Question Number 102234 by Harasanemanabrandah last updated on 07/Jul/20

Answered by OlafThorendsen last updated on 07/Jul/20

f(x) = x−elnx, x>0  f′(x) = 1−(e/x)  f′(x) = 0 ⇔ x = e  f(e) = e−elne = e−e = 0  And for x∈]e;+∞[ f′(x)>0  Then f(π) > 0  ⇔ π−elnπ>0  ⇔ π>elnπ  ⇔ πlne>elnπ  ⇔ lne^π >lnπ^e   ⇔ e^π >π^e