Question 923866

{{{f(x) = sqrt(-1/x) }}}

since in {{{(-1/x)}}}, denominator is {{{x}}} and we know denominator cannot be equal to {{{0}}}; so, {{{x<>0}}}

if we take {{{x>0}}}, then  {{{(-1/x)}}} will be negative number, and square root of  negative number gives us complex root; so, we will exclude this option

if we take {{{x<0}}}, then  {{{(-1/x)}}} will be positive number, and square root of  positive number gives us real root; so, we will go with this option


domain:

{ {{{x}}} element {{{R}}} : {{{x<0}}} }  (all negative real numbers)
(assuming a function from reals to reals)