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Piecewise-defined functions

  1. Sep 25, 2008 #1
    hi, i missed a lecture on this topic and i am really lost...there is a question that asks to write a function in the piecewise-defined form. i have no idea what that means and its not in the textbook of the course...looked everywhere online and can't find anything on it....

    they give the function in the form of [tex]f(x)=x(1-x)H(x-1)+(4x-x^2-4)H(x-2)[/tex]

    have no idea how to proceed...dont understand what H is doing in there...can someone provide some explanation and a little insight into this topic...would be appreciated.
    Last edited: Sep 25, 2008
  2. jcsd
  3. Sep 25, 2008 #2


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    You might want to check your textbook for the definition of H(x), the Heaviside step function.

    H(x) is defined as "0 is x< 0, 1 if [itex]x\ge 0[/itex]".

    From that H(x-1) is 0 if x< 1, 1 if [itex]x\ge 1[/itex] and H(x-2) is 0 if x< 2, 1 if [itex]x\ge 2[/itex].

    In this example, if x< 1, then both H(x-1) and H(x-2) are 0. If x< 1, f(x)x(1-x)(0)+ (4x- x2-4)(0)= 0. If [itex]1\le x< 2[/itex] then H(x-1)= 1 while H(x-2) is 0. f(x)= x(1-x)(1)+ (4x- x2-4)(0)= x(1-x). If [itex]x\ge 2[/itex] both H(x-1) and H(x-2) are 1 so f(x)= x(1-x)(1)+ (4x- x2-4)(1)= x(x-1)+ 4x- x2- 4= x- x2+ 4x- x2- 4= 3x- 4.

    That could also be written
    [tex]f(x)= \left[\begin{array}{c} 0 \\ x(1-x) \\ 3x- 4\end{array}\right [/tex]
  4. Sep 25, 2008 #3
    thanks, i understand...but for the third part...isn't it [tex]-2x^2+5x-4[/tex]??? i think u made a small error in the final calculation...
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