3
$\begingroup$

I am trying to solve the following problem: enter image description here

I can visualize how it looks like "approximately", it's essentially something like $\sqrt{x - 2} - 2$, with the difference that it increases faster. But based on the other parts of this problem it seems a specific shape such as a parabola, hyperbola, circle, or ... should be found to describe the graph of each function. I tried simplifying the function, with no results. Is it even possible to graph this function without taking derivatives?

I also proved that the function can be written like this: $y^2(y^2-4x) = -16$ where $y \ge2$.

$\endgroup$
5
  • 1
    $\begingroup$ It might be useful to write $\sqrt{x+2}-\sqrt{x-2}=\frac{4}{\sqrt{x+2}+\sqrt{x-2}}$ for $x\ge 2$ $\endgroup$
    – Mark Viola
    Commented Sep 14, 2018 at 16:34
  • $\begingroup$ If you only look at "small" $x$, then you visualization works. But as soon as you get aways from that, your visualization breaks down, exactly because the function 'grows faster'. $\endgroup$
    – Ingix
    Commented Sep 14, 2018 at 16:35
  • $\begingroup$ For what values of $x$ does $y$ vanish (and cross the axis)? How many such crossings are there? What is the behavior for large positive $x$? What happens near $x=2$? $\endgroup$ Commented Sep 14, 2018 at 16:35
  • $\begingroup$ @DavidG.Stork The domain is $x\ge 2$. $\endgroup$
    – Mark Viola
    Commented Sep 14, 2018 at 16:36
  • $\begingroup$ What I mean by a specific graph is this: in another part of the question we were supposed to graph $y^2 = |x^2 - x|$, the answer was a circle with two parts of a hyperbola attached to its both sides. But for this problem, I couldn't do anything better than just a sketch. $\endgroup$ Commented Sep 14, 2018 at 16:38

3 Answers 3

6
$\begingroup$

Note that the domain of $f$ is $[2,\infty)$. Furthermore, we can write

$$\begin{align} f(x)&=\sqrt{x+2}-\sqrt{x-2}\\\\ &=\frac4{\sqrt{x+2}+\sqrt{x-2}} \end{align}$$

Now, note that $f$ is clearly decreases monotonically from its maximum value $f(2)=2$ and approaches $0$ as $x\to \infty$. In fact, for large $x$, $f(x)$ behaves asymptotically like $\frac{2}{\sqrt x}$.

$\endgroup$
4
  • $\begingroup$ True, so you're saying that there is no specific shape similar to the graph of $f$? And I see why it approaches $0$ because when $x$ is large enough then $+2$ or $-2$ doesn't change the value of the square root much so it ends up acting like $\sqrt{x} - \sqrt{x} = 0$ $\endgroup$ Commented Sep 14, 2018 at 16:42
  • $\begingroup$ $f$ is convex and behaves asymptotically as $\frac2{\sqrt x}$ as $x\to \infty$. $\endgroup$
    – Mark Viola
    Commented Sep 14, 2018 at 16:44
  • $\begingroup$ Thanks, I was just confused because it seems in the rest of the questions the answers were a partial parabola, hyperbola, ... . Anyway thanks! $\endgroup$ Commented Sep 14, 2018 at 16:47
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Commented Sep 14, 2018 at 16:47
5
$\begingroup$

The graph starts at the point $(2,2)$ and stays above the $x$-axis, decreasing continuously to $0$ as $x$ goes to $\infty.$

You can see all of that from the fact that $$ \begin{align} f(x)=\frac4{\sqrt{x+2}+\sqrt{x-2}} \end{align}$$ Very smooth and nice curve, looks like the tail of $2/{\sqrt x}$

$\endgroup$
1
  • $\begingroup$ This appears to be a duplicate of my answer. What does this add? $\endgroup$
    – Mark Viola
    Commented Sep 14, 2018 at 23:06
3
$\begingroup$

The curves of $\sqrt{x\pm2}$ are identical half-parabolas with an horizontal axis.

The function to be plotted is the vertical difference between them.

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .