Hi guys. I know this is probably really simple but I'm confused with how to find the domain of an inverse function. Its question 5b on this paper
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
So i've found the inverse function to be (e^x+1)/2.
To find the domain I thought it was the range of the original function. And to find that what I did is draw the graph of y=ln(2x-1) and see that y is always greater than 1. So i made the range of this f(x)>1. This is wrong though... I'm very confused with all this domain and range stuff and help would be much appreciated xxx
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Mazza2000
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- 16-10-2017 20:49
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- 16-10-2017 20:59
(Original post by Mazza2000)
Hi guys. I know this is probably really simple but I'm confused with how to find the domain of an inverse function. Its question 5b on this paper
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
So i've found the inverse function to be (e^x+1)/2.
To find the domain I thought it was the range of the original function. And to find that what I did is draw the graph of y=ln(2x-1) and see that y is always greater than 1. So i made the range of this f(x)>1. This is wrong though... I'm very confused with all this domain and range stuff and help would be much appreciated xxx -
Mazza2000
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- 16-10-2017 21:31
(Original post by Pangol)
You are quite right that the domain of the inverse function is the range of the original function. It is the range of the original function that you don't have quite right. Can you describe how you got it? -
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- 16-10-2017 21:35
(Original post by Mazza2000)
I literally just drew the graph of y=f(x) and then saw that the assymptote of this graph is at y=1, so for the range I just said that y must be greater than 1. -
Mazza2000
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- 17-10-2017 20:04
(Original post by Pangol)
This is the graph of y = ln(2x - 1), yes? It doesn't have a horizontal asymptote. -
RDKGames
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- 17-10-2017 20:20
(Original post by Mazza2000)
y=ln(x) has an assymptote at y=1 right? Or is this wrong??
Letand you find that
, which is a perfectly valid solution.
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- 17-10-2017 21:07
(Original post by Mazza2000)
y=ln(x) has an assymptote at y=1 right? Or is this wrong?? -
Mazza2000
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- 18-10-2017 14:58
Ok cool thanks everyone who replied to this. That clears everything up. 😊😊
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