Graphics Reference
In-Depth Information
(vii) How does this lead to a contradiction, which therefore
undermines our hypothesis of the unboundedness of the
sequence (
f
(
x
))?
32 If therangeof a continuous function
f
:[
a
,
b
]
R were presumed to
be unbounded below, outline how you would establish a
contradiction.
33 A function
f
: Q
Q is defined by
f
(
x
)
1/(
x
2),
(i) show that
f
is continuous throughout its domain,
(ii) explain why the behaviour of
f
on the interval [1, 2] does not
contradict the result established in qn 31.
R
So therangeof a continuous real function
f
:[
a
,
b
]
is bounded.
Questions 31 and 32 eliminate R,(
a
,
), [
a
,
), (
,
a
) and
(
,
a
] as possiblerangs for a continuous function
f
:[
a
,
b
]
R. This
leaves open intervals, closed intervals, half-open intervals and singleton
points as the only possibilities. We know that closed intervals and
singleton points are real possibilities. Can we now eliminate the open
intervals and the half-open intervals as our investigation in qn 30 would
suggest we might?
34 (
Weierstrass
, 1860) Let
f
:[
a
,
b
]
R bea continuous function and
let
V
f
(
x
)
a
x
b
.
is an interval?
(ii) How do weknow that
V
is bounded above and below?
(iii) Can you besurethat sup
(i) How do weknow that
V
exist?
If welt sup
V
M
, then
M
f
(
x
) for all
x
[
a
,
b
]. As a
result of our graph drawing, we conjecture that there has to
bea valueof
x
at which
M
f
(
x
).
(iv) Why, given
0 must there be an
x
[
a
,
b
] such that
M
f
(
x
)
M
?
(v) Define
x
V
and inf
V
[
a
,
b
] so that
M
1/
n
f
(
x
)
M
.
Why must the sequence (
x
) contain a convergent
subsequence?
(vi) Suppose(
x
)
c
. Why must
c
[
a
,
b
]?
)) be convergent?
(viii) By considering the limit of the sequence (
f
(
x
(vii) Why must (
f
(
x
)) provethat
M
f
(
c
).
(ix) Having proved that the function
f
attains its supremum,
indicatehow to show that thefunction must also attain its
infimum.
35 Explain why function
f
:
Q
R
defined by
f
(
x
)
sin
x
does not
