Question

find the domain f(x)=1/root(16-x^2)

Answer 1

$f(x) = \frac{1}{ \sqrt{16 - {x}^{2} } }$
function is defined when ,
denominator doesn't equal to zero .
and square root always positive .
so,
$16 - {x}^{2} > 0$
( 4 - x ) ( 4 + x ) > 0
after using number line
we get, - 4 < x < 4
hence, domain ∈ ( -4, 4)