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

##### 1 Answer

#### Answer:

The domain is

#### Explanation:

First note that if

For any

So the domain is

To find the domain, let

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

Square both sides:

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

Let

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

Subtract

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

Multiply through by

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

This is in the form

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

So for our quadratic in

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

Hence:

#3-y^2 <= -6#

Hence

Hence

If

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

Hence

Hence we can deduce that the range of

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