Because $f(x) = x + \frac{6}{x}$ is similar to $g(x) = \frac{6}{x}$ when $x$ is small and tends to $h(x) = x$ when $x$ tends to infinity, we can graph $f$ as follows:

$f$ is in green and its asymptote $h(x) = x$ is the blue line. Given that graph, it is clear that, although (and by definition), one given input of $f$ will have at most one output, the inverse isn't true. In other words, one output of $f$ can have two inputs:

Here $L = 10$ (in red) intersects $f(x)$ at $x = 5 - \sqrt{19}$ and $x = 5 + \sqrt{19}$ .

Hence, $x + \frac{6}{x} = y + \frac{6}{y}$ does not imply $x = y$ .

Let $x+\dfrac {6}{x}=y+\dfrac {6}{y}=k$ .

Then $x=\dfrac {k\pm \sqrt {k^2-24}}{2},y=\dfrac {k\pm \sqrt {k^2-24}}{2}$

So, there are two possibilities :

(i) $x=y=\dfrac {k+\sqrt {k^2-24}}{2}$

or $x=y=\dfrac {k-\sqrt {k^2-24}}{2}$

(ii) $x=\dfrac {k+\sqrt {k^2-24}}{2}$ ,

$y=\dfrac {k-\sqrt {k^2-24}}{2}$ ,

$x\neq y$

Or, $x=\dfrac {k-\sqrt {k^2-24}}{2}$ ,

$y=\dfrac {k+\sqrt {k^2-24}}{2}$ ,

$x\neq y$ .

$6+\frac{6}{6}=1+\frac{6}{1}$ but $6\not=1$ .

The first equation can be rearranged to $xy^2 - (x^2 + 6)y + 6x = 0$ , which by the quadratic equation leads to $y = \frac{(x^2 + 6) \pm (x^2 - 6)}{2x}$ , which is $y = x$ or $y = \frac{6}{x}$ .

Using the second solution, some counter-examples (in the form of $(x, y)$ ) to show that $x$ and $y$ are not necessarily equal are $(1, 6)$ , $(2, 3)$ , $(3, 2)$ , $(6, 1)$ , etc.