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
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2002 winter diet, paper 1, Q9a
Started by Azie, May 13 2007 05:48 PM
13 replies to this topic
#2
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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/(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
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Posted 13 May 2007 - 06:17 PM
no, i still dnt really understand it 
#4
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Posted 13 May 2007 - 06:27 PM
can u do the normal f(g(x)) questions ?
#5
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Posted 13 May 2007 - 06:30 PM
QUOTE(xClairex @ May 13 2007, 07:07 PM) 
sorry i dont no hw to use latex so it might look a bit messy hopefully you can understand what im meaning though
You may want to read up on how uto use LaTeX, then. It's surprisingly easy to get used to
#6
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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
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Posted 13 May 2007 - 06:39 PM
i tried readin up on it but it looks 2 complicated 2 me ...ok, this mite sound like a really stupid question but how did u get the last line?
...ok, this mite sound like a really stupid question but how did u get the last line?
#8
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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
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Posted 13 May 2007 - 07:22 PM
so you get it up until the last line ye?
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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
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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*}](/latexrender/pictures/794a7d92485e23b6adee732299828c13.gif "\begin{align*}[tex]\frac{3}{1}[/tex] times [tex]\frac{x+1}{x+4}[/tex]\end{align*}")
do you understand where i get [display]
times 
[/display] from ?
#12
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Posted 13 May 2007 - 07:44 PM
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) [display] times [/display]
do you understand where i get [display] times [/display] from ?
times [/display]do you understand where i get [display]
times [/display] from ?i cant get the first line to go away so just ignore that.latex is far too complicated for me
#13
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Posted 13 May 2007 - 07:52 PM
nope ...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
...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
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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
its ok im slow at times as well i was really confused with this question before dont worry about it
