# Thread: how to finish this question..

1. ## 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
http://img168.imageshack.us/img168/1434/39293849jq8.gif

2. Originally Posted by transgalactic
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
http://img168.imageshack.us/img168/1434/39293849jq8.gif
let us do it using composition of functions..

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

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

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

$\mbox{dom}_{f \circ g} = \{ x \in \mbox{dom}_{g} \ : \ g(x) \in \mbox{dom}_f \}$
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$\mbox{dom}_g = \{x\in\mathbb{R}: x\geq 0\}$
$\mbox{dom}_h = \mathbb{R}$

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