0
$\begingroup$

I'm stuck in proving the function $$f(x) = \vert x + e^x\vert $$ has a minimum.

This is what I did:

$$f'(x) = (1+e^x)\text{sgn}(1+e^x)$$

But this function is never zero, because the exponential is always positive. So when I study the sign of the derivative, it is always increasing.

Yet the plot of the funciton shows a "sort of" a minimum.

Adds

Since there is an absolute value I calculated the difference quotient for $f(x) > 0$ and $f(x) < 0$ obtaining the function is continuous when $x+e^x > 0$ and when $x+e^x < 0$

Yet I also understood @Martin R. comment but how to find that point where $f(x)$ is discontinuous?

$\endgroup$
3
  • $\begingroup$ Hint: $x + e^x = 0$ for some $x \in (-1, 1)$. $\endgroup$
    – Martin R
    Dec 2, 2022 at 11:03
  • $\begingroup$ The function is not differentiable everywhere (where?) so you need a bit extra work. $\endgroup$
    – Michal Adamaszek
    Dec 2, 2022 at 11:05
  • $\begingroup$ Proving a function has a minimum is usually significantly easier than finding the minimum. Here your function is continuous, bounded below by $0$ and it tends to $+\infty$ for $x\to\pm\infty$. Therefore, for a given $x_0$ and $M>f(x_0)$, you can find a compact interval $[a,b]$ outside of which $f(x)>M$. Then on $[a,b]$, it's just the extreme value theorem. $\endgroup$
    – Jean-Claude Arbaut
    Dec 2, 2022 at 11:40

2 Answers 2

Reset to default
1
$\begingroup$

$g(x) = x + e^x$ is strictly increasing with $g(-1) < 0 < g(0)$. It follows that $g$ has a (unique) zero $x_0 \in (-1, 0)$.

Then $f(x) \ge 0 = f(x_0)$ for all $x \in \Bbb R$, so that $f$ has a minimum at $x=x_0$.

$f$ is strictly decreasing for $x < x_0$, and strictly increasing for $x > x_0$. But $f$ is not differentiable at $x_0$, which explains why $f'$ is nowhere zero.

$\endgroup$
3
  • $\begingroup$ Thank you, that was enlightening. So Can I simply conclude that $\min(f(x)) = f(x_0)$? Or shall I give more information (if possibile, with elementary reasonings) about the "value" of $x_0$? Which is for sure negative. $\endgroup$
    – Numb3rs
    Dec 2, 2022 at 11:18
  • $\begingroup$ @Numb3rs: If the question is “does $f$ have a minimum” then it suffices to prove the existence of an $x_0$ with $f(x) \ge f(x_0)$ for all $x$. If the question is “where does $f$ attains its minimum” then you need to give information about the location of $x_0$. In your, case $-1 < x_0 < 0$ is easy to see. One can probably express $x_0$ with the help of the Lambert W function. $\endgroup$
    – Martin R
    Dec 2, 2022 at 12:05
  • $\begingroup$ Oh, ok. Then no. That function is indeed beyond the course's program, for now. Thank you again! $\endgroup$
    – Numb3rs
    Dec 2, 2022 at 13:09
1
$\begingroup$

Your approach has a significant error. You are using the principle that a function $f$ will have a supremum (minimum or maximum) at its stationary points, where its derivative $f'$ is zero, if the derivative does exist. If the derivative does not exist at a point, you cannot use this rule. A point can still be a supremum without being a stationary point!

The absolute value function $\left|x\right|$ is not differentiable everywhere. In particular, it has a sharp change in slope at $x=0$. Using this, consider where your expression for $f'$ holds everywhere.

Hint: Split the absolute value expression into two differentiable pieces on distinct domains, one where $x+e^{x}$ is nonnegative and one where it is negative. Find the minimum of each piece and show that they coincide at the minimum of $f$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .