Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Can anyone help solve this problem:

    Explain why the graph of f(x) = (x^2 + 1)/(x^2 - 1) and its inverse function f-1(x) is also a solution to x^3 - x^2 - x -1 = 0. Show that an approximate solution is x=1.84.
    Attached Images
     
    Offline

    3
    ReputationRep:
    Why you doing math

    Posted from TSR Mobile
    Offline

    15
    ReputationRep:
    Use whatever you got for f-1(x) and set it equal to f(x), and then just rearrange.
    Offline

    17
    ReputationRep:
    (Original post by tommy_lees)
    Can anyone help solve this problem:

    Explain why the graph of f(x) = (x^2 + 1)/(x^2 - 1) and its inverse function f-1(x) is also a solution to x^3 - x^2 - x -1 = 0. Show that an approximate solution is x=1.84.
    Where have you got to so far?
    Offline

    3
    ReputationRep:
    Since they never tell you to explicitly find the inverse function, this is what I believe they want you to do:

    As you probably got from your sketch earlier, the inverse function is a reflection of the original function in the line y=x. Now for a point of intersection between both curves (what you are looking for in f(x) = f^-1 (x)) said point has to lie on both curves. Any idea on where else the point must lie? (once you figure this out, it is a simple rearrangement).
    • Study Helper
    Offline

    3
    ReputationRep:
    Study Helper
    Moved thread to Maths forum as more likely to get answers here, not that you haven't already gotten some
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Solivagant)
    Use whatever you got for f-1(x) and set it equal to f(x), and then just rearrange.

    Thanks I did that, i set f(x) = f-1(x).

    I got:

    x^4 -2x^3 + 1 = 0

    This has the correct solution of 1.84 when solved but I don't see how it relates to x^3 - x^2 - x -1 = 0


    • Study Helper
    Offline

    3
    ReputationRep:
    Study Helper
    (Original post by tommy_lees)
    Thanks I did that, i set f(x) = f-1(x).

    I got:

    x^4 -2x^3 + 1 = 0

    This has the correct solution of 1.84 when solved but I don't see how it relates to x^3 - x^2 - x -1 = 0


    Work out a factor of x^4 - 2x^3 + 1 = 0 and then if you use long division by that factor you should get that
    (Factor)*(x^3 - x^2 - x -1)=0 therefore you have either one of the two factors will equal zero and that's how it relates

    Hint: Factor
    Spoiler:
    Show
    When x=1 then the equation holds true, so divide x^4 - 2x^3 +1 by x-1 to get the other factor.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MathsNerd1)
    Work out a factor of x^4 - 2x^3 + 1 = 0 and then if you use long division by that factor you should get that
    (Factor)*(x^3 - x^2 - x -1)=0 therefore you have either one of the two factors will equal zero and that's how it relates

    Hint: Factor
    Spoiler:
    Show
    When x=1 then the equation holds true, so divide x^4 - 2x^3 +1 by x-1 to get the other factor.
    Ah ok thank you
    • Study Helper
    Offline

    3
    ReputationRep:
    Study Helper
    (Original post by tommy_lees)
    Ah ok thank you
    Happy to help
    Offline

    3
    ReputationRep:
    (Original post by tommy_lees)
    .
    I know you've managed to succeed with your question, but a way to get it out without long-division is to see that all solutions to f^-1(x) = f(x) is the same as the solutions of the intersections of the line y=x and f(x) i.e. x = f(x), rearrange the equation and you get your relation instantly!

    (That was the hint I posted above)
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: July 1, 2014
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.