What is the domain of $f (x) = \frac{1}{\sqrt{(2 - x) (6 + x)}}$ $f (x) = 1 / sqrt((2 - x)(6 + x))$ ?

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What is the domain of $f (x) = \frac{1}{\sqrt{(2 - x) (6 + x)}}$ $f (x) = 1 / sqrt((2 - x)(6 + x))$ ?

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Answer 1 · 2015-09-17T12:43:26

$x \in (- 6, 2)$ $x in (-6,2)$

Explanation:

To be able to calculate $f (x)$ $f(x)$ , we have to avoid dividing by 0 and to calculate square root of negative numbers. So,

$(\sqrt{(2 - x) (6 + x)} \neq 0 \land (2 - x) (6 + x) \geq 0) \Leftrightarrow$ $(sqrt((2-x)(6+x))!=0 ^^ (2-x)(6+x)>=0) <=>$
$(2 - x) (6 + x) > 0 \Leftrightarrow$ $(2-x)(6+x)>0 <=>$
$(2 - x > 0 \land 6 + x > 0) \lor (2 - x < 0 \land 6 + x < 0) \Leftrightarrow$ $(2-x>0 ^^ 6+x>0) vv (2-x<0 ^^ 6+x<0) <=>$
$(x < 2 \land x > - 6) \lor (x > 2 \land x < - 6) \Leftrightarrow$ $(x<2 ^^ x> -6) vv (x>2 ^^ x< -6) <=>$
$x \in (- 6, 2) \lor x \in \emptyset \Leftrightarrow$ $x in (-6,2) vv x in O/ <=>$
$x \in (- 6, 2)$ $x in (-6,2)$

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What is the domain of $f (x) = \frac{1}{\sqrt{(2 - x) (6 + x)}}$ $f (x) = 1 / sqrt((2 - x)(6 + x))$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

What is the domain of f(x)=1√(2−x)(6+x) f (x) = 1 / sqrt((2 - x)(6 + x))?

1 Answer

Explanation:

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What is the domain of $f (x) = \frac{1}{\sqrt{(2 - x) (6 + x)}}$ $f (x) = 1 / sqrt((2 - x)(6 + x))$ ?