Maths&physics
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#1
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im doing part c.

the range of the function is the domain of the inverse.

I drew out f(x), but it didnt help me with the range.
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A24_
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domain of f-1 = range of f
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monkeyman0121
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#4
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(Original post by monkeyman0121)
What are you actually stuck on, the whole question or just the finding range part?

Just from the drawing the range would probably be: Y can't equal 0
c.

ok, but thats not the answer for the domain of the inverse.
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(Original post by monkeyman0121)
So is the drawing of the original f(x) not f'(x).

If it is a drawing of a normal f(x) then the domain of f'(x) would be the range of f(x) which would be y is not equal to 0.
that is the drawing of f(x). but thats not the answer in the mark scheme
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(Original post by monkeyman0121)
Ignore what I said. I am guessing you got:

f^-1(x)= 1/2x +1/2

In this case the normal f(X) range is y is not equal to 0.

Now with this f^-1(X) The domain is this +1/2.

So therefore f^-1(x) domain is: y is not equal to 1/2.
thats wrong too
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monkeyman0121
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(Original post by Maths&physics)
thats wrong too
Ok fine. I have not brushed up on my core 3 and 4 maths I will remove these wrong answers.
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monkeyman0121
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(Original post by Maths&physics)
thats wrong too
Do you have a link to the question paper?
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(Original post by monkeyman0121)
Do you have a link to the question paper?
ive posted the question.

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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monkeyman0121
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(Original post by Maths&physics)
ive posted the question.
I know but with the link I can have a look properly, you could tell me the year of the paper if that is better.
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(Original post by monkeyman0121)
I know but with the link I can have a look properly, you could tell me the year of the paper if that is better.
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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monkeyman0121
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I sort of get it now. First of the domain of f(x) is x>1/2.

Therfore the range of the inverse function is y>1/2.

From that you know that x has to be bigger than 0 else you will get a y-value of less than 1/2. example: (1+x)/2x
x=-1 therefore: 0/-2 =0 which is smaller than 1/2 for y values.
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#15
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(Original post by monkeyman0121)
I sort of get it now. First of the domain of f(x) is x>1/2.

Therfore the range of the inverse function is y>1/2.

From that you know that x has to be bigger than 0 else you will get a y-value of less than 1/2. example: (1+x)/2x
x=-1 therefore: 0/-2 =0 which is smaller than 1/2 for y values.
im sorry but that doesn't make sense
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bump
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A24_
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https://www.examsolutions.net/exampa...-january-2012/
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(Original post by Maths&physics)
im sorry but that doesn't make sense
Ok got it now... Domain of f(x)= x>1/2.

Therfore the range is y>0. This is because 1/(2x-1) is always over 0 whatever x-value you put in that is over 1/2.

Now we all know that the inverse functions domain is equal to the normal functions range so therefore the domain of the inverse function is x>0
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RDKGames
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(Original post by Maths&physics)
im doing part c.

the range of the function is the domain of the inverse.

I drew out f(x), but it didnt help me with the range.
That sketch is not correct, but essentially the same thing anyway.

You should recognise that the range here is f(x) >0
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#20
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(Original post by monkeyman0121)
Ok got it now... Domain of f(x)= x>1/2.

Therfore the range is y>0. This is because 1/(2x-1) is always over 0 whatever x-value you put in that is over 1/2.

Now we all know that the inverse functions domain is equal to the normal functions range so therefore the domain of the inverse function is x>0
ive drawn the graph wrong. the asymptote isnt at 1. can you draw this graph and explain it? thanks
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