What are the critical values, if any, of f(x)=x+e^(-3^x)?

Aug 14, 2017

First of all, $f \left(x\right)$ is differentiable everywhere, with
$f ' \left(x\right) = 1 + {e}^{- {3}^{x}} \left(- {3}^{x}\right) \left(\ln 3\right)$
$f ' \left(x\right) = 1 - \frac{\left({3}^{x}\right) \left(\ln 3\right)}{e} ^ \left({3}^{x}\right)$
The denominator of the expression dominates as x increases in absolute value, so that f'(x) has a horizontal asymptote at y = 1.

Now this function, f', is minimal when
$f ' ' \left(x\right) = - \left(\ln 3\right) \left[\left(\ln 3\right) {e}^{- {3}^{x}} \left(- {3}^{x}\right) \left({3}^{x}\right) + \left(\ln 3\right) {e}^{- {3}^{x}} \left({3}^{x}\right)\right] = 0$
Divide by both exponentials (which are never zero), and by ${\left(\ln 3\right)}^{2}$:
$\left(- {3}^{x}\right) + 1 = 0$
${3}^{x} = 1$
$x = 0$
We know this is a minimum and not a maximum because $f ' \left(0\right) < 1$, but we may also apply the first derivative test.

$f ' \left(0\right) = 1 - \frac{\left(1\right) \left(\ln 3\right)}{e} ^ \left(1\right)$
$f ' \left(0\right) = 1 - \frac{\ln 3}{e} > 0$

Therefore f' is positive for all values of x.

Since f' is positive, f is strictly increasing on $\left(- \infty , \infty\right)$.
Therefore f has no critical points.