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Math Help - how to finish this question..

  1. #1
    MHF Contributor
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    how to finish this question..

    i need to find for what values X is defined in the given function

    i found for the sin part and for the the square root but i dont know how to combine them
    into a final answer:
    http://img168.imageshack.us/img168/1434/39293849jq8.gif
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by transgalactic View Post
    i need to find for what values X is defined in the given function

    i found for the sin part and for the the square root but i dont know how to combine them
    into a final answer:
    http://img168.imageshack.us/img168/1434/39293849jq8.gif
    let us do it using composition of functions..

    let g(x) = \sqrt{x}
    h(x) = \sin x

    f(x) = \sqrt{\sin \sqrt{x}} = (g \circ h \circ g)(x) = (g\circ (h\circ g)(x)) = g((h\circ g)(x)) = g(h(g(x)))

    --------------------------
    now, recall what the domain of composition of function is..

    \mbox{dom}_{f \circ g} = \{ x \in \mbox{dom}_{g} \ : \ g(x) \in \mbox{dom}_f \}
    --------------------------

    \mbox{dom}_g = \{x\in\mathbb{R}: x\geq 0\}
    \mbox{dom}_h = \mathbb{R}

    thus,
    \mbox{dom}_{h\circ g} = \{x\in\mathbb{R}: x\geq 0\}
    and
    \mbox{dom}_{g\circ(h\circ g)} = \{x\geq 0: \sin \sqrt{x} \geq 0\} or
    \mbox{dom}_{g\circ(h\circ g)} = \{x\geq 0: 2k\pi \leq \sqrt{x} \leq (2k+1)\pi, k\in\mathbb{Z}^+ \cup \{0\}\} or
    \mbox{dom}_{g\circ(h\circ g)} = \{x: 4k^2\pi^2 \leq x \leq (2k+1)^2\pi^2, k\in\mathbb{Z}^+ \cup \{0\}\}
    ----------------------
    note that the last inequality is possible since the square root (and square) function(s) is an(are) increasing function(s)..
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