I'm trying to find a smooth function $f$ on the real line with compact support, which has a unique minimum (at some $x_{0}\in\mathbb{R}$) and such that its third derivative there is negative, i.e. f'''(x_{0})<0. I played around with exponential, ("cut offs" of) trigonometric functions etc., but I couldn't find any. Any help is appreciated.
Looking for a smooth function with third derivative negative at minimum?
2
$\begingroup$
calculus
-
0By *smooth*, do you mean *infinitely differentiable* at every point (including the boundaries of support)? – 2011-05-11
2 Answers
3
Consider the function $f$ defined by $f(x)=(x^2-x^3-1)\exp(1/(x^2-1))$ if $|x|<1$ and $f(x)=0$ if $|x|\ge1$, and the point $x_0=0$.
2
How about $-10+10x^2-x^3+x^4$, then "round off the corners" where it crosses the x axis to make it smooth and compact support?
-
0I'll check that, as I was trying something similar. – 2011-05-11