I wanted to find a function $f(x)$ that has the following properties:

  • extreme values at $x = -1$ and $x = 0$ and $x = 1$

  • $x$-axis is asymptote

No other asymptotes or extrema.

I found the following two by integrating and trial and error:

$$f_1(x) = x^2 e^{-x^2}$$

$$f_2(x) = 2 \ln (x^2 + 1) + \ln (x^4 - x^2 + 1)$$

However, those are not rational functions. How can I find one that is actually a rational function?


If the $x$-axis has to be an asymptote, you need a denominator of higher degree than the numerator. In order to get the extreme values where you want them, suggest a rational function with sufficient parameters, e.g. suggest a function of the form: $$f(x) = \frac{ax^2+bx+c}{x^4+1}$$ Note that taking $x^4+1$ instead of $x^3+1$ (which is of sufficiently high degree as well) ensures no vertical asymptote is added.

Finding $f'(x)$ and requiring $f'(-1)=f'(0)=f'(1)=0$ yields $b=c=0$ suggesting that any function with non-zero parameter $a$ of the following form will do: $$f(x) = \frac{ax^2}{x^4+1}$$ You can verify that indeed, taking e.g. $a=1$, this function satisfies your requirements.



You must have:

$f'(-1) = f'(1) = f'(0) = 0$

So the numerator of the derivative of your function must have those 3 zeros.

$\lim_{x\to\infty}f(x) = 0$,

suggesting the denominator of the rational function must have higher degree than the numerator.


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