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.

  • $\begingroup$ 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... $\endgroup$
    – Micah
    Jul 21, 2021 at 4:51
  • $\begingroup$ 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. $\endgroup$
    – Roman Hric
    Jul 21, 2021 at 7:38

1 Answer 1


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).

  • $\begingroup$ 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? $\endgroup$
    – Dan
    Jul 21, 2021 at 5:45
  • $\begingroup$ 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. $\endgroup$ Jul 21, 2021 at 5:51
  • $\begingroup$ Oh, but I'll need to bound the previous terms too. Let me be more careful here. $\endgroup$ Jul 21, 2021 at 5:54
  • $\begingroup$ @Dan I've updated the proof of $e^t\ge1+t$. I'm sure you can find even more elsewhere, but this should suffice. $\endgroup$ Jul 21, 2021 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.