La fonction exponentielle Problème à résoudre I

fRnf(0) = 1
f0=f
0
fRf(0) = 1 f0=f
exp
•
•
fRf(0) = 1 f0=f
x∈Rf(x)×f(−x)=1
fR
x∈Rϕ(x) = f(x)f(−x)ϕR
ϕ u ×v v
ϕ0(x) = f0(x)f(−x) + f(x)f0(−x)×(−1) = f(x)f(−x)−f(x)f(−x)=0
ϕ ϕ(0) = f(0)f(−0) = 1 ×1=1 x∈R
f(x)f(−x) = ϕ(x) = ϕ(0) = 1
•a∈Rf(a) = 0 ϕ(a) = f(a)f(−a) = 0 ×
f(−a)=0 ϕ(a)=1
•f > 0a∈Rf(a)<0
fRf(0) = 1 >0f(a)<0
b0a f(b)=0
f(x)>0x∈R

f g ϕ =f
g
ϕRg(x)6= 0 x∈R
f0=f g0=g
ϕ0=f0g−fg0
g2=fg −fg g2= 0
ϕ ϕ(0) = f(0)/g(0) = 1/1 = 1 x∈Rϕ(x)=1
f/g = 1 f=g
exp Rx∈Rexp0(x) = exp(x)
exp(0) = 1
x∈Rexp(−x) = 1
exp(x)
x∈Rexp(x)>0
x∈Ry∈Rexp(x+y) = exp(x) exp(y)
x∈Ry∈Rexp(x−y) = exp(x)
exp(y)
x∈Rn∈Zexp(nx) = (exp(x))n
x∈Rexp(x) exp(−x) = 1 exp(−x) = 1
exp(x)
y∈R
ϕy(x) = exp(x+y) exp(−x)x∈R
ϕ0
y(x) = exp0(x+y) exp(−x)+exp(x+y) exp0(−x)×(−1) = exp(x+y) exp(−x)−exp(x+y) exp(−x) = 0
ϕyϕy(0) = exp(0 + y) exp(−0) = exp(y)x∈R
exp(x+y) exp(−x) = ϕy(x) = ϕy(0) = exp(y)
x∈Rexp(x+y) = 1
exp(−x)exp(y) = exp(x) exp(y)

e
exp
eexp(1)
e= exp(1)
e'2,718
n∈N
exp(n) = exp(n×1) = (exp(1))n=en
exp(−n) = 1
exp(n)=1 en=e−n
x∈Rexp(x) = ex
exp R
x∈Rexp0(x) = exp(x)>0
x
exp
−∞ +∞
00
+∞+∞
•lim
x→−∞ exp(x) = 0 y= 0 C−∞
•Cx= 0 y= exp0(0)(x−0) + exp(0) = 1 ×x+ 1 = x+ 1

•Cx= 0
ϕ(x) = ex−(x+ 1) Rϕ ϕ0(x) = ex−1
x∈R
ϕ0(x)≥0⇐⇒ ex≥1
⇐⇒ ex≥e0
⇐⇒ x≥0
x
ϕ0(x)
ϕ
−∞ 0+∞
−0+
00
x∈Rϕ(x)≥0Cexp
•Cx= 1
exp(1) = eexp0(1) = exp(1) = e
y=exp0(1)(x−1) + exp(1) = e(x−1) + e=ex
x
f(x)
−5−4−3−2−1 1 2 3 4 5
e
0
−1
1
2
3
4
5
6
T0
T1
a b
ea=eb⇐⇒ a=b
ea> eb⇐⇒ a>b
exp 

lim
x→+∞ex= +∞
lim
x→−∞ ex= 0
x∈Rex≥x+ 1 lim
x→+∞x+ 1 = +∞
lim
x→+∞ex= +∞
lim
x→+∞
1
ex= 0
1
ex=e−xlim
x→−∞
−x= +∞lim
x→−∞ ex= 0
eu
u I
(eu)0=u0eu
v= exp v0= exp
f0= (v◦u)0=u0×(v0◦u) = u0eu
f(x) = e−xx∈R
uRu(x) = −x u0(x) = −1x∈Rf(x) =
exp(u(x)) f0(x) = u0(x) exp(u(x)) = −1×exp(−x) = −e−x
lim
x→+∞
ex
x= +∞
lim
x→−∞ xex= 0