1. Exercice 20
ex − 1
f (x) =
ex + 1
e2x − 1
f (2x) = 2x
e +1
ex − 1 2
e2x − 2ex + 1
f (x)2 = x =
e +1 e2x + 2ex + 1

2. Exercice 20 (suite..)
Par ailleurs :
2ex −2 2ex −2
2f (x) ex +1 ex +1
= = e2x +2ex +1+e2x −2ex +1
1 + f (x)2 1+ e2x −2ex +1
e2x +2ex +1 e2x +2ex +1
2e x −2 e 2x + 2ex + 1
= ×
ex + 1 2e2x + 2
2(ex − 1) × (ex + 1)2
= x
(e + 1) · 2(e2x + 1)
(ex − 1)(ex + 1) e2x − 1
= = 2x = f (2x)
e2x + 1 e +1