#g(x) = sqrtx/(2x^2+x+1)#

Numerator: #sqrtx# is defined #forall x in RR: x>=0#

Denominator: #2x^2+x+1#

Discriminant of #2x^2+x+1 = 1^2-4xx2xx1=-7#

Since the discriminant #<0 -> 2x^2+x+1 !=0# for any #x in RR#

Since the numerator is defined #forall x in RR: x>=0# and the denominator is never 0 for real #x# #g(x)# is defined #forall x in RR: x>=0#

Hence, the domain of #g(x)# is #[0,+oo)#

To find the range of #g(x)# we need to find the upper and lower bounds.

By inspection, #g(0)=0#

Since, #x>=0# we can deduce that #g(x)>=0#

#:.g_min = g(0) =0#

Now consider #lim_(x->+oo) g(x)#

#= lim_(x->+oo) sqrtx/(2x^2+x+1)#

#= lim_(x->+oo) 1/((2x^2+x+1)/x^(1/2))#

#= lim_(x->+oo) 1/((2x^(3/2)+x^(1/2)+x^(-1/2)))= 1/(oo+0) =0#

NB: To find the finite upper bound of #g(x)#, if any, we would normally use calculus. However, since this question is in the Algebra section, I will solve this graphically.

Consider the graph of #g(x)# below.

graph{ sqrtx/(2x^2+x+1) [-1.145, 3.18, -1.037, 1.126]}

We can observe from the graph that #g(x)# has a maximum value of #approx 0.371# at #x=1/3#. We can further observe that #g(x)# is declining from that point.

We have already shown that #g(x) ->0# as #x-> +oo#

Hence we can deduce that #f_max = f(1/3) approx 0.371# is the absolute maximum.

Hence, the range of #g(x)# is #[0, approx 0.371]#