What is the value of $f(1999)$?

Question

Let be a function $f \colon \Bbb R\to \Bbb R$ given from:

$$f(x)=x^3+\sqrt{x^6+1}+\frac{1}{x^3-\sqrt{x^6+1}}$$

To me, it immediately seemed strange that it was necessary to calculate $f(1999)$. With a calculator everything would come easy. Since this question concerns high school students I have done the rationalization of $\frac{1}{x^3-\sqrt{x^6+1}}$ and I have found something very interesting. In fact

$$\frac{1}{x^3-\sqrt{x^6+1}}=\frac{1}{x^3-\sqrt{x^6+1}}\cdot\frac{x^3+\sqrt{x^6+1}}{x^3+\sqrt{x^6+1}}=\frac{x^3+\sqrt{x^6+1}}{-1}$$

because $(x^3-\sqrt{x^6+1})(x^3+\sqrt{x^6+1})=-1$

Thus $$f(x)=x^3+\sqrt{x^6+1}-x^3-\sqrt{x^6+1}=0$$ and $f(1999)=0$ because $f(x)=0, \forall x\in\Bbb R$. Is it correct?

Your use of "i.e." at the end is incorrect ("i.e." means "that is"; and $f(1999)=0$ is not the same, nor implies, $f(x)=0$ for all $x\in\mathbb{R}$); that should be "because", not "i.e.". Otherwise, yes, this looks correct. — Arturo Magidin, Dec 05 '22 at 19:31
https://www.wolframalpha.com/input/?i=x%5E3%2Bsqrt%28x%5E6%2B1%29%2B1%2F%28x%5E3-sqrt%28x%5E6%2B1%29%29%2C+x%3D1999 — Luis Alexandher, Dec 05 '22 at 19:37
@Sebastiano in the wolfram application in the upper right part there is a share button, the link is created there — Luis Alexandher, Dec 05 '22 at 19:44
@Sebastiano in the mobile app, just to the right of the "WolframAlpha" text — Luis Alexandher, Dec 05 '22 at 19:51
Does this answer your question? transforming a specific equation — Anne Bauval, Dec 05 '22 at 19:56
@AnneBauval The processes look the same but I was computing a function. — Sebastiano, Dec 05 '22 at 20:02
So is the duplicate (they were wrong to call it an "equation"). — Anne Bauval, Dec 05 '22 at 20:05
I added 1 character to the answer less than 5 minutes ago and it says the answer edited. Interesting. — lone student, Dec 05 '22 at 20:14
This is also not a question about mathematics education, I must suggest you avoid misusing tags again. — FShrike, Dec 05 '22 at 20:25
Sure looks god to me. My first thought was the opposite of yours, that to add them I had to put them over a common divisor so $x^3 + \sqrt{x^6+1}=\frac {(x^3 + \sqrt{x^6+1})(x^3 - \sqrt{x^6+1})}{x^3 - \sqrt{x^6+1}}$ and we know $(a+b)(a-b) = a^2 - b^2$ so we get $\frac {(x^6) - (x^6+1)}{x^3 - \sqrt{x^6+1}}$ and... well, it just falls out from there.... Oh... now I see why you did it your way. You wanted to deradicalize the denominator first. Well, that works too. (key thing is the one term is the radical conjugate of the denominator of the other and so....) — fleablood, Dec 06 '22 at 02:02

lone student · Accepted Answer · 2022-12-05T21:33:32.320

4

Your work seems correct to me. (this is stated in the comments that your work is correct)

The problem statement is essentially equivalent to:

Let $f:\mathbb R\rightarrow \mathbb R$, $$f(x):=A(x)+B(x)+\frac {1}{A(x)-B(x)}$$ where, $A^2(x)-B^2(x)=-1$. Find the value of $f(1999).$

$$ \begin{align}f(x):&=A(x)+B(x)+\frac{A(x)+B(x)}{A^2(x)-B^2(x)}\\ &=A(x)+B(x)-\left(A(x)+B(x)\right)\\ &=0~.\end{align} $$

Thus $f(x)\equiv 0$, $\forall x\in\mathbb R$, since $\operatorname{dom}f=\mathbb R.$

What we do is based on using conjugate.

Note that, when you add fractions, the conjugate will appear by itself. Indeed,

$$ \begin{align}f(x):&=A(x)+B(x)+\frac {1}{A(x)-B(x)}\\ &=\frac {\left(A(x)-B(x)\right)\cdot\left(A(x)+B(x)\right)+1}{A(x)-B(x)}\\ &=\frac{\overbrace{A^2(x)-B^2(x)}^{\color {#c00}{-1}}+1}{A(x)-B(x)}\\ &=0~. \end{align} $$

Because, $A^2(x)-B^2(x)=-1$ implies that, $A(x)±B(x)≠0$.

Finally, I want to note that this type of function is called a constant function in mathematics.

edited Dec 05 '22 at 21:33

answered Dec 05 '22 at 19:58

lone student

14,709

2

Wei hola'. Thank you very much. – Sebastiano Dec 05 '22 at 20:00
1

Provided there are no instances in which $A(x)+B(x)=0$ while $A(x)-B(x)\neq 0$ (as is the case in the function at hand). – Arturo Magidin Dec 05 '22 at 20:55
@ArturoMagidin Since $A^2(x)-B^2(x)=-1$, therefore we don't need these restrictions. – lone student Dec 05 '22 at 20:58
2

True.... ${}{}{}$ – Arturo Magidin Dec 05 '22 at 21:06

What is the value of $f(1999)$?

1 Answers1