How do you find the domain and range of `g(x) = sqrt x / (2x^2+x+1)`?

`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]`