Fonctions continues Exercice 1. (NEW)

#
f:x7−→ (cos x)1
x
Df
f0
0
f:R−→ RT > 0f T ∀x∈R, f(x+T) = f(x)
f+∞f
f f
f f
f: [0,+∞[−→ R+∞
f
f
f
f:R−→ R
x7−→ d(x, Z) = |bx+1
2c − x
f1 1
fn(x) = 1
4nf(4nx)
fn1
4n
F:x7−→
+∞
P
n=0
fn(x)
F
x∈[0,1[ x=
∞
P
k=1
xk
10kx
f: [0,1[ −→ [0,1[ f(x) =
∞
P
k=1
xk+1
10kf
g: [0,1[ −→ [0,1[ g(x) =
∞
P
k=1 x2k
102k−1+x2k−1 102kg
f(x+ 1) −f(x)> a
f:R−→ Rf(x+ 1) −f(x)−x→+∞− > a ∈R
f(n)
n−n→ ∞− > a
f(x)
x−x→ ∞− > a

max(f, g)
f, g :R−→ Rx∈Rh(x) = maxf(x), g(x)
h
f, g :R−→ R∀x, y ∈Q, f(x)< g(x)
f≤g
∀x, y ∈R, f(x)< g(x)
f:R−→ R Q f
sup(f([x, x + 1]))
f:R−→ Rg(x) = supf([x, x + 1])g
f
f, g : [a, b]−→ Rh(t) = sup{f(x) + tg(x)x∈[a, b]}
h
f: [a, b]−→ Rsup
[a,b]
f= sup
]a,b[
f
f, g : [a, b]−→ R∀x∈[a, b], f(x)> g(x)>0
k > 1f > kg
f: [0,+∞[−→ R`∈R+∞f
f(0) ` `
f(x) = g(x)
f: [0,1] −→ [0,1] x∈[0,1] f(x) = x
f, g : [0,1] −→ [0,1] f◦g=g◦f x ∈[0,1]
f(x) = g(x)f
f⇒
f:R−→ Rx f(x) = x
f:R−→ Ra f ◦f(a) = a f
1/n
f: [0,1] −→ Rf(0) = f(1)
x∈0,1
2f(x) = fx+1 2
n∈N, n ≥2x∈0,1−1
nf(x) = fx+1 n
f∀x∈0,3
5f(x)6=f(x+2 5)
a > 0∀b∈]0, a],∃x∈[0,1−b]f(x) = f(x+b)
f([a, b]) ⊂g([a, b])
f, g : [a, b]−→ R∀x∈[a, b],∃y∈[a, b]f(x) = g(y)

x∈[a, b]f(x) = g(x)
+⇒
f:R−→ Rf
∀a, b ∈Ra < b, ∀y f(a)f(b),∃x∈[a, b]f(x) = y.
f
f⇒f=
I f :I−→ Rf(I)
f⇒ |f(x)| ≤ a+b|x|
f:R−→ Ra, b ∈R∀x∈R,|f(x)| ≤
a+b|x|
ε= 1 |f(x)−f(0)|
f:D−→ Rg:R−→ Rg◦f
sin(t2)
t7→ sin(t2)R
f f(n)>+∞
f:R−→ Rf(n)−n→ ∞− >+∞f(x)
−x→ ∞− >+∞
f(x+y) = f(x) + f(y)
f:R−→ R∀x, y ∈R, f(x+y) = f(x) + f(y)a=f(1)
∀x∈Qf(x) = ax
f∀x∈Rf(x) = ax
f∀x∈Rf(x) = ax
f(x2) = f(x)
f: [0,1] −→ R∀x∈[0,1], f(x2) = f(x)
f(x+y)f(x−y) = f2(x)f2(y)
f:R−→ R∀x, y ∈R, f(x+y)f(x−y) =
f2(x)f2(y)
∗
f[a, b]Rδ ω(δ) = sup{|f(x)−f(y)| |x−
y| ≤ δ}ω(δ) 0 δ0ω
∗
f, g : [0,1] −→ [0,1] f◦g=g◦f x ∈[0,1]
f(x) = g(x)
f∗

ϕ ϕ ≤f f :Rn−→ R
k > 0f ϕ :Rn−→ Rk f
k0>0k≥k0ϕkk
f
(f(nx)) ∗
f:R−→ Rx > 0 (f(nx))n∈N
f

fn(x)1
2×4nF(x)
F
a hn=1
4nm∈NDm=fm(a+hn)−fm(a)
Cm=fm(a−hn)−fm(a)
•m≥n hn=k
4mCm=Dm= 0
•m < n m + 1 ≤n I]k−1
4m,k 4m[a∈I k ∈Z
a+hn−(a−hn)=2hn=2
4n<1 4n−1≤1 4ma+hna−hnI a +hn∈I
F(a+hn)−F(a)
hn
=1
hn
+∞
P
k=0
Dk=1
hn
n−1
P
k=0
Dk
fk±1IDk
hn
=±1F(a+hn)−F(a)
hn
n+∞F a
f(x)−x0 1
f−g a f
g(a)> a g(a)f
b=f(a) (f(a)−a)+(f(b)−b)=0 x7−→ f(x)−x a b
k∈N∗a∈Rfk(a) = a(f(a)−a)+(f2(a)−f(a))+· · ·+(fk(a)−fk−1(a)) = 0
f(x)−xmin(a, f(a), . . . , fk−1(a)) max(a, f(a), . . . , fk−1(a))
a∈Rf a anan> a |f(an)−f(a)| ≥ ε
f(a)±ε
f f(0) = ±1f
∀p∈N, f(px) = ±fp2(x)⇒f(x) = ±λx2
ω(δ)−δ→0+−>0⇔f