Anybody could answer this with a calculator. What I am looking for is an analytical explanation that proves the result.
2006-12-01 11:08:09 · 3 answers · asked by babs 1 in Science & Mathematics ➔ Mathematics
Consider the function f(x) = x^(1/x).
Then ln f = (1/x) ln x
Differentiating:
(1/f) df/dx = (-1/x^2) ln x + 1/x^2.
So df / dx = (x^1/x)[(-1/x^2) ln x + 1/x^2].
= (x^1/x)(1/x^2)(1 - ln x)
Now, this is < 0 when ln x > 1, and > 0 when ln x < 1.
Therefore f is increasing for 0
Therefore the maximum of f occurs at x = e
So e^(1/e) > pi^(1/pi)
Raise both sides to the power of e*pi:
e^pi > pi^e.
2006-12-01 11:15:35 · answer #1 · answered by stephen m 4 · 7⤊ 0⤋
pi^e
use FX 100 methods
2006-12-01 11:08:54 · answer #2 · answered by Anonymous · 0⤊ 4⤋
e^pi=23.140692632779269005729086367949
pi^e=22.459157718361045473427152204544
e^pi is bigger.
2006-12-01 11:20:05 · answer #3 · answered by yupchagee 7 · 0⤊ 3⤋