Question

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

Answer 1

Domain: `(-oo, oo)`

Range: `[-2/3sqrt(3), 1)`

Explanation:

Note that:

`x^2+x+1 = (x+1/2)^2+3/4 > 0" "` for all real values of `x`

So `sqrt(x^2+x+1)` is well defined and non-zero for any `x in RR`

Hence `f(x)` is well defined for any `x in RR`

So its implicit domain is `RR`

To find the range, let:

`y = f(x) = x/sqrt(x^2+x+1)`

Then:

`y sqrt(x^2+x+1) = x`

Squaring both sides:

`y^2(x^2+x+1) = x^2`

So:

`(y^2-1)x^2+y^2x+y^2=0`

This is a quadratic polynomial equation in `x` of the form:

`ax^2+bx+c = 0`

with

`{ (a = y^2-1), (b=y^2), (c=y^2) :}`

Its discriminant `Delta` is given by the formula:

`Delta = b^2-4ac`

`color(white)(Delta) = (y^2)^2-4(y^2-1)y^2`

`color(white)(Delta) = y^4-4y^4+4y^2`

`color(white)(Delta) = y^2(4-3y^2)`

When `y=0` then `Delta=0` so the quadratic in `x` has one solution.

Otherwise `y^2 > 0` so in order for the quadratic in `x` to have real solutions we require:

`4-3y^2 >= 0`

That is:

`4 >= 3y^2`

So:

`y^2 <= 4/3`

and:

`abs(y) <= sqrt(4/3) = sqrt((4*3)/9) = 2/3sqrt(3)`

This is a sufficient condiction for the polynomial to have a solution, but we also require the sign of `x` to match that of `y`. This is where we run into the limitation of squaring an equation that we did earlier - which can introduce spurious solutions.

Note that in order to have positive solutions, the signs of the coefficients `a`, `b` and `c` must differ. Hence we need `y^2-1 < 0`. So `abs(y) < 1`.

So the range of `f(x)` is `[-2/3sqrt(3), 1)`

graph{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}

Answer 2

Domain: `(-oo, oo)`

Range: `[-2/3sqrt(3), 1)`

Explanation:

Here's a method using some pre-calculus and calculus...

Note that `x^2+x+1 = (x+1/2)^2+3/4 > 0` for all real values of `x`

Hence for any `x in RR` we have:

`sqrt(x^2+x+1) > 0`

Hence `f(x)` is well defined for all `x in RR`

So the domain of `f(x)` is `RR`

Note that:

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

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

So there are horizontal asymptotes `y=-1` as `x->-oo` and `y=1` as `x->oo`.

`d/(dx) x/sqrt(x^2+x+1) = d/(dx) x (x^2+x+1)^(-1/2)`

`color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x^2+x+1)^(-1/2) - x 1/2(x^2+x+1)^(-3/2)(2x+1)`

`color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x^2+x+1)^(-3/2)((x^2+x+1) - 1/2x(2x+1))`

`color(white)(d/(dx) x/sqrt(x^2+x+1)) = (x+2)/(2(x^2+x+1)^(3/2))`

So there's one turning point at `x=-2`

We find:

`f(-2) = (-2)/sqrt((-2)^2+(-2)+1) = -2/sqrt(3) = -2/3sqrt(3)`

Hence the range of `f(x)` is:

`[-2/3sqrt(3), 1)`

graph{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}