Corrigé du contrôle 4

m f R
f(x) = −x+m2, x < 1
−x2+2x +m , x ≥1
x→1
x<1
f(x)x→1
x>1
f(x)
x→1
x<1
f(x)=−1+m2
x→1
x>1
f(x) = 1+m
m f R
f x 6=1R1
f−1+m2=1+m⇔m2−m−2=0
∆=9=32
x1=1−3
2= −1 x2=1+3 2=2
fRm= −1 m =2
m f 1
f m /∈{−1;2}f
m∈{−1;2}f
( ) : x3−x2=1
fRf(x) = x3−x2−1 f −∞
+∞R
f
fx→−∞ f(x) = x→−∞ x3= −∞x→+∞ f(x) = x→+∞ x3= +∞
x∈Rf0(x) = 3x2−2x =x(3x −2)
02
3

x
f0(x)
f
−∞02
3+∞
+0−0+
−∞−∞
−1−1
−31
27
−31
27
+∞+∞
αR
α 10−2
( ) ⇔f(x) = 0
f−∞;2
3f(0)=−1 x ∈∞;2 3f(x)≤−1
f(x) = 0−∞;2
3
2
3; +∞f
f(2
3)=−31 27 x→+∞
f(x)=+∞0∈[f(2
3); x→+∞
f(x)[
f(x) = 0 α 2
3; +∞
αR
f(1,465)< 0 f(1,466)> 0 1,465 < α < 1,466 α ≈1,47 10−2
uRh
h(x) = pu(x)
x
u
−∞3+∞
−∞−∞
55
4
4
1
0
h[1; +∞[
h(x)x u(x)≥0⇔x≥1
h[1; +∞[
h+∞1
x→1u(x) = u(1) = 0 →0
√=√0=0
x→1h(x) = 0

h y =2
x→+∞
u(x) = 4
→4
√=2x→+∞
h(x) = 2
y=2
h+∞
u h [1; +∞[
u[1; +∞[u>0 ]1; +∞[h
x>1
h0(x) = u0(x)
2pu(x)
x > 1 2pu(x)> 0 h0(x)u0(x)
u h [1; +∞[
f] − 1; +∞[
f(x) = x
x+13
( ;~ ı ,~ )Cf
x→−1
x>−1
f(x)x→+∞ f(x)
x→−1
x>−1
x= −1x→−1
x>−1
x+1=0+
x→−1
x>−1
x
x+1= −∞
x→−∞
x3= −∞x→−1
x>−1
f(x)=−∞
x→+∞
x
x+1=x→+∞
x
x=x→+∞
1=1
x→1x3=1x→+∞
f(x) = 1
f0(x)
x∈] − 1; +∞[f0(x) = 3x2
(x+1)4
x∈] − 1; +∞[,f(x) = (u(x))3
u(x) = x
x+1
u0(x) = 1×(x+1) − x×1
x2=1 x2

f0(x) = 3×u0(x)×(u(x))2=3×1
x2×x x+12
=3x2
(x+1)4
f0(x) ] − 1; +∞[f
x∈] − 1; +∞[,(x+1)4> 0 f0(x)3x2
3x2≥0] − 1; +∞[3x2=0⇔x2=0⇔x=0
x
g0(x)
g
−1+∞
+0+
−∞
11
y∈Rf(x) = y
y
•y∈] − ∞;1[
y∈]x→−1
x>−1
f(x); x→+∞ f(x)[ f
]1; +∞[f(x) = y] − 1; +∞[
•y∈[1; +∞[f(x) = y
] − 1; +∞[
f]−1; +∞[x∈]−1; +∞[f(x)∈]x→−1
x>−1
f(x); x→+∞ f(x)[
f(x)∈] − ∞;1[
f(x) = y
a > −1aCa
a > −1aCa
y=f0(a)(x−a) + f(a)
y=3a2
(a+1)4(x−a) + a a+13
∈a⇔−af0(a) + f(a) = 0

(0;0)
∈a⇔0=f0(a)(0−a) + f(a)
⇔−af0(a) + f(a) = 0
aa
a−af0(a) + f(a) = 0
] − 1; +∞[
−af0(a) + f(a) = 0⇔−a3a2
(a+1)4+a a+13
=0
⇔−3a3
(a+1)4+a3 (a+1)3=0
⇔−3a3
(a+1)4+a3(a+1) (a+1)4=0
⇔−3a3+a3(a+1)
(a+1)4=0
⇔−3a3+a3(a+1) = 0
⇔a3(−3+a+1) = 0
⇔a3(a−2) = 0
⇔a3=0 a −2=0
⇔a=0 a =2
g] − 1; +∞[
g(x) = x−2
x+1
Cgg( ;~ ı ,~ )
x→+∞
f(x) − g(x)C Cg
f(x) − g(x)