Algebra help! Inverse and composition functions

BlitzBall · 2012-01-17 08:49:36

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.

Last edited by BlitzBall (2012-01-17 08:50:29)

bobbym · 2012-01-17 08:52:53

Hi BlitzBall;

Before the links, what ones are you having problems with specifically?

BlitzBall · 2012-01-17 09:21:28

bobbym wrote:

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 .

Last edited by BlitzBall (2012-01-17 09:22:07)

bobbym · 2012-01-17 09:37:48

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?

BlitzBall · 2012-01-17 09:50:31

bobbym wrote:

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.

bobbym · 2012-01-17 09:58:05

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?

BlitzBall · 2012-01-17 10:00:40

bobbym wrote:

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.

bobbym · 2012-01-17 10:05:25

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

BlitzBall · 2012-01-17 10:14:23

bobbym wrote:

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?

bobbym · 2012-01-17 10:18:29

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

BlitzBall · 2012-01-17 10:29:33

bobbym wrote:

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.

bobbym · 2012-01-17 10:39:10

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

BlitzBall · 2012-01-17 10:50:25

bobbym wrote:

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

bobbym · 2012-01-17 10:53:10

If you understood the last one we can try another one.

We do the same first step.

Okay?

BlitzBall · 2012-01-17 10:58:25

bobbym wrote:

If you understood the last one we can try another one.
We do the same first step.
Okay?

Yep. I get that.

bobbym · 2012-01-17 11:05:33

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.

BlitzBall · 2012-01-17 11:11:24

bobbym wrote:

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.

bobbym · 2012-01-17 11:14:02

Now you can clean up the third step:

Are you okay with that?

BlitzBall · 2012-01-17 11:19:54

bobbym wrote:

Now you can clean up the third step:
Are you okay with that?

Yep. Is that the answer?

bobbym · 2012-01-17 11:23:13

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!

BlitzBall · 2012-01-17 11:30:04

bobbym wrote:

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?

bobbym · 2012-01-17 11:33:21

No, it would be one, because 4 * ( 1 / 4 ) = 1. So let's clean up this last step.

BlitzBall · 2012-01-17 11:41:11

bobbym wrote:

No, it would be one, because 4 * ( 1 / 4 ) = 1. So let's clean up this last step.

Okay. So, doing *4 on the right-side leaves x on its own, and the fraction is removed? where is 1/4 coming from?

bobbym · 2012-01-17 11:46:57

x/ 4 is the same as x * ( 1 / 4 ). The fours cancel in either case and just leave the x.

Now we swap the x to the other side.

BlitzBall · 2012-01-17 11:48:31

bobbym wrote:

x/ 4 is the same as x * ( 1 / 4 ). The fours cancel in either case and just leave the x.
Now we swap the x to the other side.

Ohh I see. Surely, that is the answer??

Math Is Fun Forum

#1 2012-01-17 08:49:36

Algebra help! Inverse and composition functions

#2 2012-01-17 08:52:53

Re: Algebra help! Inverse and composition functions

#3 2012-01-17 09:21:28

Re: Algebra help! Inverse and composition functions

#4 2012-01-17 09:37:48

Re: Algebra help! Inverse and composition functions

#5 2012-01-17 09:50:31

Re: Algebra help! Inverse and composition functions

#6 2012-01-17 09:58:05

Re: Algebra help! Inverse and composition functions

#7 2012-01-17 10:00:40

Re: Algebra help! Inverse and composition functions

#8 2012-01-17 10:05:25

Re: Algebra help! Inverse and composition functions

#9 2012-01-17 10:14:23

Re: Algebra help! Inverse and composition functions

#10 2012-01-17 10:18:29

Re: Algebra help! Inverse and composition functions

#11 2012-01-17 10:29:33

Re: Algebra help! Inverse and composition functions

#12 2012-01-17 10:39:10

Re: Algebra help! Inverse and composition functions

#13 2012-01-17 10:50:25

Re: Algebra help! Inverse and composition functions

#14 2012-01-17 10:53:10

Re: Algebra help! Inverse and composition functions

#15 2012-01-17 10:58:25

Re: Algebra help! Inverse and composition functions

#16 2012-01-17 11:05:33

Re: Algebra help! Inverse and composition functions

#17 2012-01-17 11:11:24

Re: Algebra help! Inverse and composition functions

#18 2012-01-17 11:14:02

Re: Algebra help! Inverse and composition functions

#19 2012-01-17 11:19:54

Re: Algebra help! Inverse and composition functions

#20 2012-01-17 11:23:13

Re: Algebra help! Inverse and composition functions

#21 2012-01-17 11:30:04

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#22 2012-01-17 11:33:21

Re: Algebra help! Inverse and composition functions

#23 2012-01-17 11:41:11

Re: Algebra help! Inverse and composition functions

#24 2012-01-17 11:46:57

Re: Algebra help! Inverse and composition functions

#25 2012-01-17 11:48:31

Re: Algebra help! Inverse and composition functions

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