show that the equation x^9-x^5+x^4+x^2+1=0 has only one real root which is negative
Answers
Answer:
there cannot be two (or more) zeros.
Step-by-step explanation:
First if we look at the limits of our function we see, that:
lim
x
→
−
∞
f
(
x
)
=
−
∞
, and
lim
x
→
+
∞
f
(
x
)
=
+
∞
, so the function has at least one real root.
To show, that the function has only one real root we have to show, that it is monotonic in the whole
R
To show it we have to calculate the derivative
f
'
(
x
)
f
'
(
x
)
=
20
x
4
+
3
x
2
+
2
Now we have to solve:
20
x
4
+
3
x
2
+
2
=
0
To do this we substitute
t
=
x
2
to transform the equation to a quadratic one:
20
t
2
+
3
t
+
2
=
0
This experssion is positive for all
x
e
R
, because
Δ
=
−
151
<
0
, so the function
f
(
x
)
is increasing in the whole domain
R
.
So we can conclude, that it has only one real root.
QED
Answer link
Jim H
Sep 29, 2015
See the explanation section.
Explanation:
Let
f
(
x
)
=
1
+
2
x
+
x
3
+
4
x
5
and note that for every
x
,
x
is a root of the equation if and only if
x
is a zero of
f
.
f
has at least one real zero (and the equation has at least one real root).
f
is a polynomial function, so it is continuous at every real number. In particular,
f
is continuous on the closed interval
[
−
1
,
0
]
.
f
(
−
1
)
=
1
−
2
−
1
−
4
=
−
8
and
f
(
0
)
=
1
0
is between
f
(
−
1
)
and
f
(
0
, so the Intermediate Value Theorem tells us that there is at least one number
c
in
(
−
1
,
0
)
with
f
(
c
)
=
0
.
This
c
is a zero of
f
and a root of the equation.
f
cannot have two (or more) zeros
Suppose that
f
had two (or more) zeros, call them
a
and
b
. So
f
(
a
)
=
0
=
f
(
b
)
f
is continuous on the closed interval
[
a
,
b
]
(it's still a polynomial) and
f
is differentiable on the open interval
(
a
,
b
)
(
f
'
(
x
)
=
2
+
3
x
2
+
20
x
4
exists for all
x
in the interval.)
and
f
(
a
)
=
f
(
b
)
so by Rolle's Theorem (or by the Mean Value Theorem) there is a
c
in
(
a
,
b
)
with
f
'
(
c
)
=
0
However,
f
'
(
x
)
=
2
+
3
x
2
+
20
x
4
can never be
0
.
(Look at each term. Both
20
x
4
and
3
x
2
are at least
0
and the constant adds
2
. So,
f
'
(
x
)
≥
2
.)
That means that no
c
with
f
'
(
c
)
=
0
can exist.
Conclusion
If there were two (or more) zeros, then we could make
f
'
(
c
)
=
0
.
But, clearly, we cannot make
f
'
(
c
)
=
0
, so there cannot be two (or more) zeros.