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2002 winter diet, paper 1, Q9a - HSN forum

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2002 winter diet, paper 1, Q9a


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#1 Azie

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Posted 13 May 2007 - 05:48 PM

The question is: "The function f, defined on a suitable domain is given by f(x) = 3/(x+1). Find an expression for h(x) where h(x) = f(f(x))." i managed 2 do the 1st step so that h(x) = f(3/x+1) but then wat do i do? :s

#2 xClairex

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Posted 13 May 2007 - 06:07 PM

sorry i dont no hw to use latex so it might look a bit messy hopefully you can understand what im meaning though

3/(3/x+1) + 1
= 3/(3+x+1)/(x+1)
=3/(x+4)/(x+1)

= 3/1 * (x+1)/(x+4)

= 3(x+1)/(x+4)


#3 Azie

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Posted 13 May 2007 - 06:17 PM

no, i still dnt really understand it sad.gif

#4 xClairex

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Posted 13 May 2007 - 06:27 PM

can u do the normal f(g(x)) questions ?


#5 Pete

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Posted 13 May 2007 - 06:30 PM

QUOTE(xClairex @ May 13 2007, 07:07 PM) View Post
sorry i dont no hw to use latex so it might look a bit messy hopefully you can understand what im meaning though

You may want to read up on how uto use LaTeX, then. It's surprisingly easy to get used to smile.gif

\\f(x) = \frac{3}{x+1}
\\h(x) = f(f(x))
\\h(x) = f(\frac{3}{x+1})
\\= \frac{3}{(\frac{3}{x+1}) + 1}
\\= \frac{3(x+1)}{x+4}

#6 Azie

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Posted 13 May 2007 - 06:30 PM

yeah, most of them, but this 1 is really annoyin me, an im a really superstitious person an all of that so i believe that a question similar 2 it will cum up in the exam if i dnt solve it jus now

#7 Azie

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Posted 13 May 2007 - 06:39 PM

i tried readin up on it but it looks 2 complicated 2 me sad.gif ...ok, this mite sound like a really stupid question but how did u get the last line?

#8 Allan_r_123

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Posted 13 May 2007 - 07:16 PM

I think the last line is the one that causes confusion, as I have a bit of bother with it myself ;P

#9 xClairex

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Posted 13 May 2007 - 07:22 PM

so you get it up until the last line ye?


#10 Azie

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Posted 13 May 2007 - 07:26 PM

yeah kinda, i definately get up 2 h(x) = f(3/x+1) but then its from there that i get stuck an dnt no what 2 do ...i kinda understand the next line now, not that id ever think of it without being shown it, but the last line totally confuses me :s

#11 xClairex

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Posted 13 May 2007 - 07:39 PM



\begin{align*}[tex]\frac{3}{1}[/tex] times [tex]\frac{x+1}{x+4}[/tex]\end{align*}

do you understand where i get [display]\frac{3}{1} times \frac{x+1}{x+4}[/display] from ?


#12 xClairex

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Posted 13 May 2007 - 07:44 PM



\begin{align*}[tex]\frac{3}{1}[/tex] times [tex]\frac{x+1}{x+4}[/tex]\end{align*}

10;\begin{align*}[tex]\frac{3}{1}[/tex] times [tex]\frac{x+1}{x+4}[/tex]\end{align*} ' alt=' \begin{align*}[tex]\frac{3}{1}[/tex] times [tex]\frac{x+1}{x+4}[/tex]\end{align*} ' align=absmiddle>otetop'>QUOTE(xClairex @ May 13 2007, 08:39 PM) View Post
[display]\frac{3}{1} times \frac{x+1}{x+4}[/display]

do you understand where i get [display]\frac{3}{1} times \frac{x+1}{x+4}[/display] from ?


i cant get the first line to go away so just ignore that.latex is far too complicated for me


#13 Azie

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Posted 13 May 2007 - 07:52 PM

nope sad.gif ...i told u i was slow, i bet its really simple as well but i dunno y i cant get it or y i cant c how u got that :s

#14 xClairex

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Posted 13 May 2007 - 08:23 PM

add me on msn maybe it will be easier if i use the handwriting tool that way i can draw arrows etc on and maybe explain better
its ok im slow at times as well i was really confused with this question before dont worry about it






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