Results 1 to 5 of 5

Math Help - Another inverse function issue...

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    100

    Another inverse function issue...

    Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

    Find the two values of x such that fg(x) = gf(x), giving each answer in the form p + q \sqrt{3}

    I have tried solving it by equating by solving it for both of the functions but that's not working. How should I proceed with it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by struck View Post
    Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

    Find the two values of x such that fg(x) = gf(x), giving each answer in the form p + q \sqrt{3}
    Well...

    f(g(x))=(2x-1)^2+6(2x-1)=\text{expand everything}=4x^2+8x-5

    g(f(x))=2(x^2+6x)-1=\dots=2x^2+12x-1


    Now solve for x in :
    4x^2+8x-5=2x^2+12x-1

    that is 2x^2-4x-4=0

    x^2-2x-2=0

    in order to do that, find the discriminant, and use your head... ^^
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2007
    Posts
    100
    Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

    \frac {2 \pm \sqrt{4 + 8}}{2}

    \frac {2 \pm \sqrt{3} \sqrt {4}}{2}

    \frac {2 \pm 2 \sqrt{3}}{2}

    The answer I am supposed to get is 1 \pm \sqrt{3} but I am getting 2 \pm \sqrt {3}. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by struck View Post
    Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

    \frac {2 \pm \sqrt{4 + 8}}{2}

    \frac {2 \pm \sqrt{3} \sqrt {4}}{2}

    \frac {2 \pm 2 \sqrt{3}}{2}

    The answer I am supposed to get is 1 \pm \sqrt{3} but I am getting 2 \pm \sqrt {3}. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.
    Yeah, by discriminant, I meant what you did.
    And yes, it would have been better to show the steps avoiding some useless lines


    Anyway, you've got the answer !

    \frac{2\pm 2 \sqrt{3}}{2} \neq 2 \pm \sqrt{3}
    I don't know how you got that !

    \frac{2 \pm 2 \sqrt{3}}{2}=\frac 22 \pm \frac{2 \sqrt{3}}{2}=1 \pm \sqrt{3}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2007
    Posts
    100
    OMG. My brain wasn't working well I guess. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Curious issue with an inverse Laplace Transform...
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 18th 2011, 04:05 AM
  2. function composition and inverse function
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: November 10th 2009, 12:18 PM
  3. Replies: 2
    Last Post: September 22nd 2009, 08:29 PM
  4. Replies: 4
    Last Post: March 17th 2008, 09:45 PM
  5. Step function Graphing issue
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 30th 2007, 07:24 AM

Search Tags


/mathhelpforum @mathhelpforum