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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : CompositionOpérations sur les fonctions/Exercices/Composition », n'a pu être restituée correctement ci-dessus.
On note
h
{\displaystyle h}
R
∖
{
0
,
1
}
{\displaystyle \mathbb {R} \setminus \{0,1\}}
h
(
x
)
=
1
−
1
x
{\displaystyle h\left(x\right)=1-{\frac {1}{x}}}
h
{\displaystyle h}
R
∖
{
0
,
1
}
{\displaystyle \mathbb {R} \setminus \{0,1\}}
h
∘
h
(
x
)
{\displaystyle h\circ h\left(x\right)}
h
∘
h
∘
h
(
x
)
{\displaystyle h\circ h\circ h\left(x\right)}
Soit
f
:
R
∖
{
0
,
1
}
→
R
{\displaystyle f:\mathbb {R} \setminus \{0,1\}\to \mathbb {R} }
x
∈
R
∖
{
0
,
1
}
{\displaystyle x\in \mathbb {R} \setminus \{0,1\}}
f
(
x
)
+
f
∘
h
(
x
)
=
a
x
+
b
{\displaystyle f\left(x\right)+f\circ h\left(x\right)=ax+b}
x
{\displaystyle x}
f
(
x
)
{\displaystyle f\left(x\right)}
h
(
x
)
{\displaystyle h\left(x\right)}
h
∘
h
(
x
)
{\displaystyle h\circ h\left(x\right)}
f
∘
h
(
x
)
{\displaystyle f\circ h\left(x\right)}
f
∘
h
∘
h
(
x
)
{\displaystyle f\circ h\circ h\left(x\right)}
f
∘
h
∘
h
∘
h
(
x
)
{\displaystyle f\circ h\circ h\circ h\left(x\right)}
En déduire que pour tout
x
∈
R
∖
{
0
,
1
}
{\displaystyle x\in \mathbb {R} \setminus \{0,1\}}
f
(
x
)
=
a
h
∘
h
(
x
)
−
a
h
(
x
)
+
a
x
+
b
2
{\displaystyle f\left(x\right)={\frac {ah\circ h\left(x\right)-ah\left(x\right)+ax+b}{2}}}
Vérifier que réciproquement, la fonction
g
:
R
∖
{
0
,
1
}
→
R
{\displaystyle g:\mathbb {R} \setminus \{0,1\}\to \mathbb {R} }
g
(
x
)
=
a
h
∘
h
(
x
)
−
a
h
(
x
)
+
a
x
+
b
2
{\displaystyle g\left(x\right)={\frac {ah\circ h\left(x\right)-ah\left(x\right)+ax+b}{2}}}
g
(
x
)
+
g
∘
h
(
x
)
=
a
x
+
b
{\displaystyle g\left(x\right)+g\circ h\left(x\right)=ax+b}
Solution
Pour tout
x
∈
R
∖
{
0
,
1
}
{\displaystyle x\in \mathbb {R} \setminus \{0,1\}}
h
(
x
)
≠
0
{\displaystyle h\left(x\right)\neq 0}
1
x
≠
1
{\displaystyle {\frac {1}{x}}\neq 1}
x
≠
1
{\displaystyle x\neq 1}
h
(
x
)
≠
1
{\displaystyle h\left(x\right)\neq 1}
1
x
≠
0
{\displaystyle {\frac {1}{x}}\neq 0}
x
×
1
x
=
1
≠
0
{\displaystyle x\times {\frac {1}{x}}=1\neq 0}
h
∘
h
(
x
)
=
1
−
1
1
−
1
x
=
1
−
x
x
−
1
=
1
1
−
x
{\displaystyle h\circ h\left(x\right)=1-{\frac {1}{1-{\frac {1}{x}}}}=1-{\frac {x}{x-1}}={\frac {1}{1-x}}}
h
∘
h
∘
h
(
x
)
=
1
1
−
h
(
x
)
=
1
1
x
=
x
{\displaystyle h\circ h\circ h\left(x\right)={\frac {1}{1-h\left(x\right)}}={\frac {1}{\frac {1}{x}}}=x}
f
∘
h
(
x
)
=
a
x
+
b
−
f
(
x
)
{\displaystyle f\circ h\left(x\right)=ax+b-f\left(x\right)}
f
{\displaystyle f}
f
∘
h
∘
h
(
x
)
=
a
h
(
x
)
+
b
−
f
∘
h
(
x
)
=
a
h
(
x
)
+
b
−
(
a
x
+
b
−
f
(
x
)
)
=
a
h
(
x
)
−
a
x
+
f
(
x
)
{\displaystyle f\circ h\circ h\left(x\right)=ah\left(x\right)+b-f\circ h\left(x\right)=ah\left(x\right)+b-\left(ax+b-f\left(x\right)\right)=ah\left(x\right)-ax+f\left(x\right)}
f
∘
h
∘
h
∘
h
(
x
)
=
a
h
∘
h
(
x
)
−
a
h
(
x
)
+
f
∘
h
(
x
)
=
a
h
∘
h
(
x
)
−
a
h
(
x
)
+
a
x
+
b
−
f
(
x
)
{\displaystyle f\circ h\circ h\circ h\left(x\right)=ah\circ h\left(x\right)-ah\left(x\right)+f\circ h\left(x\right)=ah\circ h\left(x\right)-ah\left(x\right)+ax+b-f\left(x\right)}
f
∘
h
∘
h
∘
h
(
x
)
=
f
(
x
)
{\displaystyle f\circ h\circ h\circ h\left(x\right)=f\left(x\right)}
Si (pour tout
x
∈
R
∖
{
0
,
1
}
{\displaystyle x\in \mathbb {R} \setminus \{0,1\}}
g
(
x
)
=
a
h
∘
h
(
x
)
−
a
h
(
x
)
+
a
x
+
b
2
{\displaystyle g\left(x\right)={\frac {ah\circ h\left(x\right)-ah\left(x\right)+ax+b}{2}}}
g
∘
h
(
x
)
=
a
x
−
a
h
∘
h
(
x
)
+
a
h
(
x
)
+
b
2
{\displaystyle g\circ h\left(x\right)={\frac {ax-ah\circ h\left(x\right)+ah\left(x\right)+b}{2}}}
g
(
x
)
+
g
∘
h
(
x
)
=
a
x
+
b
{\displaystyle g\left(x\right)+g\circ h\left(x\right)=ax+b}
