Constructing a function $f$ with $f (0) = 0$, $\lim_{x \to \infty}f(x) = 0$, $f^{'} (0) = 0$, and $\lim_{x \to \infty}f^{'}(x) = 0$.

Question

I am trying to define a function $f:\mathbb{R} \to \mathbb{R}$ that has the appearance of a bell curve, but has the following properties: \begin{align*} f (0) = 0, \\ \lim_{x \to \infty}f(x) = 0, \\ f^{'} (0) = 0, \\ \lim_{x \to \infty}f^{'}(x) = 0. \end{align*}

The closest I've gotten to finding this is with $f(x) = (1 - \tanh(x))(1 - (\tanh(x) - 1)^2)$, but this fails since this would mean $f^{'} (0) = 2$. I've plotted this function in the figure below.

Attempt at defining <span class= $f$." />

score 2 · Answer 1 · answered Feb 04 '23 at 16:24

2

You can take $f(x)=\dfrac{x^2}{x^4+1}$. Since $f'(x)=-\dfrac{2x\left(x^4-1\right)}{\left(x^4+1\right)^2}$, $f'(0)=\lim_{x\to\infty}f'(x)=0$. And it is clear that $f(0)=\lim_{x\to\infty}f(x)=0$. You can see its graph below.

answered Feb 04 '23 at 16:24

José Carlos Santos

427,504

3

A similar example is to take $f(x)=x^2 e^{-x^2}$, which behaves as $x^2$ for small $x$ but vanishes rapidly at infinity. – Semiclassical Feb 04 '23 at 16:28
@Semiclassical Indeed. – José Carlos Santos Feb 04 '23 at 16:29

Constructing a function $f$ with $f (0) = 0$, $\lim_{x \to \infty}f(x) = 0$, $f^{'} (0) = 0$, and $\lim_{x \to \infty}f^{'}(x) = 0$.

1 Answers1