What is the domain and range of the function #f(x) = sqrt(x^2+9/x^2+3)# ?

1 Answer
Nov 8, 2016

Answer:

The domain is #(-oo, 0) uu (0, oo)# and the range #[3, oo)#

Explanation:

#f(x) = sqrt(x^2+9/x^2+3)#

First note that if #x=0# then the denominator of #9/x^2# is zero, resulting in an undefined value. So #x=0# is not part of the domain.

For any #x != 0#, we have #x^2 > 0#, #9/x^2 > 0# and #3 > 0#, hence the radicand #x^2+9/x^2+3 > 0#, resulting in a well defined Real square root.

So the domain is #RR "\" { 0 } = (-oo, 0) uu (0, oo)#

To find the domain, let #y = f(x)# and solve for #x#, then analyse what values of #y# give solutions:

#y = sqrt(x^2+9/x^2+3)#

Square both sides:

#y^2 = x^2+9/x^2+3#

Let #t = x^2# (noting that we require #t >= 0#

#y^2 = t+9/t+3#

Subtract #y^2# from both ends to get:

#0 = t+9/t+(3-y^2)#

Multiply through by #t# and transpose to get:

#t^2+(3-y^2)t+9 = 0#

This is in the form #at^2+bt+c = 0# with #a=1#, #b=(3-y^2)# and #c=9#. It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (3-y^2)^2-4(1)(9) = (3-y^2)^2-6^2#

So for our quadratic in #t# to have Real solutions (let alone positive ones), we require:

#(3-y^2)^2 >= 6^2#

Hence:

#3-y^2 <= -6#

Hence #y^2 >= 9#

Hence #y >= 3# (since we require #y > 0#)

If #y = 3# then our quadratic in #t# becomes:

#0 = t^2-6t+9 = (t-3)^2#

Hence #x^2 = t = 3# and #x = +-sqrt(3)#

Hence we can deduce that the range of #f(x)# is #[3, oo)#

graph{(y-sqrt(x^2+9/x^2+3))(y-3) = 0 [-10.12, 9.88, -1.56, 8.44]}