# domain C3 question

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#1
Here's a link to the question https://pmt.physicsandmathstutor.com...0Functions.pdf

June 2011 Q4b

Why isn't the domain X is any real number
Last edited by JacobBob; 1 year ago
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1 year ago
#2
(Original post by JacobBob)
Here's a link to the question https://pmt.physicsandmathstutor.com...0Functions.pdf

Jan 2012 Q7b

Why isn't the domain X is any real number
When x = 1 the denominator is zero and the function undefined.
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#3
(Original post by Muttley79)
When x = 1 the denominator is zero and the function undefined.
Oh god, I'm sorry, I meant June 2011 Q4b
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1 year ago
#4
(Original post by JacobBob)
Oh god, I'm sorry, I meant June 2011 Q4b
If you were given the function without any other information then yes the domain would be .

But is defined based on . The domain of is restricted to which gives you a range of . This must mean that the domain of is .

The domain of is always the range of and vice-versa. So if you are asked for the domain/range of an inverse function then you should always get these from the domain/range of the original function.
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#5
(Original post by Notnek)
If you were given the function without any other information then yes the domain would be .

But is defined based on . The domain of is restricted to which gives you a range of . This must mean that the domain of is .

The domain of is always the range of and vice-versa. So if you are asked for the domain/range of an inverse function then you should always get these from the domain/range of the original function.
In 2012 Jan 7c
I found the range of fx to be y<0 so I took the domain of f inverse x as x<0
But the answer in the mark scheme is x>0 so I'm confused

I found the range by "forming the function" using the domain they had given me

2x-1>0
and then I flipped it to get y<0
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1 year ago
#6
(Original post by JacobBob)
In 2012 Jan 7c
I found the range of fx to be y<0 so I took the domain of f inverse x as x<0
But the answer in the mark scheme is x>0 so I'm confused

I found the range by "forming the function" using the domain they had given me

2x-1>0
and then I flipped it to get y<0
y<0 is incorrect for the range of f(x). E.g. I can input x = 1 into f(x) and I get f(1)=1. Have another think about it.
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