Analyse et algèbre pour les sciences - IMJ-PRG

f:−π
2,π 2→Rf(x) = sin(x)
f f−1
f−1[−1,1]
f−1lim
x→(−1)+f−1(x) = f−1(−1) = −π
2lim
x→1−f−1(x) = f−1(1) = π
2
f0(x) = cos(x)x∈−π
2,π 2(f−1)0(x) = 1
√1−x2x∈]−1,1[
f I f
f−1f(I)a∈I y ∈f(I)y6=f(a)y f
x=f−1(y)
x=a y =f(a)x6=a
f−1(y)−f−1(f(a))
y−f(a)=x−a f(x)−f(a).
x−a
f(x)−f(a)x→af(x)−f(a) x−ax→a f−1
f(a)f a (f−1)0(f(a)) = 1
f0(a)
arctan0(tan(x)) = 1
tan0(x)=1
1 + tan2(x).
arctan0(y) = 1
1+y2
arctan(1) = π
4arctan0(1) = 1 2
a b f
f(x) = 
0x < −2
ax +b−2≤x≤0
x3
sin 1
x1 x+ ln(x)
x > 0
f(0) = b f(−2) = −2a+b
lim
x→0+x3ln(x)=0
x∈R−1≤sin(x)≤1
−x2≤x2sin 1
x≤x2.
lim
x→0+x2sin 1
x= 0
x > 0
0≤f(x)≤
x3ln(x) +
x2sin 1
x
.
lim
x→0+f(x) = 0

f0 lim
x→0+f(x) = f(0) = b b = 0
f−2 lim
x→(−2)−f(x) = f(−2) = −2a+b0 = −2a+b
a=b/2 = 0
f(x)−f(0)
x=x2
sin 1
x1 x+ ln(x)
.
lim
x→0+x2ln(x) = 0 lim x→0+xsin 1
x= 0
lim
x→0+
f(x)−f(0)
x= 0
a=b= 0 lim
x→0−
f(x)−f(0)
x= 0 f0f0(0) = 0
A B RA⊂B
A B
sup(B)B x ∈A x ∈B x ∈A x ≤sup(B) sup(B)
Asup(A)Asup(A)≤sup(B)
inf(B)Ainf(A)A
inf(B)≤inf(A)
a b a < b f : [a, b]→Rf(a) = f(b)f
[a, b] [a, b]m f
M f
m=a m =b M =a M =b x ∈[a, b]f(a) = f(b)≤f(x)≤f(a) = f(b)
f x ∈]a, b[f0(x)=0 c∈]a, b[
m∈]a, b[f0(m) = 0 c=m
M∈]a, b[f0(M) = 0 c=M