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Piece wise function

  1. Apr 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that f(R)=R , g(R)=R be defined respectively by

    [tex]f(x)=\begin{cases} \sin x+2 & \text{if } x>1 \\ \cos x-2 & \text{if } x\leq 1\end{cases}[/tex]

    [tex]g(x)=\begin{cases}2x+3& \text{if } x>0 \\ x^2 & \text{if } x\leq 0 \end{cases} [/tex]

    Find f o g and g o f


    2. Relevant equations



    3. The attempt at a solution

    i have no idea to begin except for the obvious substitution of g(x) into the function f(x) . I am not sure how to adjust the domain .
     
  2. jcsd
  3. Apr 23, 2010 #2

    rock.freak667

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    For x>1, wouldn't f(x) = sinx +2 and g(x)=2x +3 ?
     
  4. Apr 23, 2010 #3

    HallsofIvy

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    You need to be careful of the domains. If [itex]x\le 0[/itex], [itex]g(x)= x^2[/itex] but then [itex]x^2< 1[/itex] if x> -1 and [itex]x^2> 1[/itex] if x< -1.

    If x> 0, [itex]g(x)= 2x+ 3[/itex] and, since x is positive, that is always larger than 3 which is larger than 1.

    You need to divide into three intervals: for x<-1, [itex]g(x)= x^2> 1[/itex] so [itex]f(g(x))= f(x^2)= sin(x^2)+ 2[/itex]. for [itex]-1\le x\le 0[/itex], [itex]g(x)= x^2\le 1[/itex] so [itex]f(g(x))= f(x^2)= cos(x^2)- 1[/itex]. For x> 0, g(x)= 2x+ 3 and f(g(x))= f(2x+3)= sin(2x+3)+ 2.

    Now, you do g(f(x))
     
  5. Apr 23, 2010 #4
    thanks !
     
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