Rido12 said:
Graphing the function on paper, without using derivatives and finding critical points and all that. :D
Well, I don't know how accurate you need it to be, but you can try this. The function is smooth for $x \geq 2$, being a sum of two smooth functions (I'm using the word "smooth" in the non-technical sense of "well-behaved" here) hence to graph it, it is probably sufficient to calculate a few points on the curve and interpolate smoothly between them. If you want to, you can show that it is strictly decreasing first to further justify interpolating nearby points.
Good choices are $x$ values which make the square root terms easy to compute. Obviously, $x + 2$ and $x - 2$ cannot both be perfect squares for $x \ne 2$, so that's not helpful, but they don't have to be integers, rationals will do too. That is, you want to find $x$ values such that:
$$x + 2 = \frac{a^2}{b^2}$$
$$x - 2 = \frac{c^2}{d^2}$$
for integers $a, b, c, d > 0$. Note that this implies:
$$x = \frac{1}{2} \left ( \frac{a^2}{b^2} + \frac{c^2}{d^2} \right ) = \frac{a^2 d^2 + c^2 b^2}{2 b^2 d^2}$$
And, of course, the sufficiency criterion:
$$\frac{a^2}{b^2} - \frac{c^2}{d^2} = 4 ~ ~ ~ \iff ~ ~ ~ a^2 d^2 - b^2 c^2 = 4 b^2 d^2$$
We don't need all of the solutions, just sufficiently many to plot the curve. So we will assert $b = d$ in order to easily find solutions without losing too many of them, and we see that this simplifies the criterion to:
$$a^2 - c^2 = (2b)^2 ~ ~ ~ \iff ~ ~ ~ (a - c)(a + c) = (2b)^2$$
That second expression lends itself well to finding $(a, b, c)$ tuples by hand, because all we need is to select any $2b$, factor its square into a product of any two even numbers $p$, $q$ and derive $a$ and $c$ from that. In other words, let $(2b)^2 = pq$ with $p, q$ even, then:
$$a = \frac{p + q}{2} ~ ~ ~ \text{and} ~ ~ ~ c = \frac{p - q}{2}$$
For instance, let $2b = 16$, so $(2b)^2 = 256 = 32 \times 8$, and we get:
$$a = \frac{32 + 8}{2} = 20, ~ ~ ~\text{and} ~ ~ ~ c = \frac{32 - 8}{2} = 12$$
Which gives us $a = 20$, $b = 8$, $c = 12$, $d = 8$, that is:
$$x = \frac{a^2}{b^2} - 2 = \frac{17}{4} = 4.25$$
And, indeed:
$$\sqrt{17/4 + 2} - \sqrt{17/4 - 2} = \sqrt{\frac{20^2}{8^2}} - \sqrt{\frac{12^2}{8^2}} = \frac{20}{8} - \frac{12}{8} = 1$$
So $(17/4, 1)$ is on the curve. Thus we've computed a point on the curve without leaving the rationals and having to compute a square root! Note we can, and should, refine this method to be more efficient, and give us more control over what $x$ is going to be close to (after all, we want to plot the curve, so we want to plot points on the curve at regular intervals, not all clumped together). Observe that we can select $b = p^k$ for prime $p$ and any $k \geq 1$, which gives us the generic factorization:
$$(2b)^2 = 4 p^{2k} = (2 p^{r_1})(2 p^{r_2}) ~ ~ \text{with} ~ r_1 + r_2 = 2k$$
Which satisfies the criterion. We then get $a$ (and, for completeness, $c$) equal to:
$$a = \frac{2 p^{r_1} + 2 p^{r_2}}{2} = p^{r_1} + p^{r_2} ~ ~ ~ \text{and} ~ ~ ~ c = \frac{2 p^{r_1} - 2 p^{r_2}}{2} = p^{r_1} - p^{r_2}$$
And therefore:
$$x + 2 = \frac{\left ( p^{r_1} + p^{r_2} \right )^2}{p^{2k}} = \left ( \frac{p^{r_1} + p^{r_2}}{p^k} \right )^2 ~ ~ ~ \iff ~ ~ ~ x = \left ( \frac{p^{r_1} + p^{r_2}}{p^k} \right )^2 - 2$$
This let's us find a few points on the curve with $x$ in the approximate interval $\left ( 2, p^{2(k - 1)} + \epsilon \right )$ by careful selection of $r_1$ and $r_2$ subject to $r_1 + r_2 = 2k$, that is, have them nearly equal for the lower bound, and $r_2 \gg r_1$ for the upper bound. Increasing $k$ only increases range, whereas increasing $p$ increases resolution a bit, and this should let you find a few points with $2 \leq x \leq 10$ suitable for graphing. It might be possible to optimize this further, but this looks good enough for a start.
What I've just described is a relatively fast method for locating rational points on your curve, and hence graph it without needing to perform a single square root operation (assuming you can perform simple division to retrieve $x$).
Or you could use a calculator and plot a few random points. That works, too.
