Turn on thread page Beta

C3 (AEA Ext.) Proof of functions watch

Announcements
    • Thread Starter
    Offline

    16
    ReputationRep:
    Given that  f(x)= \frac {ax+b}{cx+d}, with x \neq - \frac{c}{d} , show that,
    a) if f(x) is an even function, then ad=bc and hence f(x)=k, where k is constant.

    I can show that ad=bc, since f(x)=f(-x) when the function is even, however I become stuck trying to prove that there will be a constant.

    Would it be that for ax+b/cx+d to be even, that a and c must be 0? Therefore ad=bc=0 but b and d can be any value, leaving just a constant or is there another way to show this?
    Offline

    18
    ReputationRep:
    (Original post by TheTallOne)
    Given that  f(x)= \frac {ax+b}{cx+d}, with x \neq - \frac{c}{d} , show that,
    a) if f(x) is an even function, then ad=bc and hence f(x)=k, where k is constant.

    I can show that ad=bc, since f(x)=f^-1(x) when the function is even, however I become stuck trying to prove that there will be a constant.
    Why does f(x)=f^{-1}(x) when the function is even?

    Would it be that for ax+b/cx+d to be even, that a and c must be 0? Therefore ad=bc=0 but b and d can be any value, leaving just a constant or is there another way to show this?
    No. What happens if a=b=c=d=1?
    • Thread Starter
    Offline

    16
    ReputationRep:
    Sorry, it's f(x)=f(-x)

    and I could add if a=c and b=d
    Offline

    18
    ReputationRep:
    Think about it the other way: What does it mean for \frac{ax+b}{cx+d} to be a constant k? (i.e. what does that tell you about a,b,c,d)?
    • Thread Starter
    Offline

    16
    ReputationRep:
    That the coefficient of x must be 0 when ax+b/cx+d is simplified.
    Offline

    18
    ReputationRep:
    (Original post by TheTallOne)
    That the coefficient of x must be 0 when ax+b/cx+d is simplified.
    That's true, and it's one way of finding an answer. But to a large extent I think you're going in circles here - if you couldn't see how to answer the question before, I'm not sure you'll be able to now.

    I was more thinking that if you know \frac{ax+b}{cx+d} = k, then what's the obvious thing to do next?
    • Thread Starter
    Offline

    16
    ReputationRep:
     ax+b= ckx + dk
     x= \frac {dk-b}{a-ck}
    ???
    Sorry- I went for lunch.
    Offline

    18
    ReputationRep:
    You're expecting it to be true for all x, so trying to solve for x isn't a terribly good plan.

    You have ax+b = k(cx+d), and this is true for all x.

    So what happens when x= 0?
    What happens when x=1? (Given what you learned from looking at x=0).

    Of course, this is all arguing backwards from the answer, so you now have to adapt your reasoning so that you are arguing forwards from ad = bc. (Hint for that: work out what k must be first).
    • Thread Starter
    Offline

    16
    ReputationRep:
     when x=0, b=dk
     k=b/d
    When x=1,
     a+b=k(c+d)

    So
     a+b= \frac{b}{d} (c+d)
     a+b= \frac{bc}{d} +b

    Goes to....
    a=bc/d
    and ad=bc

    Thanks
 
 
 
Turn on thread page Beta
Updated: July 19, 2007

University open days

  • Manchester Metropolitan University
    Postgraduate Open Day Postgraduate
    Wed, 14 Nov '18
  • University of Chester
    Chester campuses Undergraduate
    Wed, 14 Nov '18
  • Anglia Ruskin University
    Ambitious, driven, developing your career & employability? Aspiring in your field, up-skilling after a career break? Then our Postgrad Open Evening is for you. Postgraduate
    Wed, 14 Nov '18
Poll
Should Banksy be put in prison?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.