple as there being no
solutions in that interval!

I still don't know about the horizontal asymptotes and sketching but at
least I'm working on the problem and (hopefully) getting closer.

Ivan.
From: Ivan on 8 Mar 2005 08:00
"Michael Jýrgensen" <ccc59035(a)vip.cybercity.dk> wrote in
news:422d994c$0$306$edfadb0f(a)dread12.news.tele.dk:
>
> "Ivan" <ivan999(a)remove34this.iprimus.com.au> wrote in message
> news:Xns9613BE6525929ivan999remove34thisi(a)203.134.67.67...
>> p(x)=[abs(x-1)+x-2] / [abs(x-1)-3x]

> As so often is the case, it looks more difficult than it really is :-)

*smiles* or maybe it looks just as difficult as it is - see below

>
> There is really not very much to think about. The key is to get rid of
> the abs() functions. In your problem, you have two instances of
> abs(x-1). So the first step is to divide into two cases: x>1 and x<1.
> In each interval you will have an expression for p(x), written without
> abs() functions. Note that the two expressions for p(x) will not be
> the same.
>
> Now that the abs() is gone, you're back on safe ground again.

Yes, your explanation makes perfect sense and I thought I understood it!
Perhaps I'm missing some understanding about abs(x) ?

If x<1, (x-1) is negative so abs(x-1)= -(x-1)= 1-x
In this case, p(x)=[(1-x)+x+2]/[(1-x)-3x] = 3/(1-4x)

If x>, (x-1) is positive so abs(x-1)=(x-1)
In this case, p(x)=[(x-1)+x+2]/[(x-1)-3x] = (2x+1)/(-2x-1)=-1!

Plotting these two functions against the original shows quite a substantial
difference between the two.

I'm sorry if I'm being stupid about this, despite lots of practise and
reading I have never really understood abs(x).

Ivan.
>
> Hope that helps.

Every little bit helps, I'm just hoping that you're not throwing pearls
before the swine :(

>
> -Michael.
>
>
>

From: Michael J�rgensen on 8 Mar 2005 08:23

"Ivan" <ivan999(a)remove34this.iprimus.com.au> wrote in message
news:Xns9613D6B6EEA3Fivan999remove34thisi(a)203.134.67.67...
> "Michael Jýrgensen" <ccc59035(a)vip.cybercity.dk> wrote in
> news:422d994c$0$306$edfadb0f(a)dread12.news.tele.dk:
> >
> > "Ivan" <ivan999(a)remove34this.iprimus.com.au> wrote in message
> > news:Xns9613BE6525929ivan999remove34thisi(a)203.134.67.67...
> >> p(x)=[abs(x-1)+x-2] / [abs(x-1)-3x]
>
> If x<1, (x-1) is negative so abs(x-1)= -(x-1)= 1-x
> In this case, p(x)=[(1-x)+x+2]/[(1-x)-3x] = 3/(1-4x)

You've copied the function incorrectly. It should be "-2", not "+2", in the
numerator.

> If x>, (x-1) is positive so abs(x-1)=(x-1)
> In this case, p(x)=[(x-1)+x+2]/[(x-1)-3x] = (2x+1)/(-2x-1)=-1!

Ditto here.

-Michael.


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