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Math Help - Combination of Functions

  1. #1
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    Combination of Functions

    In each of the following cases find:
     f \circ g
    g \circ f
    f \circ f
    g \circ g




    c)

    f(x) = \sqrt{x}

    g(x) = x^2


    For  f \circ g
    = f(g(x))

    = f(x^2)

    = \sqrt{x^2}

    How would you simplify the answer?


    For g \circ f
    = g(f(x))

    = (\sqrt {x})^2

    How would you simplify the answer?



    Could you please check the following to see if I did them correctly and show me where I messed up?


    d)
    f(x) = \frac {1}{x + 1}

    g(x) = \frac {x}{x -1}



    For g \circ f

    = g (\frac {x}{x + 1})

    = \frac {(\frac{x}{x + 1})}{( \frac{x}{x + 1} - 1)}

    = \frac {(\frac {x}{x + 1})}{(\frac{1}{x + 1})}

    = \frac {x}{x + 1} (x + 1)

    = \frac {x^2 + x}{x + 1}



    For g \circ g
    <br />
= g (\frac {x}{x - 1})

    = \frac {(\frac {x}{x - 1})}{(\frac {x}{x -1} - 1)}

    = (\frac {x}{x - 1}) (x - 1)

    = \frac {x^2 - x}{x - 1}
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  2. #2
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    Quote Originally Posted by Macleef View Post
    = \sqrt{x^2}

    How would you simplify the answer?
    \sqrt{x^2} = |x|

    Quote Originally Posted by Macleef View Post
    For g \circ f
    = g(f(x))

    = (\sqrt {x})^2

    How would you simplify the answer?
    (\sqrt{x})^2 = x <-- Since x can't be negative anyway, we don't need to worry about negative answers.


    Quote Originally Posted by Macleef View Post
    Could you please check the following to see if I did them correctly and show me where I messed up?


    d)
    f(x) = \frac {1}{x + 1}

    g(x) = \frac {x}{x -1}



    For g \circ f

    = g (\frac {x}{x + 1})
    f(x) = \frac{1}{x + 1}
    You are off by a factor of x.


    Quote Originally Posted by Macleef View Post
    For g \circ g
    <br />
= g (\frac {x}{x - 1})

    = \frac {(\frac {x}{x - 1})}{(\frac {x}{x -1} - 1)}

    = (\frac {x}{x - 1}) (x - 1)

    = \frac {x^2 - x}{x - 1}
    You can write this more simply as x;~x \neq 1

    -Dan
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