Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 114

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 127

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 136

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 137

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 141
Winter Diet 2002 Paper 1 Question 10 - HSN forum

Jump to content


Winter Diet 2002 Paper 1 Question 10


12 replies to this topic

#1 Rabeeto

    Newbie

  • Members
  • Pip
  • 3 posts
  • Gender:Male

Posted 15 May 2006 - 11:46 AM

Ive got as far as f'(x)= 2 - 18(x-4)^-2

the question is;

A function f is defined by f(x)= 2x + 3 + 18/(x-4), x is not equal to 4

#2 loved-up-loon

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 457 posts
  • Interests:Well I like a WIDE variety of things! haha i sound like a freak!<br />Anyways, I LOVE!!! music! ooh sooo much!!<br />haha i also like shopping and ehhm, drama and talking and swimmin(woo look at me aint i fit (not)) and ehhh thats all folks! haha! man i laugh alot! and i can talk for scotland! dont put iot past me i rambol on a lot!!
  • Gender:Female

Posted 15 May 2006 - 01:09 PM

Statinary points. (dy/dx=0)
nature table
And then see when its increasing.
Grrrrrr!!

#3 Rabeeto

    Newbie

  • Members
  • Pip
  • 3 posts
  • Gender:Male

Posted 15 May 2006 - 01:27 PM

I know that you need to do that, thanks anyway. I just cant seem to get to the right answer, could someone maybe show me how to get to the right answer from where I left off?

#4 SncZ

    Site Swot

  • Members
  • PipPipPipPip
  • 114 posts
  • Location:Edinburgh
  • Gender:Male

Posted 15 May 2006 - 01:57 PM

the way i do it:

differentiate the function to get :

f'(x) = 2- 18/(x-4)2 which you got

the clue to this question is realising that the function is increasing when the gradient is positive. ( which is why we differentiated the function )

2- 18/(x-4) 2 > 0
let 2- 18/(x-4) 2 = o to find where it crosses the x axis
2 - 18/(x-4) 2 = 0
therefore.gif 2 = 18/(x-4) 2

cross multiply to get 2 (x-4)2 = 18

expand the brackets

2(x2-8x+16)=18
2x2-16x+32=18
2x2-16x+14=0
2(x2-8x+7)=0
(x-7)(x-1) =o

therefore.gif x=7,x=1

You know that the function is 'u' shaped as the x2 coefficient is +ve ( remember the old saying, if x2 is positive it has a smiley face i.e u shaped ? )

and they way i was taught, is that the only way to solve a quadratic inequation is to draw a graph... so draw a graph and mark on the points where the function crosses the x axis (0,1) (0,7) and the function is 'u' shaped so draw the graph passing through those two points, but label the axis x (horiz) and f'(x)(vert)

now... the function is +ve when the function is above the x axis which is when x<1 and when x>7 which is the same as 1<x<7

and finally remember to state:
as powerx.gif implies.gif infinity.gif , y implies.gif infinity.gif
powerx.gif implies.gif - infinity.gif , y implies.gif - infinity.gif

well thats the way i was told how to do them, i dunno how loved up loon does it .. why would you need the stationary points ?

#5 loved-up-loon

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 457 posts
  • Interests:Well I like a WIDE variety of things! haha i sound like a freak!<br />Anyways, I LOVE!!! music! ooh sooo much!!<br />haha i also like shopping and ehhm, drama and talking and swimmin(woo look at me aint i fit (not)) and ehhh thats all folks! haha! man i laugh alot! and i can talk for scotland! dont put iot past me i rambol on a lot!!
  • Gender:Female

Posted 15 May 2006 - 02:17 PM

Oh no sorry I was thinking of a different question. yeah that sounds about right, i think i did it the way i said before and just got totally confused! Ive just got a really bad memory! sorry!
Grrrrrr!!

#6 Rabeeto

    Newbie

  • Members
  • Pip
  • 3 posts
  • Gender:Male

Posted 15 May 2006 - 02:21 PM

thanks alot for that, really appreciatte it

#7 ad absurdum

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 390 posts
  • Location:Cambridge
  • Interests:Muzak.
  • Gender:Male

Posted 15 May 2006 - 05:16 PM

SncZ's answer is perfect, but I thought I better point out that it is apparently not enough to say that the co-efficient of x2 is positive, therefore it's a happy parabola and there is a minimum turning point, I lost a mark in my prelim for this. Apparently you have to take the derivative of the function (in this case the derivative of the derivative, i.e. the second derivative) and prove that it is a minimum turning point by use of nature table (or you could take yet another derivative and do it that way if you felt like giving the marker a sore head tongue.gif). Maybe my teacher was just being harsh, but I guess it's always a good idea to put in more instead of less, if you've got time in the exam it's always best to make sure I think.
HMFC - Founded 1874, beefing the Cabbage since 1875

#8 Joel

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 271 posts
  • Gender:Not Telling

Posted 15 May 2006 - 05:26 PM

I was wondering, after expressing a quadratic in completed square form, how does one find the coordinates of the Turning Point? I know it's only worth one mark in the exam but I can't figure out what the formula is and it's bugging me.

#9 loved-up-loon

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 457 posts
  • Interests:Well I like a WIDE variety of things! haha i sound like a freak!<br />Anyways, I LOVE!!! music! ooh sooo much!!<br />haha i also like shopping and ehhm, drama and talking and swimmin(woo look at me aint i fit (not)) and ehhh thats all folks! haha! man i laugh alot! and i can talk for scotland! dont put iot past me i rambol on a lot!!
  • Gender:Female

Posted 15 May 2006 - 05:29 PM

Say its (x-2) power2.gif + 3

I no the x coordinate would be 2 and i think the y would be 3. But if your not sure, even though its only one mark, you could differentate it. but that could take up alot of time.
Grrrrrr!!

#10 ad absurdum

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 390 posts
  • Location:Cambridge
  • Interests:Muzak.
  • Gender:Male

Posted 15 May 2006 - 05:30 PM

QUOTE(Joel @ May 15 2006, 06:26 PM) View Post

I was wondering, after expressing a quadratic in completed square form, how does one find the coordinates of the Turning Point? I know it's only worth one mark in the exam but I can't figure out what the formula is and it's bugging me.
I'm not sure about a formula, but if you think of the the minimum turning point as the value which brings about the smallest value of y, then your value of x is going to be the value that makes the bracket zero (as any other value will make the bracket overall greater than zero). The value of y is then the constant added on.
For example if you have (x+a)2 + b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.
HMFC - Founded 1874, beefing the Cabbage since 1875

#11 Joel

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 271 posts
  • Gender:Not Telling

Posted 15 May 2006 - 05:37 PM

QUOTE(ad absurdum @ May 15 2006, 06:30 PM) View Post
I'm not sure about a formula, but if you think of the the minimum turning point as the value which brings about the smallest value of y, then your value of x is going to be the value that makes the bracket zero (as any other value will make the bracket overall greater than zero). The value of y is then the constant added on.
For example if you have (x+a)2 + b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.
Yeah, i'm not having a problem with the x-coordinate. It's just the y-coordinate that i'm having trouble with.

Here's an example:



\begin{align*}f(x) = 2(x + 2)^2 - 11\end{align*}

Find the coordinates of the TP of this function.

I thought this would be (-2, -11) but infact it is (-2, 11), can anyone explain this?

#12 ad absurdum

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 390 posts
  • Location:Cambridge
  • Interests:Muzak.
  • Gender:Male

Posted 15 May 2006 - 05:45 PM

QUOTE(Joel @ May 15 2006, 06:37 PM) View Post

QUOTE(ad absurdum @ May 15 2006, 06:30 PM) View Post
I'm not sure about a formula, but if you think of the the minimum turning point as the value which brings about the smallest value of y, then your value of x is going to be the value that makes the bracket zero (as any other value will make the bracket overall greater than zero). The value of y is then the constant added on.
For example if you have (x+a)2 + b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.
Yeah, i'm not having a problem with the x-coordinate. It's just the y-coordinate that i'm having trouble with.

Here's an example:



\begin{align*}f(x) = 2(x + 2)^2 - 11\end{align*}

Find the coordinates of the TP of this function.

I thought this would be (-2, -11) but infact it is (-2, 11), can anyone explain this?

I'm sure that you are right, I think it may just be a mistake in the answers or something...Try putting x=-2 into the function, the ouput is -11 so I can't see where the +11 is coming from huh.gif
HMFC - Founded 1874, beefing the Cabbage since 1875

#13 Joel

    Top of the Class

  • Members
  • PipPipPipPipPip
  • 271 posts
  • Gender:Not Telling

Posted 15 May 2006 - 05:50 PM

QUOTE(ad absurdum @ May 15 2006, 06:45 PM) View Post
I'm sure that you are right, I think it may just be a mistake in the answers or something...Try putting x=-2 into the function, the ouput is -11 so I can't see where the +11 is coming from huh.gif
Neither can I, the question was from a specimen question paper on the SQA website so I doubt there is a mistake. When I try other similar questions I keep on getting them wrong aswell.






1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users