1
Tutorial 2 Answers + Comments
Q1
This question uses the Intermediate Value Theorem. In it's most general form, we have
Théorème 0.1
Let
I
be an interval and let
f
:
I
7→
R
be continuous on
I
. If
a,b
∈
I
and
if
k
∈
R
satis es
f
(
a
)
< k < f
(
b
)
, then there exists a point
c
∈
I
between
a
and
b
such that
f
(
c
) =
k.
In this question, we are localising the roots of the polynomial
p
(
x
) =
x
3

3
x
2
+
c
, where
c >
2
.
Note that in general, a polynomial of odd degree has always at least one root. This is
guaranteed by the Intermediate Value Theorem since for a polynomial
p
of odd degree, the
dominant term is of the form
x
2
n
+1
, and when
x
→ ∞
p
→ ∞
, and when and when
x
→
+
∞
,
p
→
+
∞
. Taking
k
= 0
in the Intermediate Value Theorem (generalised), we
have thus a point
m
∈
R
such that
p
(
m
) = 0
. Note that this theorem guarantees
at least
one
real root, but
NOT exactly one
real root.
Another way to show that there exists at least one real root is to recall that if the roots
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 Spring '11
 phoon
 Fundamental Theorem Of Algebra, Intermediate Value Theorem, lim, Complex number

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