# Find the minimum value of $xe^x$ without calculus

Recently I have been interested in finding minimum value of various functions without calculus, as a challenge.

How can we find the minimum value of $$xe^x$$ without calculus? Assume the limit or series definition of $$e$$.

If you use Calculus, you would differentiate $$f(x) = xe^x$$ and find that the only stationary point is $$x = -1$$. Since $$f(-1) = e^{-1}, \ \ \ \lim_{x\to -\infty} f(x) = 0, \ \ \ \lim_{x\to +\infty} f(x) = +\infty,$$ we conclude that $$e^{-1}$$ is the minimum of $$xe^x$$.

However, I am not able to find the stationary point of this function without calculus.

• Maybe show that $\displaystyle xe^x+\frac{1}{e}$ is nonnegative by writing it as $\displaystyle \sum_{n=0}^\infty \frac{x^{n+1}+(-1)^n}{n!}$? But this seems to lead to a bunch of messy case analysis and I haven't gotten it to actually work... Jul 21, 2021 at 4:51
• I get the spirit of the question but, paradoxically, if you assume the limit or series definition of $e$ then you're already using calculus. Jul 21, 2021 at 7:38

This answer tries to find the minimizer of $$xe^x$$.

Obviously the minimum is attained when $$x<0$$. Substituting $$u=-x$$, it suffices to minimize $$-ue^{-u}$$ when $$u>0$$, or equivalently maximize $$ue^{-u}$$.

Hopefully you know $$\ln$$ is monotonic (which should follow by monotonicity of $$e^x$$ if you define $$\ln x$$ as the inverse of $$e^x$$), so it suffices to maximize $$\ln(ue^{-u})=\ln u-u$$ for $$u>0$$.

We know (see below) $$e^t\ge1+t$$ for all real $$t$$ (equality at $$t=0$$), so substituting $$t=\ln u$$, we get $$u\ge 1+\ln u$$, i.e. $$\ln u-u\le-1$$ with equality at $$u=1$$. This immediately implies the minimum of $$xe^{-x}$$ is $$-e^{-1}$$, at $$x=-1$$.

EDIT: Here's a proof of $$e^t\ge 1+t$$ without calculus, expanded from this answer. First, note Bernoulli's inequality $$(1+x)^n\ge 1+nx$$ (where $$n>0,n\in\mathbb Z,x\ge-1$$) can be proven without calculus (only induction). Now for any fixed $$x$$, $$\lim_{n\to\infty}\frac xn=0$$, so when $$n$$ is sufficiently large, $$\frac xn>-1$$. Hence for all but finitely many $$n$$, $$(1+\frac xn)^n\ge1+x$$. Taking the limit $$n\to\infty$$, we now have $$e^x\ge1+x$$ as limit preserves non-strict inequalities (and ignores finitely many starting terms).

• Without calculus, how do we know that the tail of the series expansion of $e^t$ (i.e. the part without $1+t$), is non-negative?
– Dan
Jul 21, 2021 at 5:45
• It is obvious when $t\ge0$. When $t<0$, by Archimedean property there is an integer $N>-t$. So when $n\ge N, |t^n/n!|$ will be decreasing. Now the tail becomes an alternating series with strictly decreasing magnitude. Jul 21, 2021 at 5:51
• Oh, but I'll need to bound the previous terms too. Let me be more careful here. Jul 21, 2021 at 5:54
• @Dan I've updated the proof of $e^t\ge1+t$. I'm sure you can find even more elsewhere, but this should suffice. Jul 21, 2021 at 6:05