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#76 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 11:30:04
No, not yet but we are close. You have to get x by itself.
Now since x is being divided by 4 the inverse would be to multiply by 4. Remember, both sides!
Oh right. I must remember that ''4(y + 6)''. Yup I understand that. With the *4 on the right-side, the denominator of the fraction would be gone, or zero?
#77 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 11:19:54
Now you can clean up the third step:
Are you okay with that?
Yep. Is that the answer?
#78 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 11:11:24
Add 6 to both sides, whatever you do to one side of an equation you have to do to the other to keep it balanced.
Okay. You always have to put +6 on both sides.
#79 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 10:58:25
If you understood the last one we can try another one.
We do the same first step.
Okay?
Yep. I get that.
#80 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 10:50:25
Remember you are just trying to get x all by itself on the right hand side. You are solving for x.
f(x) = x-4
y = x-4
y + 4 = x
Now we swap sides which is only for looks:
x = y + 4
Put f^(-1) (y ) in front replacing the x and we are done.
f^(-1) (y ) = y + 4
Yea I get that. That was in my assignment lol. But when it comes to invert multiplication to division, and vice-versa. That's confusing 
#81 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 10:29:33
Hi;
What I mean is
y (+4)= x-4 (+4)
I took and added +4 to both sides, y (+4) is just y +4 but x - 4 + 4 is (-4) + (4) = 0
Okay. I think I get that.
#82 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 10:14:23
f(x) = x-4
y = x-4
Okay next step!
Add 4 to both sides because subtraction ( MIF has a table on that page showing inverses ) is the inverse of addition.
y + 4 = x
I think I get that. You just inverted x to y, and + to - . But what do you mean "Add 4 to both sides"? Do you mean Adding +4 on to -4 which makes 0?
#83 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 10:00:40
Hi;
I didn't do it right. I just pasted that long example from your website and had no idea how to do it that way
Great reply! Okay, let's do one.
f(x) = x-4
Now the great thing about what MIF has done has put the process of getting the inverse in the form of an algorithm. A step by step procedure that terminates eventually. No thinking allowed, to start work like a computer, just do what you are told to do. Let's do it like that.
First step put y for f(x)
y = x-4
I hope you are okay up to here?
Yep.
#84 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 09:50:31
Hi;
But I don't get it.
Maybe, but you did it right. Except you left out MIF's last step. Take a look at my signature, there is a lot of truth in that. Understanding like love comes later.
For the composition problems what answers are you getting?
I didn't do it right. I just pasted that long example from your website and had no idea how to do it that way.
The answers I am getting from composition function are:
(g º f) = 3x^2+9
(f º g) = (3+9)^2
(f º f) = 6x^2
(g º g) = x^2 +18
It's just that putting g in f and etc gets me muddled up, when I looked at the answers it's all wrong. Noticed that all of it's wrong.
#85 Re: Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 09:21:28
Hi BlitzBall;
Before the links, what ones are you having problems with specifically?
For the inverse function:
For example, inverse this f(x) = 2x+3. I only get this by moving these around, and inverse Addition to subtraction symbols only. I also looked at the algebra technique:
(Taken from Mathsisfun website)
The function: f(x)= 2x+3
Put "y" for "f(x)": y= 2x+3
Subtract 3 from both sides: y-3 = 2x
Divide both sides by 2: (y-3)/2 = x
Swap sides: x= (y-3)/2
But I don't get it.
Composition of functions:
Say f(x) = 3x and g(x) = x+9
The answers I get through --> (g º f)(x), (f º g)(x), (f º f)(x), (g º g)(x) = Are wrong. Apart from (f º f).
I think the problem I'm having is algebra .
#86 Help Me ! » Algebra help! Inverse and composition functions » 2012-01-17 08:49:36
- BlitzBall
- Replies: 121
Hi guys,
I looked at your website on Inverse and Composing functions. When I look at it I get it, but doing it myself is difficult. I'm quite new to algebra and have less experience in them.
Can anyone give me any links on algebra that can quickly help me understand inverse and composition functions well? I have a test soon and must get this section solved 
Thanks.
#87 Re: Help Me ! » Pigeonhole principle help » 2012-01-09 12:55:27
Oh I see. I absolutely understood the example you shown me. But in the exam, I am not allowed to use a calculator and I would find it hard to divide and reach the answer 2 if I was at this question. I have been trying hard to understand this pigeonhole principle and when I look at the questions, it's understandable, but when I look at the answers, I don't know why it's +1 to my answer. Like in the example I showed, I said it should be 3, but ends up as 2, so close 
Not sure if you have heard 'extended pigeonhole principle' where you use |A|-->K|B|.
By the way, I got the answer right from that link you gave me! 
Yea, I'm sure the question is about birthdays.
#88 Help Me ! » Pigeonhole principle help » 2012-01-09 07:50:00
- BlitzBall
- Replies: 3
Hi guys, it's me again.
I'm a little confused about this question and the pigeonhole technique.
Here an example:
There are 15 students in a classroom. Show that at least two of them have the same birthday in the same month of the year.
They way I worked it out is basically A = 15 students and B = 12 months because f:A-->B and |A| > |B|.
So, if i allocate one student to 12 pigeonholes, I am left with 3 students who will have the same birthday with either of the 12 students.
The ACTUAL answer is two students, but I don't understand how you can get that.
#89 Re: Help Me ! » What is x^2+4x=12 ?? » 2012-01-08 12:38:42
Basically that is correct.
YEAHHHHH!
#90 Re: Help Me ! » What is x^2+4x=12 ?? » 2012-01-08 06:43:32
hi BlitzBall and bobbym,
Welcome to the forum, BlitzBall.
Everyone (book included) is 'right'.
The difficulty lies here:
x^2+4x=12
x^2+4x-12
(x+6)(x-2) or (x-6)(x+2)You started out with an equation and then lost the equals sign.
Here's the complete answer:
Don't forget the zero!
so
Now, if two factors multiply to make zero, one or the other must be zero.
so
Check these values.
Hope that wraps it up.
Bob
Hey bob, thanks for your reply. I think i got this because I looked at a website through BBC GCSE about Solving quadratic equations by factorising.
I think I understand this, I pick (x+6)(x-2) as it meets 12 and 4. Then I have to make it equal to zero. So, in the first bracket (x+6) i would have to subtract by -6 which makes zero! And same with the second bracket. Is that the right way of how to do it??
#91 Re: Help Me ! » What is x^2+4x=12 ?? » 2012-01-07 16:18:32
Hi BlitzBall;
Welcome to the forum.
You are right.
You can stop the argument my expanding.
(x+6)(x-2).
Have they taught you how?
Nope. Well, I'm at university and I have absolutely forgotten about algebra since Secondary school. I looked at the book and the answer is -6 and +2.
I will reply back to you tomorrow because its 4:17am 
Thanks for your help
:)
#92 Help Me ! » What is x^2+4x=12 ?? » 2012-01-07 16:09:34
- BlitzBall
- Replies: 9
This is the part where I got stuck on...
x^2+4x=12
x^2+4x-12
(x+6)(x-2) or (x-6)(x+2)
Which one is right? And why? I'm leaning towards +6 and -2 because it meets 4x. But others say its -6 and +2 
Also, is it okay if you can explain it simpler? Because I have got a lot of other topics that I am stuck on.
Thank you.