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# Help needed for C3 range/domain question. watch

1. Hi, I need help with a range/domain question I came across. Its C3 solomon paper F question 7 for anyone interested.

I manage to do part (a) which is to show that f(x) = (3x+2) / (x-1) x ∈ R, x < 1.

Then on part (b) I manage to get f^-1 (x) which is (x+2) / (x-3) the question then ask for its domain which I got stuck on. I tried looking at the mark scheme but it makes no sense to me.

So I was hoping if anyone can walk me through finding the domain please?

Thanks.
2. The domain of the inverse is just the range of the function. So in this example, as the range of f(x) is x ∈ R, x < 1, this is the domain you want.
3. Yes I was taught that the domain of the inverse is just the range of the function but this does not seem to be the case. I was looking at the mark scheme trying to figure out how they got this but I still have no clue.

Here it is:

f(x) = (3x+2) / (x-1)

f(x) = [3(x-1)+5] / (x-1) therefore = 3 + [5/(x-1)]

x < 1 therefore f(x) < 3 therefore domain of f^-1 (x) is x ∈ R, x < 3

I don't get this at all....
f(x) = (3x+2) / (x-1)

f(x) = [3(x-1)+5] / (x-1) therefore = 3 + [5/(x-1)]
They did this by noticing that 3x+2 = 3(x-1) + 5 so that they could simplify the fraction.

x < 1 therefore f(x) < 3 therefore domain of f^-1 (x) is x ∈ R, x < 3
This is because whenever x<1, the fraction 5/(x-1) is negative, and so 3 + 5/(x-1) must be less than 3. This is the range of the original function and is therefore the domain of the inverse.

I don't get this at all....
I've put an explanation of each step in green in the quote.
5. (Original post by nuodai)
I've put an explanation of each step in green in the quote.
This is amazing! Thanks a lot. Just a quick question though, is the noticing part where they did 3x+2 = 3(x-1) + 5 so that they could simplify the fraction etc a guessing thing or is there actually a way to do it? so why use (x-1)?

I understand it now but I can see that in an exam if this came up I won't be able to use the (x-1) at the top to simply it.

Thanks.
This is amazing! Thanks a lot. Just a quick question though, is the noticing part where they did 3x+2 = 3(x-1) + 5 so that they could simplify the fraction etc a guessing thing or is there actually a way to do it? so why use (x-1)?
Because is the denominator of the fraction, so it will cancel when you simplify it leaving no terms on the numerator.

So if you had , for example, you could write it as , which simplifies to .

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