Find the domain of this function

Question

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.

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

If $x$ is real, then $x^2>-18$. You can't take the square root of both sides that way... — Thomas Andrews, Feb 03 '15 at 03:53
Hard to tell, since you haven't said how you are going about solving it. I'd say $T(x)$ is defined only if $(3-x)^2+1^2\geq 0$ and $x^2+2^2\geq 0$. — Thomas Andrews, Feb 03 '15 at 03:56
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 — John Dias, Feb 03 '15 at 04:01
Put it in the question. People wanting to answer your question should not have to read the comments to get the complete question. — Thomas Andrews, Feb 03 '15 at 04:05
I don't see the problem. The quantities in the square roots are both sums of squares and are thus both positive. — MasB, Feb 03 '15 at 04:18
so what would i put as the domain sqrt(-18) is a nonreal answer? — John Dias, Feb 03 '15 at 04:25
@JohnDias the domain of the function is the reals. Any x satisfies your function. — MasB, Feb 03 '15 at 04:28
@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. — MasB, Feb 03 '15 at 04:41

score 2 · Answer 1 · edited Dec 01 '17 at 21:52

2

$\sqrt{x^{2}+4}$ $\Longrightarrow$ $x^2+ 4\geq0,\;$ 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 $\mathbb{R}$

edited Dec 01 '17 at 21:52

amWhy

209,954

answered Oct 17 '17 at 11:26

Kislay Tripathi

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Find the domain of this function

1 Answers1