Question 937709
{{{L(u) = sqrt(3u^2 )+4}}}

since we have a product {{{3u^2 }}} under radical sign,note that we cannot have a negative value under the square root sign or we will end up with a complex number, but {{{u^2 }}} will give as positive value for all values of {{{u}}} whether is positive or negative number; whatever is under the root sign will be greater than or equal to {{{0}}} and we will have a solution 

for {{{u^2 =0}}}=>{{{L(u) = 4}}}, for other values of {{{u^2 }}} we will have {{{L(u) }}}

so, the domain is: 

{{{R}}}  (all real numbers)

({{{-infinity}}},{{{infinity}}})

({{{-infinity<=u<=infinity}}})