# Maths C3 - Working out what Domains get mapped back to themselves...??

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#2

(Original post by

Ok so I came across a question yesterday which says...

How do I work this out if the answers aren't whole numbers? What is the process of working this out?

Thanks in advance

**Philip-flop**)Ok so I came across a question yesterday which says...

**Find the elements of the domain that get mapped to themselves by the function**How do I work this out if the answers aren't whole numbers? What is the process of working this out?

Thanks in advance

How do you work it out algebraically? (I.e setting up an equation and solving...)

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#3

**Philip-flop**)

Ok so I came across a question yesterday which says...

**Find the elements of the domain that get mapped to themselves by the function**How do I work this out if the answers aren't whole numbers? What is the process of working this out?

Thanks in advance

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#4

**Philip-flop**)

Ok so I came across a question yesterday which says...

**Find the elements of the domain that get mapped to themselves by the function**How do I work this out if the answers aren't whole numbers? What is the process of working this out?

Thanks in advance

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Ok so I managed to work the answer out. Here are my workings

Use quadratic formula to work out values of x of the function that cross/lie on the y=x line

so... or

Oh my god. YES of course!! Thank you. I temporarily forgot that Domains that are mapped back to themselves are points of the function/graph that crosses/lies on the y=x line!!

Thank you so much!! I was having yet another daft moment! :P I've managed to show my workings above.

Thank you! It all makes sense now. Check out my answer above

Use quadratic formula to work out values of x of the function that cross/lie on the y=x line

so... or

(Original post by

If the elements map to themselves, then it gives x = y. Therefore substitute x = h(x) and solve as a quadratic.

**Dapperblook22**)If the elements map to themselves, then it gives x = y. Therefore substitute x = h(x) and solve as a quadratic.

(Original post by

Well it's anything on the y=x line so you can work it out algebraically as SeanFM said.

**RDKGames**)Well it's anything on the y=x line so you can work it out algebraically as SeanFM said.

(Original post by

You do it algebraically, not manually. Then the non-whole numbers shouldn't be a problem.

How do you work it out algebraically? (I.e setting up an equation and solving...)

**SeanFM**)You do it algebraically, not manually. Then the non-whole numbers shouldn't be a problem.

How do you work it out algebraically? (I.e setting up an equation and solving...)

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#7

(Original post by

Ok. So I was wondering if any of you guys could also help me out with this question...

I know that... but why is the domain surely it would be ??

Because the range of the function

What have I done wrong?

**Philip-flop**)Ok. So I was wondering if any of you guys could also help me out with this question...

I know that... but why is the domain surely it would be ??

Because the range of the function

**f(x)**is**f(x)<-3**so surely the domain of the inverse function**f^-1(x)**would be**x<-3**too?What have I done wrong?

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#8

**Philip-flop**)

Ok. So I was wondering if any of you guys could also help me out with this question...

I know that... but why is the domain surely it would be ??

Because the range of the function

**f(x)**is

**f(x)<-3**so surely the domain of the inverse function

**f^-1(x)**would be

**x<-3**too?

What have I done wrong?

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I thought since

**f(x)**crosses the y-axis at

**y=-3**and that the

**domain**is any real number less than zero hence the

**x<0**, that the range of f(x) can only have values as long as they are in the section of the graph that is below zero or

**x<0**.

Can someone please explain this to me. I'm so confused and feel really stupid

(Original post by

The range of f(x) is f(x)>-3, not <-3

**AlexLawrence1453**)The range of f(x) is f(x)>-3, not <-3

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#10

(Original post by

Ok I am definitely being stupid

I thought since

Can someone please explain this to me. I'm so confused and feel really stupid

Can you please explain. I'm so lost it's unreal

**Philip-flop**)Ok I am definitely being stupid

I thought since

**f(x)**crosses the y-axis at**y=-3**and that the**domain**is any real number less than zero hence the**x<0**, that the range of f(x) can only have values as long as they are in the section of the graph that is below zero or**x<0**.Can someone please explain this to me. I'm so confused and feel really stupid

Can you please explain. I'm so lost it's unreal

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#11

**Philip-flop**)

Ok I am definitely being stupid

I thought since

**f(x)**crosses the y-axis at

**y=-3**and that the

**domain**is any real number less than zero hence the

**x<0**, that the range of f(x) can only have values as long as they are in the section of the graph that is below zero or

**x<0**.

Can someone please explain this to me. I'm so confused and feel really stupid

Can you please explain. I'm so lost it's unreal

Look at the values for for

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(Original post by

Sketch the function and it'll make sense. It's parabola after all, all negative values become positive due to the squared term.

**RDKGames**)Sketch the function and it'll make sense. It's parabola after all, all negative values become positive due to the squared term.

On the plus side. I managed to work out why the

**range**of f(x) is

**f(x)>-3**which means the

**domain**of

**f^-1(x)**is

**x>-3**

Sorry if I frustrated any of you guys with my stupidity

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Can I ask how you guys managed to figure this out without physically drawing the graph out?

Seriously though, thanks for being so patient with me. I really appreciate it

Seriously though, thanks for being so patient with me. I really appreciate it

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#14

(Original post by

Can I ask how you guys managed to figure this out without physically drawing the graph out?

**Philip-flop**)Can I ask how you guys managed to figure this out without physically drawing the graph out?

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#15

(Original post by

Can I ask how you guys managed to figure this out without physically drawing the graph out?

Seriously though, thanks for being so patient with me. I really appreciate it

**Philip-flop**)Can I ask how you guys managed to figure this out without physically drawing the graph out?

Seriously though, thanks for being so patient with me. I really appreciate it

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(Original post by

I look at what type of function it is and get the general shape in my head. Since it's a parabola and I'm working with the range, I simply find the minimum point which is easy here because it's at x=0. So since x cannot be zero, and it's everything less than that, the only solution is that the range is everything more than and not equal to -3. Works quicker in my head than I explained.

**RDKGames**)I look at what type of function it is and get the general shape in my head. Since it's a parabola and I'm working with the range, I simply find the minimum point which is easy here because it's at x=0. So since x cannot be zero, and it's everything less than that, the only solution is that the range is everything more than and not equal to -3. Works quicker in my head than I explained.

(Original post by

I saw the x^2 term and recognised that this can only be positive in the real numbers, so any term for x will always necessarily make f(x) greater than or equal to -3. Since x can't be zero, it is always greater than.

**AlexLawrence1453**)I saw the x^2 term and recognised that this can only be positive in the real numbers, so any term for x will always necessarily make f(x) greater than or equal to -3. Since x can't be zero, it is always greater than.

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#17

(Original post by

Yeah. I need to learn how to visualise these kind of problems in my head better tbh! But there is no harm in drawing the graph out I guess, I mean in definitely helped me out in this example haha :P

**Philip-flop**)Yeah. I need to learn how to visualise these kind of problems in my head better tbh! But there is no harm in drawing the graph out I guess, I mean in definitely helped me out in this example haha :P

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(Original post by

Really what I did should have been secondary to the geometric interpretation. Having an understanding of the shapes of these graphs will help with intuitive understanding of the properties.

**AlexLawrence1453**)Really what I did should have been secondary to the geometric interpretation. Having an understanding of the shapes of these graphs will help with intuitive understanding of the properties.

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#19

**Philip-flop**)

Ok so I came across a question yesterday which says...

**Find the elements of the domain that get mapped to themselves by the function**How do I work this out if the answers aren't whole numbers? What is the process of working this out?

Thanks in advance

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#20

(Original post by

If f(x) maps to itself, it means that f(x)=x. So you are solving x=(2x+1)/(x-2)

**an_atheist**)If f(x) maps to itself, it means that f(x)=x. So you are solving x=(2x+1)/(x-2)

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