Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By skeeter

Math Help - Can anyone help?

  1. #1
    Newbie
    Joined
    Jun 2012
    From
    san diego
    Posts
    14

    Can anyone help?

    Find each function value.g(x)=x3-2x

    (a) g(3) = 1
    (b) g(10) = 2
    (c) g(-3) = 3
    (d) g(7) = 4
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member PaulAdrienMauriceDirac's Avatar
    Joined
    Oct 2012
    From
    T'bilisi
    Posts
    33
    Thanks
    1

    Re: Can anyone help?

    Quote Originally Posted by impressu2 View Post
    Find each function value.g(x)=x3-2x

    (a) g(3) = 1
    (b) g(10) = 2
    (c) g(-3) = 3
    (d) g(7) = 4
    To find value of a function at x=a, where a is some constant, you replace every x in that function with a.

    g(x)=x^3-2x

    (a) g(3)=3^3-2*3=27-6=21.

    (b) g(10)=10^3-2*10=1000-20=980.

    So your answers are incorrect, but I hope you understand now, can you do rest of the problem (parts (c) and (d))?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Sep 2012
    From
    Sweden
    Posts
    250
    Thanks
    6

    Re: Can anyone help?

    he had exact same question early and why do he give the answer "1,2,3,4" plz i hope you dont try that you want us to solve ur homework...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2012
    From
    san diego
    Posts
    14

    Re: Can anyone help?

    it's not the same question, and I not looking for answers only, I am trying to really understand it. As for the 1234 numbers, I am not putting those, I think it's a pc error.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013

    Re: Can anyone help?

    you've been given ample instruction in evaluating function values. thread closed.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum