Consider the set of functions: \begin{equation} f_n(t) := t^n e^{(\frac{c}{t^n})}, \end{equation} where $c$ is a non-zero real constant.
I know that for $n=1$ $f_1(t)$ is convex on $(0,\infty)$ and when I graphed memebers of this family they all "look like" they're convex on the $(0,\infty])$.
My question is: is this indeed true? If not which members are convex?
On
Hint: as the function is $C^\infty$ you can check if the second derivative is always positive or not. Now, the second derivative is given by $$f_n''(x)=e^{c x^{-n}} n x^{-2-n} (c^2 n-c (-1+n) x^n+(-1+n) x^{2 n}),$$ and as the first terms are positive you only need to check if the expression in the parenthesis is always positive
Taking the second derivative, we find that $f_n$ (for $n > 0$) is convex when
$$ (n-1) t^{2n}-c (n-1)t^n +c^2 n \ge 0$$
The left side is a quadratic $Q(s)$ in $s = t^n$.
If $n > 1$, the minimum of the quadratic occurs at $s = c/2$, with $Q(c/2) = (3n+1) c^2/4 > 0$. Thus it is indeed convex.
On the other hand, if $0 < n < 1$, the left side is negative, making the function concave, for sufficiently large $t$.