Sia
f
⁡
(
x
)
=
m
2
x
2
+
m
1
x
+
m
0
e
g
⁡
(
x
)
=
n
2
x
2
+
n
1
x
+
n
0
data la natura delle funzioni e poichè
g
'
⁡
(
x
)
∖
=
0
in [a,b] si può applicare il teorema di Cauchy.
f
⁡
(
b
)
=
m
2
b
2
+
m
1
b
+
m
0
e
f
⁡
(
a
)
=
m
2
a
2
+
m
1
a
+
m
0
da cui:
f
⁡
(
b
)
-
f
⁡
(
a
)
=
m
2
b
2
+
m
1
b
+
m
0
-
(
m
2
a
2
+
m
1
a
+
m
0
)
=
m
2
⋅
(
b
2
-
a
2
)
+
m
1
⋅
(
b
-
a
)
g
⁡
(
b
)
=
n
2
b
2
+
n
1
b
+
n
0
e
g
⁡
(
a
)
=
n
2
a
2
+
n
1
a
+
n
0
da cui:
g
⁡
(
b
)
-
g
⁡
(
a
)
=
n
2
b
2
+
n
1
b
+
n
0
-
(
n
2
a
2
+
n
1
a
+
n
0
)
=
n
2
⋅
(
b
2
-
a
2
)
+
n
1
⋅
(
b
-
a
)
Si ricava che:
f
⁡
(
b
)
-
f
⁡
(
a
)
g
⁡
(
b
)
-
g
⁡
(
a
)
=
m
2
⋅
(
b
2
-
a
2
)
+
m
1
⋅
(
b
-
a
)
n
2
⋅
(
b
2
-
a
2
)
+
n
1
⋅
(
b
-
a
)
=
m
2
⋅
(
b
+
a
)
+
m
1
n
2
⋅
(
b
+
a
)
+
n
1
Applicando Cauchy:
f
⁡
(
b
)
-
f
⁡
(
a
)
g
⁡
(
b
)
-
g
⁡
(
a
)
=
f
'
⁡
(
c
)
g
'
⁡
(
c
)
=
2
m
2
c
+
m
1
2
n
2
c
+
n
1
dal confronto si evince che deve essere:
2
c
=
b
+
a
→
c
=
a
+
b
2