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Thread: f(x) and g(x), need to solve (fg)(x) and (gf)(x)

  1. #1
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    f(x) and g(x), need to solve (fg)(x) and (gf)(x)

    f(x) = cuberoot(x+1) and g(x) = cosx

    how do I do this?



    Here is the note I took in class:

    Let f and g be two functions having the domain D. The product of f and g is that function which assigns to x is a subset of D the number f(x)g(x)

    fg
    x ----> (fg)(x) = f(x)g(x)

    Did I copy this incorrectly? I thought (fg)(x) = f[g(x)] not f(x)g(x)

    help! Thanks
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  2. #2
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    Quote Originally Posted by bloodyfinger View Post
    f(x) = cuberoot(x+1) and g(x) = cosx

    how do I do this?



    Here is the note I took in class:

    Let f and g be two functions having the domain D. The product of f and g is that function which assigns to x is a subset of D the number f(x)g(x)

    fg
    x ----> (fg)(x) = f(x)g(x)

    Did I copy this incorrectly? I thought (fg)(x) = f[g(x)] not f(x)g(x)

    help! Thanks

    Hi bloodyfinger,

    Usage varies sometimes by textbook and author. When in doubt, ask for clarification.

    My opinion...I would look at $\displaystyle (fg)(x)$ as the product $\displaystyle f(x) \cdot g(x)$

    It should have been written as $\displaystyle (f \cdot g)(x)$. This way the meaning is clear.

    If they had intended to indicate the composition of functions, they would've used $\displaystyle [f \circ g](x)]$ to mean $\displaystyle f[g(x)]$

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  3. #3
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    Oh shoot, my mistake. I didn't realize there was a difference between $\displaystyle \circ$ and $\displaystyle \cdot$ .

    The original question actually asks me to find $\displaystyle [f \circ g](x)]$ and $\displaystyle [g \circ f] (x)] $.

    How do I do this if $\displaystyle f(x) = \sqrt[3](x+1)$ and $\displaystyle g(x) = cosx $
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  4. #4
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    Quote Originally Posted by bloodyfinger View Post
    Oh shoot, my mistake. I didn't realize there was a difference between $\displaystyle \circ$ and $\displaystyle \cdot$ .

    The original question actually asks me to find $\displaystyle [f \circ g](x)]$ and $\displaystyle [g \circ f] (x)] $.

    How do I do this if $\displaystyle f(x) = \sqrt[3](x+1)$ and $\displaystyle g(x) = cosx $
    $\displaystyle f[g(x)] = f[\cos{x}] = \sqrt[3]{\cos{x} + 1}$

    you do $\displaystyle g[f(x)]$
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  5. #5
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    Quote Originally Posted by skeeter View Post
    $\displaystyle f[g(x)] = f[\cos{x}] = \sqrt[3]{\cos{x} + 1}$

    you do $\displaystyle g[f(x)]$
    Is there any way to simplify $\displaystyle f[g(x)] = \sqrt[3]{\cos{x} + 1}$ or $\displaystyle g[f(x)] = cos\sqrt[3]{x+1} $ ?
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  6. #6
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    Quote Originally Posted by bloodyfinger View Post
    Is there any way to simplify $\displaystyle f[g(x)] = \sqrt[3]{\cos{x} + 1}$ or $\displaystyle g[f(x)] = cos\sqrt[3]{x+1} $ ?
    looks like both are in "simple" form to me.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    looks like both are in "simple" form to me.
    That was shockingly easy.... I have a hard time believing the question was that simple... haha

    Thanks so much though!!
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