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
Help on Qs - HSN forum

Jump to content


Help on Qs


7 replies to this topic

#1 Rocky

    Site Swot

  • Members
  • PipPipPipPip
  • 239 posts
  • Gender:Male

Posted 05 September 2005 - 07:15 PM

If you can help, it'd be really appreciated. Thanx.

1- 2 Functions are defined on suitable domains and are given as f(x)= x + 3 & g(x)= x squared - 1
(i) Find an expression for the function H where H(X)= g( f(x) )
(ii) Find the values of A given that H(a)= f(A) + 1

2- On a suitable set of REAL numbers functions F and G are defined by
f(x)= 1/x + 2 **this on left all a fraction** AND g(x)= 1/x (- 2) **1st part of it is fraction the minus 2 isn't in the fraction**
Find f( g(x) ) in its simplest form.

#2 dfx

    Fully Fledged Genius

  • Members
  • PipPipPipPipPipPipPip
  • 1,955 posts
  • Gender:Male

Posted 05 September 2005 - 07:29 PM

1. (i)

For h(x) = g(f(x):

you first take g(x) = x power2.gif - 1

then g(f) = f power2.gif - 1

then g(f(x) = f(x) power2.gif - 1

= (x+3) power2.gif - 1

therefore.gif h(x)=g(f(x)=x power2.gif + 6x + 8 implies.gif h(x) = x power2.gif + 6x + 8

Not too sure about (ii), I'll get back to you on it.

2.

Exactly same method

For f(g(x)):

f(x) = 1/(x+2)

Then f(g) = 1/(g+2)

Then f(g(x)) = 1/(g(x) + 2)

= 1/[(1/x)-2] <- you forgot to put the + 2 back in this line [edit from Dave]

=1/[(1/x) - (2x/x)

=1/(1-2x)/x

=x/(1-2x)

^ I've done this in a hurry so someone check the algebra please. Cheers.

#3 Dave

    Ruler (but not owner) of hsn

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 4,252 posts
  • Location:kilmarnock(ok kilmaurs)
  • Interests:programming, exercising, brass band, using this board
  • Gender:Male

Posted 05 September 2005 - 07:32 PM

H(X) = g(f(x))

right so the function g(x) = x2 - 1

so if we sub f(x) for the x in the function above we get

g(f(x)) = (x+3)2 - 1 = H(X)

ii)

the value inside the brackets there the "a" is just the variable in the function

so we get

H(a) = (a+3)2 - 1 = (a+3) + 1
a2 +6a + 9 - 1 = a + 4
a2 + 5a + 6 = 0
(a+6)(a-1) = 0
a = -6 and a = 1

2)

f(g(x)) so thats the function of f where all the variable we sub in the function g(x)

= 1/ (1/x(-2)) +2) = x -1

i think

NB: dfx if you look at my wee correction there and get back to me

If i am not here i am somewhere else



#4 dondon

    HSN Legend

  • Members
  • PipPipPipPipPipPipPipPip
  • 2,888 posts
  • Location:Glasgow
  • Gender:Female

Posted 05 September 2005 - 07:34 PM

i)

its hard to explain but

H(x)=g(f(x))
=g(x+3)
= (x+3) power2.gif -1
= x power2.gif +6x+8

so you put in the f(x) function then put that into the g(x) fuction, but put it into each place where there is an x it's easier to see what to do if you see it more like:
g(x)=x[^2] -1

then put in x+3 where there is an x.

ii)
i think that you do this

H(a)=f(A)+1
=(a+3)+1
=a+4

but i could be wrong

2)
i don't know if i'm doing that one wrong but it comes out with x/0 which doesn't exist but you should just be able to do the same thing as the first question but its slightly more comlicated. And remember when you divide a fraction by a fraction you turn one upside down and multiply them, thats where i think i went wrong.


#5 Dave

    Ruler (but not owner) of hsn

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 4,252 posts
  • Location:kilmarnock(ok kilmaurs)
  • Interests:programming, exercising, brass band, using this board
  • Gender:Male

Posted 05 September 2005 - 07:38 PM

dondon for part to you are looking to find the value of a

i think wot i have done is correct but please look and see and tell me if its not (it happens a lot lol)

If i am not here i am somewhere else



#6 dondon

    HSN Legend

  • Members
  • PipPipPipPipPipPipPipPip
  • 2,888 posts
  • Location:Glasgow
  • Gender:Female

Posted 05 September 2005 - 07:40 PM

yip i didn't realise that, you can tell my brain is really slowing down

#7 Rocky

    Site Swot

  • Members
  • PipPipPipPip
  • 239 posts
  • Gender:Male

Posted 08 September 2005 - 06:44 PM

Thanx dave, dfx and dondon for the help.

#8 dfx

    Fully Fledged Genius

  • Members
  • PipPipPipPipPipPipPip
  • 1,955 posts
  • Gender:Male

Posted 08 September 2005 - 08:30 PM

You're welcome.

Error well spotted Dave, cheers. smile.gif





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users