Question 1204249
{{{f(x) = 1/(1 + 2/(1 + 3/x))}}}.....{{{x}}} is in denominator, so {{{x<>0}}}


{{{f(x) = 1/(1 + 2/((x + 3)/x))}}}


{{{f(x) = 1/(1 + 1/(1 + 2x/(x + 3)))}}}.....{{{x+3}}} is in denominator, so {{{x<>-3}}}


{{{f(x) = 1/((3(x + 1))/(x + 3))}}}


{{{f(x) = (x + 3)/(3 (x + 1))}}}....{{{x+1}}} is in denominator, so {{{x<>-1}}}


domain:

{ {{{x}}} element {{{R }}}: {{{x<>-3}}} and {{{x<>-1}}} and {{{x<>0}}} }


interval:

({{{-infinity}}},{{{-3}}}) U ({{{-3}}},{{{-1}}}) U ({{{-1}}},{{{0}}}) U ({{{0}}},{{{infinity}}})


the sum of those three numbers is:

{{{-3+(-1)+0=-4}}}