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Math Help - F(g(x))

  1. #1
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    F(g(x))

    Can someone please confirm my answer on this question.

    F(x)=sqrt(x), G(x)=cubedrt(1-x)

    First I got sqrt(cubedrt(1-x)= sqrt(1-x)^1/3 =[(1-x)^1/3]^1/2= (1-x)^1/5?
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  2. #2
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    Re: F(g(x))

    Or is it \displaystyle \sqrt{\sqrt[3]{1-x}} = ((1-x)^{\frac{1}{3}})^\frac{1}{2}= (1-x)^{\frac{1}{6}} ??
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    Re: F(g(x))

    of course ...thats what i meant so would the domain be the domain of g and the domain of f before the combination? Also, I am just starting this calculus class and there is all these fancy symbols and ways to write the domain and range. What would you recommend is the simplest form of expressing domain and range?
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    Re: F(g(x))

    Quote Originally Posted by jmanna98 View Post
    of course ...thats what i meant so would the domain be the domain of g and the domain of f before the combination?
    Maybe not, what is the domains of f and the domain of g?


    Quote Originally Posted by jmanna98 View Post
    Also, I am just starting this calculus class and there is all these fancy symbols and ways to write the domain and range. What would you recommend is the simplest form of expressing domain and range?
    It like using the (,),[,] notations. Do you know them?
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  5. #5
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    Re: F(g(x))

    The book says fog is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (fog)(x) is defined whenever both g(x) and f(g(x)) are defined. This throws me off.

    I do like the open and closed interval notation. Although, how do we specify if we are talking about domain or range?

    thanks for all the help by the way
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  6. #6
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    Re: F(g(x))

    Quote Originally Posted by jmanna98 View Post
    The book says fog is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (fog)(x) is defined whenever both g(x) and f(g(x)) are defined. This throws me off.
    I do like the open and closed interval notation. Although, how do we specify if we are talking about domain or range?
    Because \displaystyle \sqrt{\sqrt[3]{1-x}} = ((1-x)^{\frac{1}{3}})^\frac{1}{2}= (1-x)^{\frac{1}{6}}
    it is clear that the domain of f\circ g is x\le 1.
    In interval notation that is (-\infty,1].
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