# Domain of a function

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#1
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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2 years ago
#2
domain of f-1 = range of f
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2 years ago
#3
Deleted
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#4
(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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2 years ago
#5
Deleted
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#6
(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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2 years ago
#7
Deleted
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#8
(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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2 years ago
#9
(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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2 years ago
#10
(Original post by Maths&physics)
thats wrong too
Do you have a link to the question paper?
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#11
(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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2 years ago
#12
(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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#13
(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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2 years ago
#14
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
(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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#16
bump
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2 years ago
#17
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2 years ago
#18
(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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2 years ago
#19
(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
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#20
(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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