Posted 10 October 2005 - 05:52 PM

**1-** A function is defined by g(x)=x(power to 3) +a, where *a* is a constant.

When g(x) is divided by x-3, the remainder is 36. Find *a*, and hence solve

g(-2x)=-18.

**2-** f(x)=x(squared)-1 *and* g(x)=2x+7

The function *h* is such that whenever

f(a)=b, and g(b)=c, then h(a)=c.

**(a)** If b=8, find values for *a* & *c*.

**(b)** Find a formula for *h-1 (x)*, the inverse of *h(x)* in terms of *x*.

Posted 11 October 2005 - 03:13 PM

Well guess this must be the wrong forum to get help....but wait it says 'Problem Questions' here. So wonder why no reply, hmmm.

Posted 12 October 2005 - 02:52 PM

QUOTE(Rocky @ Oct 11 2005, 03:13 PM)

Well guess this must be the wrong forum to get help....but wait it says 'Problem Questions' here. So wonder why no reply, hmmm.

Everyone on their October holiday??

Just spotted your question but am just running out of the door. Will past worked solutions later if no one else gets back to you first.

Posted 12 October 2005 - 02:57 PM

to do the first one

use the remainder thereum you know the one where you have a remainder at the end

well this remainder will be made to = 36

then its a simple solving one

next question

to find c use g(b) = c as we know the value of b

f(a) = b which means f(a) = 8

x^{2} - 1 = 8

solve it

inverse is the same as changing the subject of the formula from h(x) = something, to x = something

Posted 12 October 2005 - 04:40 PM

Hey soz, Mr.H, as october week is next week for me. Plus worked solutions would be helpful if it's no problem to you and Dave for the remainder Q, do I make the factor 3, as x-3 is in the Q?

Posted 12 October 2005 - 04:47 PM

I have got a=9, so it says hence solve:

g(-2x)= -18, so now what do I do?

Posted 12 October 2005 - 04:53 PM

yes the factor is 3

you solve the equation

g(-2x) = -18

8x^{2} + 9 = -18

etc etc

Posted 12 October 2005 - 04:59 PM

How did you turn "g(-2x)" into "8x

2p + 9"?????

Posted 12 October 2005 - 05:14 PM

Come on, listen I ain't trying to annoy you Dave, though you could help by explaining better. Then I won't need to keep posting for help, if you posted the relative information in one post. As I don't know a clue how you are doing some things.

Posted 12 October 2005 - 05:59 PM

Here is question 1:

http://www.mathsroom.co.uk/hsnposts/rocky1.pdfWhile you read that Rocky, I'll have a go at question 2.

Posted 12 October 2005 - 06:16 PM

Posted 12 October 2005 - 06:30 PM

And question 2:

http://www.mathsroom.co.uk/hsnposts/rocky2.pdfHope this helps.

Where did you get these questions btw?

Posted 12 October 2005 - 07:31 PM

Thanks also it was 2 Qs that were in my homework sheet, which I couldn't do.

Posted 12 October 2005 - 07:38 PM

QUOTE(Rocky @ Oct 12 2005, 07:31 PM)

... it was 2 Qs that were in my homework sheet, which I couldn't do.

If you had said that at the beginning I wouldn't have given you full solutions!!

I hope I get them right then!

Posted 12 October 2005 - 09:51 PM

QUOTE(Rocky @ Oct 12 2005, 06:14 PM)

Come on, listen I ain't trying to annoy you Dave, though you could help by explaining better. Then I won't need to keep posting for help, if you posted the relative information in one post. As I don't know a clue how you are doing some things.

We're here to help you with specific problems you're having, not to do your homework for you.

Posted 12 October 2005 - 10:11 PM

well yeah thats kinda wot i was trying to get at the "p" u asked about was a typo btw rocky

Posted 13 October 2005 - 03:34 PM

Oh right dave, soz about that. Plus Ally, I did my Maths H.W. but had problems with these 2 Qs so asked for help.