Question

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

Answer 1

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]}