MHB How to Graph \( y = \sqrt{x+2} - \sqrt{x-2} \) Without Calculus?

[Dethrone]

How do I graph this without calculus?

$$y=\sqrt{x+2}-\sqrt{x-2}$$

Any hints?

 

[Nono713]

How do I graph this without calculus?

$$y=\sqrt{x+2}-\sqrt{x-2}$$

Any hints?

I wasn't aware graphing a function required the use of calculus. Could you clarify what you are after?

 

[Dethrone]

Graphing the function on paper, without using derivatives and finding critical points and all that. :D

 

[Jameson]

Hmm, if limits are allowed then I think multiplying by the conjugate divided by itself will be useful when taking the $\displaystyle \lim_{x \rightarrow \infty} f(x)$ and $\displaystyle \lim_{x \rightarrow -\infty}f(x)$. I would start with that and then plot some points in between to try to get a basic shape of the graph. Don't know if that's what the question is looking for exactly, but it's a start. :)

 

[Dethrone]

Good suggestion! but the question is in the "absolute value" section, so I think maybe that is involved somewhere...at this point in the book, limits are not introduced yet. (Doh)

 

[Nono713]

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.

 

[I like Serena]

How do I graph this without calculus?

$$y=\sqrt{x+2}-\sqrt{x-2}$$

Any hints?

Hey Rido! ;)

Start with the graph for $y=\sqrt x$.
That one should be familiar.

Shift $y=\sqrt x$ to the left by 2 units to get the graph of $y=\sqrt{x+2}$.
Shift $y=\sqrt x$ to the right by 2 units to get the graph of $y=\sqrt{x-2}$.
Subtract them to get the graph for $y=\sqrt{x+2}-\sqrt{x-2}$.

 

[Ackbach]

I would also add that finding the domain first is probably a good idea. In this case, the domain is $[2,\infty)$. No need to try plotting any points outside this interval.

 

[mathbalarka]

Are you trying to find the nature of the graph? If so, here are a few hints for you :

• $y = \sqrt{x+2} - \sqrt{x-2}$ is not real if $x < 2$, so it has domain $[2, \infty]$ on $\Bbb R$. At $x = 2$, $y = 2$.
• Furthermore, since both of the terms $\sqrt{x+2}$ and $\sqrt{x-2}$ asymptotically grows like $\sqrt{x}$, $y = y(x)$ is expected to tend towards $0$ monotonically with the growth of $x$.
• This function is globally convex in it's domain. [Hint : use Jensen's inequality]