Question

[5] Lets show that e π > πe . (e π is known as Gelfond’s constant)

(a) [2] Find the local maximum of f(x) = ln(x) x for x > 0

(b) [1] Find the global maximum of f(x) = ln(x) x on [1, e2 ]. (hint: 2 e 2 < 1 e because 2 < e)

(c) [1] Observe that π ∈ [1, e2 ]. (hint: 1 < π < 4 < e2 because 2 < e)

Answer 1

Answer

We have to show that

e

(a)

Given that

Intwherer >0 f (x)

As f(x) is continuous for x > 0, we have to find the critical point(s) by setting the derivative of f(x) to 0.

1 In(x) x 1 n() In(a) d r()d H

1 In 0 1 n(r) = 0 In(r ) 1 re 2 f'(x)

Thus, x = e is the only critical point.

for x < e, f'(x) > 0.

for x > e, f'(x) < 0.

f'(x) changes its sign from positive to negative at x = e.

Hence, by the first derivative test, we have a local maximum at x = e.

At x = e, the local maximum is

(e)Ine)

(b)

To find the global maximum of a function, we need to compare the local maximum(s) of the function and the function value at the endpoints of the interval.

At x = e, the local maximum is

(e)n(e)

The given interval is

$[1,e^2]$

At x = 1,

(1)In(1) = 0

At x = e²,

In(e2) f(e2) e 2 2

2e <1 E e V V IN 1

Hence,

$0< \frac{2}{e^2}<\frac{1}{e}$

The global maximum of f(x) on the interval [1,e²] is

(e)Ine)

(c)

on the interval [1,e²], the function f(x) attains global maximum value at x = e.

$1<\pi<4<e^2$

Hence,

$f(\pi)<f(e)$

In(e) n(T

e In(<T In(e)

$\Rightarrow \ln(\pi^e)<\ln(e^{\pi})$

$\Rightarrow \pi^e<e^{\pi}$

${\color{Blue} \Rightarrow e^{\pi}>\pi^{e}}$