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th
th R R
x∈Rth0(x) = 1
ch2(x)>0R
th R
th(x) = ex−e−x
ex+e−x=1−e−2x 1 + e−2xlim
x→+∞e−2x= 0 lim x→+∞th(x)=1
th(x) = e2x−1
e2x+ 1 lim
x→−∞ e2x= 0 lim x→−∞ th(x) = −1
th R]−1,1[
th RI=]0,1[
x∈Rth0(x) = ch2(x)−sh2(x)
ch2(x)= 1 −th2(x)
∀x∈R, th0(x)=1−th2(x)
Argth ]−1,1[ x∈]−1,1[ −x∈]−1,1[
x∈]−1,1[ y=Argth(x)th(y) = x th(−y) = e−y−ey
e−y+ey=−th(y)
Argth(−x) = Argth(−th(y)) = Argth(th(−y)) = −y=−Argth(x)
Argth
th Rth0(x) = 1
ch2(x)R
Argth th(R) =] −1,1[ x∈]−1,1[
Argth0(x) = 1
th0(Argth(x)) =1 1−th2(Argth(x)) =1 1−x2
Argth ]−1,1[ ∀x∈]−1,1[, Argth0(x) = 1
1−x2
x∈]−1,1[ y=Argth(x)
x=th(y) = ey−e−y
ey+e−y=e2y−1 e2y+ 1 (e2y+ 1)x=e2y−1e2y(x−1) = −1−x e2y=1 + x 1−x
x∈]−1,1[ 1 + x
1−x>0 2y= ln 1 + x 1−x
∀x∈]−1,1[, Argth(x) = 1
2ln 1 + x 1−x
x6= 0 J y0+3
xy=1 x(1 −x2)
•y0+3
xy= 0
a(x) = 3
xa J A(x) = 3 ln(x)
x7→ ke−3 ln(x)k∈RJRx7→ k
x3

•
fpfp(x) = k(x)
x3k J
f0
p(x) = k0(x)x3−3x2k(x)
x6=k0(x) x3−3k(x) x4
fp
f0
p(x) + 3
xfp(x) = 1 x(1 −x2)
k0(x)
x3−3k(x) x4+3 x
k(x)
x3=1 x(1 −x2)
k0(x)
x3=1 x(1 −x2)
k0(x) = x2
1−x2=−x2 x2−1=−x2−1+1 x2−1=−x2−1 x2−1+1 x2−1=−1−1 x2−1=−1 + 1 1−x2
k(x) = −x+Argth(x)
S=J→R:x7→ k−x+Argth(x)
x3, k ∈R
f(x) = C C ∈R R
∀x∈R, f(2x) = 2f(x)
1 + f(x)2C=2C 1 + C2
C(1 + C2)=2C C(1 + C2−2) = 0 C(C2−1)2
C= 0 C= 1 C=−1
f≡0f≡1f≡ −1
f f(0) = 2f(0)
1 + f(0)2
f(0) −1
∀x∈R, f(x) =
2fx
2
1 + fx
22
∀x∈R:f(x)−1 =
2fx
2−(1 + fx 22)
1 + fx
22=−−2fx
2+1+fx 22
1 + fx
22=−fx
2−12
1 + fx
22≤0
∀x∈R, f(x)≤1
∀x∈R:f(x) + 1 = fx
2−12
1 + fx
22≥0
∀x∈R,−1≤f(x)

∀x∈R,−f(2x) = −2f(x)
1 + f(x)2f
=2(−f(x))
1+(−f(x))2
=2(−f)(x)
1+(−f)(x)2
f−f
th R
∀x∈R,2th(x)
1 + th(x)2= 2
ex−e
−x
ex+e
−x
1 + ex−e
−x
ex+e
−x2
= 2 (ex+e−x)2ex−e
−x
ex+e
−x
(ex+e−x)2+ (ex−e−x)2
= 2(ex+e−x)(ex−e−x)
2e2x+ 2e−2x
=e2x−e−2x
e2x+e−2x
=th(2x)
th
lim
n→+∞
x0
2n= 0 lim
n→+∞2n= +∞
f0 0 un=fx0
2nlim
n→+∞fx0
2n=f(0) = 1
lim
n→+∞un= 1
∀n∈N, un=fx0
2n=f2×x0 2n+1 =
2fx0
2n+1 
1 + fx0
2n+1 2f
un+1 =x0
2n+1
∀n∈N, un=2un+1
1 + u2
n+1
∀n∈N,2
1 + u2
n+1
>0un=un+1 ×2
1 + u2
n+1
unun+1
∀n∈N:
un+1 −un=un+1 −2un+1
1 + u2
n+1
=un+1 1−2
1 + u2
n+1 =un+1 1 + u2
n+1 −2
1 + u2
n+1 =un+1 u2
n+1 −1
1 + u2
n+1 
•u0>0n∈Nun+1 un
n∈N, un>0un+1 >0
• ∀n∈Nun+1 =fx0
2n+1 x∈R−1≤f(x)≤1−1≤fx0 2n+1 ≤1
0≤fx0
2n+1 2≤1u2
n+1 −1≤0
1 + u2
n+1 >0u2
n+1 −1
1 + u2
n+1 ≤0

un+1 −un≤0
(un)
•u0>0 (un)∀n∈Nun≤u0=f(x0)
f(x0)≥1f(x0)6=f(0) = 1 f(x0)<1
n∈Nun<1 (un) lim
n→+∞un= 1
•u0= 0 (un) lim
n→+∞un= 1
•u0<0n∈Nun<0 lim
n→+∞un= 1
f−f f(0) = −1 (−f)(0) = 1
f−f
f(0) = 1
x0f(x0)6=f(0)
f(0) = 1 f(0) = −1fR
PR= (O, −→ i , −→ j). O0(5; 2)
−→
u=−1
2−→
i−√3
2−→
j−→ v=√3
2−→
i−1
2−→
j .
2
R0= (O0,−→ u , −→ v)
M(x, y)R
R0
DPR2x+ 2y−5=0.
D R D R0
A(1,4) R R0
CA2
AD
B(0,1 + √3) C R
TCB