Un livre de Wikilivres.
Soit
f
{\displaystyle f\,}
R
{\displaystyle \mathbb {R} }
f
(
x
)
=
(
x
−
1
)
(
2
−
e
−
x
)
{\displaystyle f(x)=(x-1)(2-e^{-x})\,}
C
{\displaystyle C\,}
(
O
,
i
→
,
j
→
)
{\displaystyle (O,{\overrightarrow {i}},{\overrightarrow {j}})\,}
1)
a) Déterminer les limites de
f
{\displaystyle f\,}
+
∞
{\displaystyle +\infty \,}
−
∞
{\displaystyle -\infty \,}
D
{\displaystyle D\,}
y
=
2
x
−
2
{\displaystyle y=2x-2\,}
C
{\displaystyle C\,}
C
{\displaystyle C\,}
(
D
)
{\displaystyle (D)\,}
2) Soit la fonction
g
{\displaystyle g\,}
R
{\displaystyle \mathbb {R} }
g
(
x
)
=
x
e
−
x
−
2
e
−
x
+
2
{\displaystyle g(x)=xe^{-x}-2e^{-x}+2\,}
g
′
(
x
)
{\displaystyle g'(x)\,}
R
{\displaystyle \mathbb {R} }
lim
x
→
+
∞
g
(
x
)
{\displaystyle \lim _{x\to +\infty }\ g(x)}
g
(
0
)
{\displaystyle g(0)\,}
g
{\displaystyle g\,}
R
{\displaystyle \mathbb {R} }
3)
f
′
(
x
)
{\displaystyle f'(x)\,}
f
′
(
x
)
{\displaystyle f'(x)\,}
R
{\displaystyle \mathbb {R} }
f
′
(
0
)
{\displaystyle f'(0)\,}
f
{\displaystyle f\,}
R
{\displaystyle \mathbb {R} }
4) Déterminer les coordonnées du point
A
{\displaystyle A\,}
C
{\displaystyle C\,}
(
T
)
{\displaystyle (T)\,}
C
{\displaystyle C\,}
(
D
)
{\displaystyle (D)\,}
http://paquito.amposta.free.fr/symboles/symboles.htm http://fr.wikipedia.org/wiki/Notation_(math%C3%A9matiques)
u
n
=
1
+
1
1
!
+
1
2
!
+
.
.
.
+
1
n
!
=
1
+
∑
k
=
1
n
1
k
!
{\displaystyle u_{n}=1+{\frac {1}{1!}}+{\frac {1}{2!}}+...+{\frac {1}{n!}}=1+\sum _{k=1}^{n}{\frac {1}{k!}}}
1
1
≤
1
1
{\displaystyle {\frac {1}{1}}\leq {\frac {1}{1}}}
1
2
≤
1
2
{\displaystyle {\frac {1}{2}}\leq {\frac {1}{2}}}
1
6
≤
1
4
{\displaystyle {\frac {1}{6}}\leq {\frac {1}{4}}}
1
24
≤
1
8
{\displaystyle {\frac {1}{24}}\leq {\frac {1}{8}}}
.
.
.
{\displaystyle ...\,}
1
n
!
≤
1
2
n
−
1
{\displaystyle {\frac {1}{n!}}\leq {\frac {1}{2^{n-1}}}}
En sommant, en obtient :
∑
k
=
1
n
1
k
!
≤
1
+
1
2
+
1
4
+
1
8
+
.
.
.
+
1
2
n
−
1
{\displaystyle \sum _{k=1}^{n}{\frac {1}{k!}}\leq 1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+...+{\frac {1}{2^{n-1}}}}
⇔
u
n
≤
2
+
1
2
+
1
4
+
1
8
+
.
.
.
+
1
2
n
−
1
≤
3
{\displaystyle \Leftrightarrow u_{n}\leq 2+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+...+{\frac {1}{2^{n-1}}}\leq 3}
3) démontrons que
u
n
{\displaystyle u_{n}\,}
u
n
{\displaystyle u_{n}\,}
u
n
+
1
−
u
n
=
1
+
1
1
!
+
1
2
!
+
.
.
.
+
1
n
!
+
1
(
n
+
1
)
!
−
1
−
1
1
!
−
1
2
!
−
.
.
.
−
1
n
!
=
1
(
n
+
1
)
!
≥
0
{\displaystyle u_{n+1}-u_{n}=1+{\frac {1}{1!}}+{\frac {1}{2!}}+...+{\frac {1}{n!}}+{\frac {1}{(n+1)!}}-1-{\frac {1}{1!}}-{\frac {1}{2!}}-...-{\frac {1}{n!}}={\frac {1}{(n+1)!}}\geq 0}
u
n
{\displaystyle u_{n}\,}
On a la loi expo définie par
N
(
t
)
=
N
0
e
−
λ
t
{\displaystyle N(t)=N_{0}e^{-\lambda t}\,}
A
(
t
)
=
−
d
N
d
t
{\displaystyle A(t)=-{\frac {dN}{dt}}}
A
(
t
)
=
−
d
(
N
0
e
−
λ
t
)
d
t
=
−
(
−
λ
N
0
e
−
λ
t
)
=
λ
N
0
e
−
λ
t
{\displaystyle A(t)=-{\frac {d(N_{0}e^{-\lambda t})}{dt}}=-(-\lambda N_{0}e^{-\lambda t})=\lambda N_{0}e^{-\lambda t}}
A
(
t
)
=
λ
N
(
t
)
{\displaystyle A(t)=\lambda N(t)\,}
A
0
=
λ
N
0
{\displaystyle A_{0}=\lambda N_{0}\,}
A
(
t
)
=
A
0
e
−
λ
t
{\displaystyle A(t)=A_{0}e^{-\lambda t}\,}
