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# Functions help! watch

1. 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
2. (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
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?
3. (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?
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.
4. (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.
This is the graph of y = ln(2x - 1), yes? It doesn't have a horizontal asymptote.
5. (Original post by Pangol)
This is the graph of y = ln(2x - 1), yes? It doesn't have a horizontal asymptote.
y=ln(x) has an assymptote at y=1 right? Or is this wrong??
6. (Original post by Mazza2000)
y=ln(x) has an assymptote at y=1 right? Or is this wrong??
Has one at

Let and you find that , which is a perfectly valid solution.
7. (Original post by Mazza2000)
y=ln(x) has an assymptote at y=1 right? Or is this wrong??
As mentioned above, y=ln(x) only has a vertical asymptote. There is no restriction on the value of ln(x) as x ranges over all real values.
8. Ok cool thanks everyone who replied to this. That clears everything up. 😊😊

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