function

Welcome to ScienceForums.Net!

Welcome to ScienceForums.Net! We welcome science discussion at all levels — from beginners to researchers, covering topics from biology to computer science, and much more. Registration is fast and free, and allows you to post on the forums, so register now and join the discussions!

After you've registered, come in and introduce yourself, or visit the forum index. If you need any help registering, posting, or if you just have some questions about our site, please feel free to contact us at staff at scienceforums dot net.

Start new topics and reply to others
Subscribe to topics and forums to get automatic updates
Create a ScienceForums.Net Blog!

Homework Help Rules

A simple reminder to all: this is the "Homework Help" forum, not the "Homework Answers" forum. We will not do your work for you, only point you in the right direction. Posts that do give the answers may be removed.

Page 1 of 1

You cannot start a new topic
You cannot reply to this topic

function word problem Rate Topic:

#1 8 February 2012 - 12:21 PM Vastor

Baryon

word problem? probably... k, let see

function f is defined by $f(x) = \frac{kx - 4}{x - 1}, x \neq 1$
find the range of values of k such that $f(x) = x$ has no solutions

my trials,

none, lol...

anyone can give idea what this question mean by "solution" ? :mellow:

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.

Posts: 146 | Joined: 13-May 11
Back to top of the page up there ^
MultiQuote
Reply

#2 8 February 2012 - 12:42 PM imatfaal

Primate

Edit IGNORE this- didnt read question carefully. very sorry

If by solutions you are looking for values of x such that y = 0 which I am pretty sure is the idea. Then I can think of one fairly trivial value of k that will allow no value of x that will give y=0 . I can also think of one more interesting value. I cannot think of a range of values of k - just two distinct values. I will give it some more thought.

In the meantime - think about (for different values of k)
1. what these curves look like in general (and is there any k that makes a curve that doesn't look the same!)
2. you know what happens at x=1 - but what happens a little bit before and after
3. what is the value when x is very large

This post has been edited by imatfaal: 8 February 2012 - 01:38 PM

A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:
there shallow draughts intoxicate the brain, and drinking largely sobers us again.
- Alexander Pope
feel free to click the green [+] ---->

Posts: 1,759 | Joined: 28-September 10
Back to top of the page up there ^
MultiQuote
Reply

#3 8 February 2012 - 01:14 PM Vastor

Baryon

imatfaal, on 8 February 2012 - 12:42 PM, said:

If by solutions you are looking for values of x such that y = 0 which I am pretty sure is the idea. Then I can think of one fairly trivial value of k that will allow no value of x that will give y=0 . I can also think of one more interesting value. I cannot think of a range of values of k - just two distinct values. I will give it some more thought.

In the meantime - think about (for different values of k)
1. what these curves look like in general (and is there any k that makes a curve that doesn't look the same!)
2. you know what happens at x=1 - but what happens a little bit before and after
3. what is the value when x is very large

too much calculus for a "chapter 1: function"? xD
oh, btw, this coming from exercise book, answer : -5 < k < 3

1. shape, *calculate on graph calculator*, herm, some sort of "reciprocal"
2. yeah, it's undefined, before = y reaching infinity, after = y moving from the infinity, the timeline of "after" and "before" is based on "x"
3. not really learned calculus deep enough to conclude anything here...

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.

Posts: 146 | Joined: 13-May 11
Back to top of the page up there ^
MultiQuote
Reply

#4 8 February 2012 - 01:37 PM imatfaal

Primate

Hmmm. Ignore my above - I was misreading question. Will open my eyes and redo from start

Posts: 1,759 | Joined: 28-September 10
Back to top of the page up there ^
MultiQuote
Reply

#5 8 February 2012 - 01:53 PM ajb

Physics Expert

You have to think about

$x= \frac{kx - 4}{x-1}$ ,

assuming that $x \neq 1$ and I assume real.

Can you rearrange this into a form you recognise?

"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

My homepage.

Posts: 4,446 | Joined: 04-June 06
Back to top of the page up there ^
MultiQuote
Reply

#6 8 February 2012 - 01:59 PM Vastor

Baryon

ajb, on 8 February 2012 - 01:53 PM, said:

You have to think about

$x= \frac{kx - 4}{x-1}$ ,

assuming that $x \neq 1$ and I assume real.

Can you rearrange this into a form you recognise?

huh? x actually not equal 1 and for that reason it's real, why should I assume it anymore?

I'm missing your point...

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.

Posts: 146 | Joined: 13-May 11
Back to top of the page up there ^
MultiQuote
Reply

#7 8 February 2012 - 02:12 PM imatfaal

Primate

OK the book is correct - I will take you through it

$f(x) = \frac{kx - 4}{x - 1}, x \neq 1$

now we are equating f(x) with x

$x = \frac{kx - 4}{x - 1}$

simplifiy

$x (x-1) = (kx-4)$
$x^2-x = kx-4$
$x^2 -x -kx +4 = 0$
$x^2 -(1+k)x +4 = 0$

now we have a quadratic - we know when this is solvable by using the quadratic formula

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

See if you can take it from here - if not I will come back after this meeting I have to go to

Posts: 1,759 | Joined: 28-September 10
Back to top of the page up there ^
MultiQuote
Reply

#8 8 February 2012 - 02:45 PM Vastor

Baryon

imatfaal, on 8 February 2012 - 02:12 PM, said:

with my high-school algebra style(plug and solve), I can't really perform any formula for this... but from what I understand,
doesn't the exercise mean by "no solution" that

$b^2 - 4ac < 0$ ???

ok, got it, thnx, but...

when I got here

$k^2 + 2k -15 < 0$
$(k-3)(k+5) < 0$
$(k-3) < 0$ & $(k+5) < 0$
$k < 3$ & $k < -5$

but the answer -5 < k < 3

I always "cheat" just to fit it into the answer
$(k-3)(-k-5) < 0$
$(k-3) < 0$
$k < 3$
&
$(-k-5) < 0$
$-k < 5$
$k > -5$

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.

Posts: 146 | Joined: 13-May 11
Back to top of the page up there ^
MultiQuote
Reply

#9 8 February 2012 - 03:04 PM imatfaal

Primate

Think about the shape of a graph made by

$y=(k-3)(k+5)$

for which ranges is y positive and which range is y negative

Posts: 1,759 | Joined: 28-September 10
Back to top of the page up there ^
MultiQuote
Reply

#10 8 February 2012 - 03:25 PM Vastor

Baryon

so, it's just basically the direct reasoning "from words to mathematic"?!

usually, with equation, there always a way to derive, but this not really works with range it seems...

btw, what a silly revision book, lucky I'm form 5 already,(revised what I learn from form 4 chapter for sake of proficiency)
it's kind of lame to include chapter 2 into this chapter, especially when the reader is form 4, and with a confusing word problem
(there is exercise tell "find the value of x which is mapped onto itself" wtf?!) good revision book for one who good at math, but not really for freshman.
anyway, thnx to this book, I search on how to solve absolute value equation yesterday, which come out in exam today :lol:

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.

Posts: 146 | Joined: 13-May 11
Back to top of the page up there ^
MultiQuote
Reply

#11 8 February 2012 - 03:43 PM ajb

Physics Expert

Vastor, on 8 February 2012 - 01:59 PM, said:

huh? x actually not equal 1 and for that reason it's real, why should I assume it anymore?

I'm missing your point...

As you see from the workings out of others, x could be a complex number.

"In physics you don't have to go around making trouble for yourself - nature does it for you" Frank Wilczek.

My homepage.

Posts: 4,446 | Joined: 04-June 06
Back to top of the page up there ^
MultiQuote
Reply

#12 8 February 2012 - 04:14 PM Vastor

Baryon

ajb, on 8 February 2012 - 03:43 PM, said:

As you see from the workings out of others, x could be a complex number.

oh, so that's what you mean by "real", ha3, well, I thought non-real = undefined(x = 1, etc...), but talking about my "non-existant" knowledge of imaginary number, I'm just self- teaching about it, never taught at school yet... think I just leave the case from me...

This post has been edited by Vastor: 8 February 2012 - 04:15 PM

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.