Answer:
Here's one way to do it.
Step-by-step explanation:
Let
f
(
x
)
=
x
3
−
2
x
2
+
3
x
.
(Needed because the intermediate value theorem is a theorem about functions .)
Observe that the equation
x
3
−
2
x
2
+
3
x
=
5
has a root (a solution) exactly when
f
(
x
)
=
5
So the question now is to show that for at least one number
c
, in
[
1
,
2
]
, we get
f
(
c
)
=
5
.
f
is continuous on
[
1
,
2
]
(Because it is a polynomial and they are continuous everywhere.)
f
(
1
)
=
2
and
f
(
2
)
=
6
5
is between
f
(
1
)
and
f
(
2
)
, so
by the intermediate value theorem, there is at least one number
c
, in
[
1
,
2
]
, for which
f
(
c
)
=
5
.
That is, the original equation has a solution.
Step-by-step explanation: