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Math Help - 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 (fg)(x) as the product f(x) \cdot g(x)

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

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

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

    How do I do this if f(x) = \sqrt[3](x+1) and 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 \circ and \cdot .

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

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

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

    you do g[f(x)]
    Is there any way to simplify f[g(x)] = \sqrt[3]{\cos{x} + 1} or  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 f[g(x)] = \sqrt[3]{\cos{x} + 1} or  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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