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

Nov 8, 2016

The domain is $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$ and the range $\left[3 , \infty\right)$

Explanation:

$f \left(x\right) = \sqrt{{x}^{2} + \frac{9}{x} ^ 2 + 3}$

First note that if $x = 0$ then the denominator of $\frac{9}{x} ^ 2$ is zero, resulting in an undefined value. So $x = 0$ is not part of the domain.

For any $x \ne 0$, we have ${x}^{2} > 0$, $\frac{9}{x} ^ 2 > 0$ and $3 > 0$, hence the radicand ${x}^{2} + \frac{9}{x} ^ 2 + 3 > 0$, resulting in a well defined Real square root.

So the domain is $\mathbb{R} \text{\} \left\{0\right\} = \left(- \infty , 0\right) \cup \left(0 , \infty\right)$

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

$y = \sqrt{{x}^{2} + \frac{9}{x} ^ 2 + 3}$

Square both sides:

${y}^{2} = {x}^{2} + \frac{9}{x} ^ 2 + 3$

Let $t = {x}^{2}$ (noting that we require $t \ge 0$

${y}^{2} = t + \frac{9}{t} + 3$

Subtract ${y}^{2}$ from both ends to get:

$0 = t + \frac{9}{t} + \left(3 - {y}^{2}\right)$

Multiply through by $t$ and transpose to get:

${t}^{2} + \left(3 - {y}^{2}\right) t + 9 = 0$

This is in the form $a {t}^{2} + b t + c = 0$ with $a = 1$, $b = \left(3 - {y}^{2}\right)$ and $c = 9$. It has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(3 - {y}^{2}\right)}^{2} - 4 \left(1\right) \left(9\right) = {\left(3 - {y}^{2}\right)}^{2} - {6}^{2}$

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

${\left(3 - {y}^{2}\right)}^{2} \ge {6}^{2}$

Hence:

$3 - {y}^{2} \le - 6$

Hence ${y}^{2} \ge 9$

Hence $y \ge 3$ (since we require $y > 0$)

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

$0 = {t}^{2} - 6 t + 9 = {\left(t - 3\right)}^{2}$

Hence ${x}^{2} = t = 3$ and $x = \pm \sqrt{3}$

Hence we can deduce that the range of $f \left(x\right)$ is $\left[3 , \infty\right)$

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