#g'(x)=3x^2+1>0# , #AA##x##in##RR# so #g# is increasing in #RR# and so is at #x_0=0#
Another approach,
#g'(x)=3x^2+1# #<=>#
#(g(x))'=(x^3+x)'# #<=>#
#g#, #x^3+x# are continuous in #RR# and they have equal derivatives, therefore there is #c##in##RR# with
#g(x)=x^3+x+c#,
#c##in##RR#
Supposed #x_1#,#x_2##in##RR# with #x_1<##x_2# #(1)#
#x_1<##x_2# #=># #x_1^3<##x_2^3# #=># #x_1^3+c<##x_2^3+c# #(2)#
From #(1)+(2)#
#x_1^3+x_1+c<##x_2^3+x_2+c# #<=>#
#g(x_1)<##g(x_2)# #-># #g# increasing in #RR# and so at #x_0=0##in##RR#