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

##### 2 Answers

#### Answer:

Domain:

Range:

#### Explanation:

Note that:

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

So

Hence

So its implicit domain is

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

#ax^2+bx+c = 0#

with

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

Its discriminant

#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

Otherwise

#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

Note that in order to have positive solutions, the signs of the coefficients

So the range of

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

#### Answer:

Domain:

Range:

#### Explanation:

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

Note that

Hence for any

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

Hence

So the domain of

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

#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

We find:

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

Hence the range of

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

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