word problem? probably... k, let see

 

function f is defined by [math]f(x) = \frac{kx - 4}{x - 1}, x \neq 1 [/math]

find the range of values of k such that [math]f(x) = x [/math] has no solutions

 

my trials,

 

none, lol...

 

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

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

Edited February 8, 201213 yr by imatfaal

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...

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

You have to think about

 

[math]x= \frac{kx - 4}{x-1}[/math],

 

assuming that [math]x \neq 1[/math] and I assume real.

 

Can you rearrange this into a form you recognise?

You have to think about

 

[math]x= \frac{kx - 4}{x-1}[/math],

 

assuming that [math]x \neq 1[/math] 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...

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

 

[math] f(x) = \frac{kx - 4}{x - 1}, x \neq 1 [/math]

 

now we are equating f(x) with x

 

[math] x = \frac{kx - 4}{x - 1} [/math]

 

simplifiy

 

[math] x (x-1) = (kx-4) [/math]

[math] x^2-x = kx-4 [/math]

[math] x^2 -x -kx +4 = 0 [/math]

[math] x^2 -(1+k)x +4 = 0 [/math]

 

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

 

[math] x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} [/math]

 

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

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

 

[math] f(x) = \frac{kx - 4}{x - 1}, x \neq 1 [/math]

 

now we are equating f(x) with x

 

[math] x = \frac{kx - 4}{x - 1} [/math]

 

simplifiy

 

[math] x (x-1) = (kx-4) [/math]

[math] x^2-x = kx-4 [/math]

[math] x^2 -x -kx +4 = 0 [/math]

[math] x^2 -(1+k)x +4 = 0 [/math]

 

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

 

[math] x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} [/math]

 

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

 

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

 

[math]b^2 - 4ac < 0 [/math] ???

 

ok, got it, thnx, but...

 

when I got here

 

[math]k^2 + 2k -15 < 0[/math]

[math](k-3)(k+5) < 0[/math]

[math](k-3) < 0[/math] & [math](k+5) < 0[/math]

[math]k < 3[/math] & [math]k < -5[/math]

 

but the answer -5 < k < 3

 

I always "cheat" just to fit it into the answer

[math](k-3)(-k-5) < 0[/math]

[math](k-3) < 0[/math]

[math]k < 3[/math]

&

[math](-k-5) < 0[/math]

[math]-k < 5[/math]

[math]k > -5[/math]

Think about the shape of a graph made by

 

[math] y=(k-3)(k+5) [/math]

 

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

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

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.

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...

Edited February 8, 201213 yr by Vastor

For a rough understanding, a real number is something that you can point to on a number line. Or, for the sake of staying on topic, on an axis.

Edited February 8, 201213 yr by the tree