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how do i find the domain of this function? I keep ending up with $0 \leq x^2 + 18$, $-18 \leq x^2$, and this give me a non-real answer $\sqrt{-18}\leq x \leq \sqrt{18}$? i set the function <= 0 so their is no more square root on the numerators, then multiply the numerators by the LCD 4, then divide then add common terms which leave me with 0 <= x^2 + 18.

$$T(x)= \frac{\sqrt{x^2 + 2^2}}2 + \frac{\sqrt{(3-x)^2 + 1^2}}4$$

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  • $\begingroup$ If $x$ is real, then $x^2>-18$. You can't take the square root of both sides that way... $\endgroup$ – Thomas Andrews Feb 3 '15 at 3:53
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    $\begingroup$ I don't see the problem. The quantities in the square roots are both sums of squares and are thus both positive. $\endgroup$ – Bernard Massé Feb 3 '15 at 4:18
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    $\begingroup$ @JohnDias the domain of the function is the reals. Any x satisfies your function. $\endgroup$ – Bernard Massé Feb 3 '15 at 4:28
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    $\begingroup$ @JohnDias, Try plugging in some numbers and see what happens. For example, 10, -23, 0.111, -1/10, 3000. You'll notice that every time the quantities in the square roots are positive so you can always calculate the square root as a real number. $\endgroup$ – Bernard Massé Feb 3 '15 at 4:41
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    $\begingroup$ $ -\infty < x < \infty $ $\endgroup$ – Bernard Massé Feb 3 '15 at 4:57
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$\sqrt{x^{2}+4}$ $\Longrightarrow$ $x^2+ 4\geq0,\;$ which is true $\forall x\in \mathbb{R}.$

Again $\sqrt{\left( 3-x\right) ^{2}+12}\Longrightarrow\left(3-x\right)^2+12\geq0,\;$ which is true $\forall x \in\mathbb{R}$

hence domain of $T(x)$ is $\mathbb{R}$

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