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

2 Answers
Nov 8, 2017

Answer:

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]}

Nov 8, 2017

Answer:

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]}