A calculus problem by Steven Jim

$A = a^a$ , $\implies \dfrac {dA}{da} = (\ln a + 1) a^a$ . When $\dfrac {dA}{da} = 0$ , $\implies \ln a + 1 = 0$ and $a = \dfrac 1e$ . Note that $\dfrac {d^2A}{da^2} = a^{a-1} + (\ln a+1)^2a^a$ , $\implies \dfrac {d^2A}{da^2} \bigg|_{a=\frac 1e} > 0$ , $\implies \min (A) = \left(\frac 1e\right)^\frac 1e$ and $\min (A) + x = \left(\frac 1e\right)^\frac 1e + \frac 1e \approx \boxed{1.060}$ .