Why is this?

[Albert Chan]

f = x * exp(-x^2) is product of linear growth and exponential decay. A peak is expected.
Peak of f also at peak of ln(f) = ln(x) - x^2

(ln(f))' = 1/x - 2x = 0, we have x^2 = 1/2, or x = √(1/2)

---

g = x^(1/x) = exp(ln(x)/x)
ln(x) does not grow as fast as x, so again g is decaying after a peak.
Peak of g also at peak of ln(g) = ln(x)/x

(ln(g))' = (x*(1/x) - 1*ln(x)) / x^2 = (1 - ln(x)) / x^2 = 0, we have ln(x) = 1, or x = e