Prove the function has a minimum

Question

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?

Answer 1

$g(x) = x + e^x$ 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.

Answer 2

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 $f$.