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How do we confirm or disprove that? And is there a name for this kind of function? $$f(x)=c(x-a)+\frac{d}{x-a}+b$$ If we restrict that $x-a>0$ and $c,d>0$, an observation is that the minimum $2\sqrt{cd}+b$ is reached when $$x=\frac{d}{\sqrt{cd}}+a.$$ This can be confirmed by differentiation. However, noticing that's also exactly when $$c(x-a)=\frac{d}{x-a},$$ I'd like to ask if there's a simpler explanation why this function reaches its minimum when two of its components are equal?

SomeShyGuy
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    Yes, it's a hyperbola, and how you confirm it depends on what you would take as proof that something is a hyperbola. How would you define hyperbola? Also, its$\ne$it's. – Gerry Myerson Mar 20 '13 at 00:04
  • @GerryMyerson: Any way is okay. As long as the proof is simple. Don't even need a serious proof, just some observation that is convincing will do. – SomeShyGuy Mar 20 '13 at 00:05

2 Answers2

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Note that $$y = x + \dfrac1x \implies xy = x^2 + 1 \implies x \underbrace{(y-x)}_{z} = 1$$ Hence, we have $xz = 1$, where $z=y-x$. This is the (rectangular) hyperbola in the $XZ$ plane with the lines $x=0$ i.e. $y$ axis (or $z$ axis) and the line $z=0$ i.e. the line $y=x$ as asymptotes.

  • Thank you. What you've shown is that the graph of this function is a linear transformation of a hyperbola. Do hyperbola have the property that they remain hyperbola under linear transformation? – SomeShyGuy Mar 20 '13 at 00:16
  • @SomeShyGuy All this means is it is a hyperbola if you look at it from appropriate coordinates/axes. –  Mar 20 '13 at 00:25
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The graph of this function is a true hyperbola, but let us say that a function $f(x)$ is "hyperbol-ish" if

  • There is a line $L: y=ax+b$ such that as $x\to \infty$ or $\to -\infty$, the function $f(x)$ is asymptotic to $L$, i.e., $$\lim_{x\to \pm \infty} f(x) - (ax+b)=0,$$ and
  • There is a vertical line $L':x=c$ such that as $x\to c^+$ or as $x\to c^-$, the function $f(x)$ is asymptotic to $L'$, i.e., $$\lim_{x\to c^\pm} f(x)=\pm\infty.$$

Let us now show that $f(x)=x+\frac{1}{x}$ is hyperbol-ish. Let $L: y=x$. Then: $$\lim_{x\to \pm\infty} f(x) - x=\lim_{x\to \pm\infty} x+\frac{1}{x}-x=\lim_{x\to \pm\infty} \frac{1}{x}=0.$$ Now, let $L': x=0$. Then: $$\lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} x+\frac{1}{x}=\infty$$ and $$\lim_{x\to 0^-} f(x)=\lim_{x\to 0^-} x+\frac{1}{x}=-\infty,$$ as desired.

Álvaro Lozano-Robledo
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  • Then why is it a true hyperbola? I think as Marvis had shown, it is indeed an affine transformation of a rectangular hyperbola, but why it is necessarily a hyperbola then? And thank you for typing so much and explaining patiently. – SomeShyGuy Mar 20 '13 at 00:33
  • I thing what would be a convincing proof is a linear transformation from a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ that only involves rotation and evenly scaling. – SomeShyGuy Mar 20 '13 at 00:36
  • @SomeShyGuy it is a true hyperbola because as Marvis points out, the graph satisfies the equation of a hyperbola. – Álvaro Lozano-Robledo Mar 20 '13 at 00:41
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    I don't think so. I just found the answer myself. After a rotation transformation $$\begin{bmatrix}\hat{x}\\hat{y}\end{bmatrix}=\begin{bmatrix}\cos{t}&-\sin{t} \ \sin{t}&\cos{t}\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix},$$ the original curve $$y=x+\frac{1}{x}$$ becomes $$\frac{\hat{x}^2}{2\sqrt{2}+2}-\frac{\hat{y}^2}{2\sqrt{2}-2}=1,$$ which satisfies the standard form of hyperbola. – SomeShyGuy Mar 20 '13 at 01:04
  • To emphasize it again, I don't think it's trivial to say that after a linear transformation, the "hyperbol-ish" shape will still be congruent to a hyperbola, especially when this transformation involves affine transformation. – SomeShyGuy Mar 20 '13 at 01:10
  • By the way, in the above rotation transformation, $$\cos{t}=\frac{1}{\sqrt{4+2\sqrt{2}}}.$$ – SomeShyGuy Mar 20 '13 at 01:13
  • I never said that a "hyperbol-ish" function is a hyperbola. You seemed to have an idea of a definition of hyperbola, but when queried by Gerry Myerson, you said that any definition was ok. Well, I just defined the asymptotic behavior of a hyperbola separately, and proved that the original function satisfies said property. That's all. But an equation of the form $xy=1$ is definitely a hyperbola, which is more generally defined as a special kind of conic section, and not only narrowly as a curve of the form $ax^2-by^2=1$. – Álvaro Lozano-Robledo Mar 20 '13 at 01:16
  • If what you wanted was an affine transformation from the graph to a hyperbola of the form $ax^2-by^2=1$, then you should have said so. – Álvaro Lozano-Robledo Mar 20 '13 at 01:17
  • Sorry I made things so unclear, but when I said "any definition," I really meant "any definition that is equivalent to the standard definition (by Wikipedia)," that is, "congruent to a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$." – SomeShyGuy Mar 20 '13 at 01:21
  • And $xy=1$ is indeed a hyperbola because it is also congruent to a standard hyperbola $$\frac{\hat{x}^2}{2}-\frac{\hat{y}^2}{2}=1$$ under the rotation $$\begin{bmatrix}\hat{x}\\hat{y}\end{bmatrix}=\begin{bmatrix}\cos{\pi /4}&-\sin{\pi /4} \ \sin{\pi /4}&\cos{\pi /4}\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix},$$ – SomeShyGuy Mar 20 '13 at 01:24