Question 325098
{{{g(3)=3/(2+3)=3/5}}} so {{{g(3)=3/5}}}



{{{g(3^(-1))=(3^(-1))/(2+3^(-1))=(1/3)/(2+1/3)=(1/3)/(6/3+1/3)=(1/3)/(7/3)=(1/3)(3/7)=3/21=1/7}}} which means that {{{g(3^(-1))=1/7}}}



So {{{g(3)+g(3^(-1))=3/5+1/7=21/35+5/35=(21+5)/35=26/35}}} which simply means that {{{g(3)+g(3^(-1))=26/35}}} where {{{g(x)=x/(2+x)}}}