Results 1 to 7 of 7

Math Help - Piece wise function help

  1. #1
    Member
    Joined
    Sep 2012
    From
    US
    Posts
    123

    Piece wise function help

    f(x) =
    x24
    x2
    if x < 2
    ax2bx + 3 if 2x < 3
    4xa + b if x ≥ 3



    OK guys so I'm having trouble with this, it includes a lot of algebra steps but I'll put some of them to show that I DID attempt it several times, not sure if I'm just tired or I messed up at some point....

    I factored out at the start of coarse...
    (x-2)(x+2)/ (x-2) = ax^2 - bx + 3 -----> (2+2) = a(2)^2-b(2)+3
    4 = 4a - 2b +3
    4 - 3 = 4a - 2b
    1 = 4a - 2b
    1+2b/4 = 4a/4
    1/4+1/2b = a

    ax^2 -bx + 3 = 4x - ab
    a(3)^2 - b(3) +3 = 4(3) - a + b
    9a + a - 3b - b = 12- 3
    10a - 4b = 9
    10(1/4 + 1/2b) - 4b = 9
    5/2 + 5b - 4b = 9
    5/2 - 1b = 9
    -1b = 13/2
    = -13/2

    1/4 + 1/2(-13/2) = a
    1/4 + (-13/4) = a
    - 3 = a

    a = -3 b = -13/2
    Last edited by Oldspice1212; February 2nd 2013 at 10:28 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,343
    Thanks
    1247

    Re: Piece wise function help

    This is extremely hard to read. Is your function \displaystyle \begin{align*} f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} \textrm{ if } x < 2 \\ a x^2 - b  x + 3 \textrm{ if } 2 \leq x < 3 \\ 4x - a + b \textrm{ if } x \geq 3 \end{cases} \end{align*}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2012
    From
    US
    Posts
    123

    Re: Piece wise function help

    Yes exactly right, sorry about that, I wasn't sure how to make it look like that so I copied the question exactly :S
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,343
    Thanks
    1247

    Re: Piece wise function help

    What are you actually trying to do with this question? Are you trying to find the values of a and b which make this function continuous?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2012
    From
    US
    Posts
    123

    Re: Piece wise function help

    Quote Originally Posted by Prove It View Post
    What are you actually trying to do with this question? Are you trying to find the values of a and b which make this function continuous?
    Yes sir, sorry for not mentioning that been up for longer then I should.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,343
    Thanks
    1247

    Re: Piece wise function help

    It should be clear that the function needs to approach 4 if you make x approach 2 from the left (why). For the function to be continuous at x = 2, it needs to approach the same value from the right. So that means \displaystyle \begin{align*} a(2)^2 - b(2) + 3 = 2 \end{align*}.

    For the function to be continuous at x = 3, we require the function to approach the same value from the left as from the right. So \displaystyle \begin{align*} a(3)^2 - b(3) + 3 = 4(3) - a + b \end{align*}.

    Simplify both of these equations and solve them simultaneously for a and b.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2012
    From
    US
    Posts
    123

    Re: Piece wise function help

    Ah I think I missed a step, I got a= 7/2 and b = 13/2 this time hows that look?

    9=10a-4b
    2=8a-4b
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expressing A Piece Wise Function
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 15th 2011, 02:00 PM
  2. continous piece wise function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 30th 2010, 08:42 AM
  3. piece wise defined function
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 24th 2010, 09:05 AM
  4. Piece wise function graphs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 14th 2009, 01:17 PM
  5. Integration of a piece wise function?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 18th 2009, 06:12 PM

Search Tags


/mathhelpforum @mathhelpforum