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Thread: Combination of Functions

  1. #1
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    Combination of Functions

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




    c)

    $\displaystyle f(x) = \sqrt{x}$

    $\displaystyle g(x) = x^2$


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

    $\displaystyle = f(x^2)$

    $\displaystyle = \sqrt{x^2}$

    How would you simplify the answer?


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

    $\displaystyle = (\sqrt {x})^2$

    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)
    $\displaystyle f(x) = \frac {1}{x + 1}$

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



    For $\displaystyle g \circ f$

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

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

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

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

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



    For $\displaystyle g \circ g$
    $\displaystyle
    = g (\frac {x}{x - 1})$

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

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

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

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

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

    $\displaystyle = (\sqrt {x})^2$

    How would you simplify the answer?
    $\displaystyle (\sqrt{x})^2 = x$ <-- 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)
    $\displaystyle f(x) = \frac {1}{x + 1}$

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



    For $\displaystyle g \circ f$

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


    Quote Originally Posted by Macleef View Post
    For $\displaystyle g \circ g$
    $\displaystyle
    = g (\frac {x}{x - 1})$

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

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

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

    -Dan
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