Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    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.
    Offline

    0
    ReputationRep:
    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    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....
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by novadragon849)
    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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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.
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by novadragon849)
    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 x-1 is the denominator of the fraction, so it will cancel when you simplify it leaving no x terms on the numerator.

    So if you had \dfrac{4x+3}{2x+1}, for example, you could write it as \dfrac{2(2x+1)+1}{2x+1}, which simplifies to 2 + \dfrac{1}{2x+1}.
 
 
 
Poll
Who is your favourite TV detective?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.