How do you find the domain and range of $\sqrt{\frac{13 x}{(x^{2}) - 1}}$ $sqrt((13x)/((x^2)-1)$ ?

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How do you find the domain and range of $\sqrt{\frac{13 x}{(x^{2}) - 1}}$ $sqrt((13x)/((x^2)-1)$ ?

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Answer 1 · 2017-05-29T14:51:39

The domain is $x \in (- 1, 0] \cup (+ 1, + \infty)$ $x in (-1,0] uu (+1,+oo)$
The range is $y \in [0, + \infty)$ $y in [0,+oo)$

Explanation:

What's uner the $\sqrt{}$ $sqrt$ sign is $\geq 0$ $>=0$

So,

$\frac{13 x}{x^{2} - 1} \geq 0$ $(13x)/(x^2-1)>=0$

$\frac{13 x}{(x + 1) (x - 1)} \geq 0$ $(13x)/((x+1)(x-1))>=0$

Let $f (x) = \frac{13 x}{(x + 1) (x - 1)}$ $f(x)=(13x)/((x+1)(x-1))$

We can build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a$ $color(white)(aaaaa)$ $- 1$ $-1$ $a a a a a$ $color(white)(aaaaa)$ $0$ $0$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+ 1$ $+1$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 1$ $x-1$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

Therefore,

$f (x) \geq 0$ $f(x)>=0$ when $x \in (- 1, 0] \cup (+ 1, + \infty)$ $x in (-1,0] uu (+1,+oo)$

The domain is $x \in (- 1, 0] \cup (+ 1, + \infty)$ $x in (-1,0] uu (+1,+oo)$

Let,

$y = \sqrt{\frac{13 x}{x^{2} - 1}}$ $y=sqrt((13x)/(x^2-1))$

When $x = - 1$ $x=-1$ , $\Rightarrow$ $=>$ , $y = + \infty$ $y=+oo$

When $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $y = 0$ $y=0$

The range is $y \in [0, + \infty)$ $y in [0,+oo)$

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How do you find the domain and range of $\sqrt{\frac{13 x}{(x^{2}) - 1}}$ $sqrt((13x)/((x^2)-1)$ ?

1 Answer

Explanation:

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How do you find the domain and range of √13x(x2)−1sqrt((13x)/((x^2)-1)?

1 Answer

Explanation:

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How do you find the domain and range of $\sqrt{\frac{13 x}{(x^{2}) - 1}}$ $sqrt((13x)/((x^2)-1)$ ?