Results 1 to 13 of 13

Math Help - Functions - Pre Calculus

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    37

    Functions - Pre Calculus

    Functions f and g are defined as follows
    <br />
f(x) = 2x + 3, x\epsilon\Re and g(x) = \frac{1}{x-1}, x\epsilon\Re , x  \ne 1



    (i) Find the expression for the inverse function f^-1 (x)

    (ii) Find the expression for the composite function gf (x)

    (iii) Solve the equation f^-1 (x) = gf(x)-1


    I got:
    (i)
     y = 2x+3

     y - 3 = 2x

     \frac{y-3}{2} = x

     \frac{x-3}{2} = y ?

    (ii)
    And I got g(f(x)) = \frac{1}{2x+2}




    Am i correct? If not, please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Correct on both questions, what did you get for the third?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    I can't do to be honest. I can only plug in the values to the equation. If anyone could post the solution, I would be grateful.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,441
    Thanks
    1862
    You already know that f^{-1}(x)= \frac{x-3}{2} and that g(f(x))= \frac{1}{2x+2}.

    To solve f^{-1}(x)= g(f(x)), set \frac{x-3}{2}= \frac{1}{2x+ 2} and solve for x.

    I recommend multiplying both sides by 2x+2= 2(x+1) to get rid of the fractions. That will result in a quadratic equation.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Quote Originally Posted by gary223 View Post
    Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math.
    I'm happy to give you a kick start.

    \frac{x-3}{2}= \frac{1}{2x+ 2}

    (x-3)(2x+2)= 2

    2x^2-4x-6=2

    2x^2-4x-8=0

    The rest is yours to finish

    here's a hint: The Quadratic Formula Explained
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Shouldn't there be a '-1' after the \frac{1}{2x+2} since its in part (iii) of the question? Or is it how you posted?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Quote Originally Posted by gary223 View Post
    (iii) Solve the equation f^-1 (x) = gf(x)-1
    It seems there should be,

    \frac{x-3}{2}= \frac{1}{2x+ 2}-1

    This thread has given you all the tools you need to solve this equation. Look forward to seeing your answer.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Quote Originally Posted by pickslides View Post
    It seems there should be,

    \frac{x-3}{2}= \frac{1}{2x+ 2}-1

    This thread has given you all the tools you need to solve this equation. Look forward to seeing your answer.
    Okay, I got:

    2x^2-4x-5=0

    (2x+1)(x-5)=0

    2x+1= 0 and x-5=0

    2x=-1 and  x=5

    x=-\frac{1}{2} and x=5
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    I have

     \frac{x-3}{2}= \frac{1}{2x+ 2}-1

     \frac{x-3}{2}+1= \frac{1}{2x+ 2}

     \frac{x-3}{2}+\frac{2}{2}= \frac{1}{2x+ 2}

     \frac{x-3+2}{2}= \frac{1}{2x+ 2}

     \frac{x-1}{2}= \frac{1}{2x+ 2}

     (x-1)(2x+2)= 2

    continue...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Quote Originally Posted by pickslides View Post
    I have

     \frac{x-3}{2}= \frac{1}{2x+ 2}-1

     \frac{x-3}{2}+1= \frac{1}{2x+ 2}

     \frac{x-3}{2}+\frac{2}{2}= \frac{1}{2x+ 2}

     \frac{x-3+2}{2}= \frac{1}{2x+ 2}

     \frac{x-1}{2}= \frac{1}{2x+ 2}

     (x-1)(2x+2)= 2

    continue...

     (x-1)(2x+2)= 2

    2x^2+2x-2x-2=2

    2x^2-2=2

    2x^2 = 0
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    Quote Originally Posted by gary223 View Post
     (x-1)(2x+2)= 2

    2x^2+2x-2x-2=2

    2x^2-2=2
    Adding 2 to both sides gives

    2x^2 = 4
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 27th 2010, 02:17 AM
  2. Calculus (Functions Help)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 16th 2010, 01:44 PM
  3. Pre-Calculus Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 15th 2010, 06:17 AM
  4. pre-calculus/functions help
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: October 1st 2009, 06:58 PM
  5. Possible Functions Calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 28th 2009, 06:44 PM

Search Tags


/mathhelpforum @mathhelpforum