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

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

Find the elements of the domain that get mapped to themselves by the function
h(x)=\frac{2x+1}{x-2}

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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Kevin De Bruyne
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(Original post by 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
h(x)=\frac{2x+1}{x-2}

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

Thanks in advance
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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Dapperblook22
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(Original post by 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
h(x)=\frac{2x+1}{x-2}

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

Thanks in advance
If the elements map to themselves, then it gives x = y. Therefore substitute x = h(x) and solve as a quadratic.
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RDKGames
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(Original post by 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
h(x)=\frac{2x+1}{x-2}

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

Thanks in advance
Well it's anything on the y=x line so you can work it out algebraically as SeanFM said.
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Philip-flop
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Ok so I managed to work the answer out. Here are my workings

h(x)=x
\frac{2x+1}{x-2}=x
2x+1=x(x-2)
x^2-4x-1=0
Use quadratic formula to work out values of x of the function that cross/lie on the y=x line
so...x=2+\sqrt{5} or x=2-\sqrt{5}


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

(Original post by RDKGames)
Well it's anything on the y=x line so you can work it out algebraically as SeanFM said.
Thank you so much!! I was having yet another daft moment! :P I've managed to show my workings above.

(Original post by 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...)
Thank you! It all makes sense now. Check out my answer above
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Philip-flop
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Ok. So I was wondering if any of you guys could also help me out with this question...

Name:  Exe2E Q8 (C3 Modular Maths).png
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I know that... f^-1(x)=\sqrt{\frac{x+3}{2}} but why is the domain x>-3 surely it would be x<-3??

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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RDKGames
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(Original post by Philip-flop)
Ok. So I was wondering if any of you guys could also help me out with this question...

Name:  Exe2E Q8 (C3 Modular Maths).png
Views: 492
Size:  3.4 KB

I know that... f^-1(x)=\sqrt{\frac{x+3}{2}} but why is the domain x&gt;-3 surely it would be x&lt;-3??

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?
The range isn't f(x)&lt;-3 though.
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username2246259
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(Original post by Philip-flop)
Ok. So I was wondering if any of you guys could also help me out with this question...

Name:  Exe2E Q8 (C3 Modular Maths).png
Views: 492
Size:  3.4 KB

I know that... f^-1(x)=\sqrt{\frac{x+3}{2}} but why is the domain x&gt;-3 surely it would be x&lt;-3??

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?
The range of f(x) is f(x)>-3, not <-3
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Philip-flop
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(Original post by RDKGames)
The range isn't f(x)&lt;-3 though.
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
(Original post by AlexLawrence1453)
The range of f(x) is f(x)>-3, not <-3
Can you please explain. I'm so lost it's unreal
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RDKGames
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(Original post by 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
Sketch the function and it'll make sense. It's parabola after all, all negative values become positive due to the squared term.
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username2246259
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(Original post by 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
Try drawing the graph
f(x) = 2x^2 - 3 Look at the values for f(x) for x&lt;0
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Philip-flop
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(Original post by RDKGames)
Sketch the function and it'll make sense. It's parabola after all, all negative values become positive due to the squared term.
Wow. How embarrassing. After drawing the "half" parabola I realise exactly where I went wrong. I feel so freaking stupid right now
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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Philip-flop
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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
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RDKGames
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(Original post by Philip-flop)
Can I ask how you guys managed to figure this out without physically drawing the graph out?
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.
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username2246259
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(Original post by 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
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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Philip-flop
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(Original post by 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.
Thank you. Imagining the shape of the graph is something I need to learn to do quicker. Especially when determining where graphs cross the x or y-axis'!
(Original post by 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.
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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username2246259
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(Original post by 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
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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Philip-flop
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(Original post by 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.
Yeah I guess I just need to find out what works best for me tbh
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an_atheist
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(Original post by 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
h(x)=\frac{2x+1}{x-2}

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

Thanks in advance
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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RDKGames
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(Original post by an_atheist)
If f(x) maps to itself, it means that f(x)=x. So you are solving x=(2x+1)/(x-2)
He has already done it, you're a bit late to the party.
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