# Domain of a function

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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.

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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(Original post by

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

**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

ok, but thats not the answer for the domain of the inverse.

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(Original post by

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.

**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.

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(Original post by

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.

**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.

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#9

(Original post by

thats wrong too

**Maths&physics**)thats wrong too

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#10

(Original post by

thats wrong too

**Maths&physics**)thats wrong too

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(Original post by

Do you have a link to the question paper?

**monkeyman0121**)Do you have a link to the question paper?

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

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#12

(Original post by

ive posted the question.

**Maths&physics**)ive posted the question.

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(Original post by

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.

**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.

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#14

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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(Original post by

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.

**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.

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#18

(Original post by

im sorry but that doesn't make sense

**Maths&physics**)im sorry but that doesn't make sense

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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#19

(Original post by

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.

**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.

You should recognise that the range here is

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(Original post by

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

**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

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