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Math Help - piece wise defined function

  1. #1
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    piece wise defined function

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

    f(x)= sin x+2 , x>1
    = cos x-2 , x<=1

    g(x)=2x+3 , x>0
    =x^2 , x<=0

    Find f o g
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  2. #2
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    (f o g)(x)=f(g(x))

    so put g(x) as x in f(x), you should get f o g defined for the intervals
    x>1,0<x \leq 1,x \leq 0
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  3. #3
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    Quote Originally Posted by Krahl View Post
    (f o g)(x)=f(g(x))

    so put g(x) as x in f(x), you should get f o g defined for the intervals
    x>1,0<x \leq 1,x \leq 0
    ok

    fg(x)= sin (2x+3) +2
    =cos (2x+3) - 2

    Do i substitute each of the g(x) in each of the function f(x) ? Could you explain on the interval ? Thats what i am most confused with . Thanks .
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  4. #4
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    we have f(g(x))

    2 functions for g(x).

    x >0 g(x)=2x+3>3 and f(x)=sin(x)+2 for x>1

    so for x>0 f(g(x))=sin(2x+3)+2

    similarly

    0 \geq x \geq -1, 0 \leq g(x)=x^2 \leq 1 and
    f(g(x))=cos(g(x))-2=cos(x^2)-2

    And

    for x<-1 g(x)=x^2>1 f(g(x))=sin(x^2)+2


    Finally

    f(g(x))= sin(x)+2 if x>0
    cos(x^2)-2 if -1 \leq x \leq 0
    sin(x^2)+2 if x<-1

    notice we have covered the whole real line with x. The important part is to use restrictions of x so that g(x) fits in the inequalities for f(x). if i had used
    0 \geq x > -1 instead of 0 \geq x \geq -1 then the next would be g(x)=x^2 \geq 1 which doesn't fit the inequalities for f(x).
    [I rushed this there may be mistakes ]
    Last edited by Krahl; April 24th 2010 at 10:15 AM.
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