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10 Sketch the graph of a continuous function f : A R
for which the
rangeof thefunction is not an interval.
What would you conjecture about the domain A of thefunction in
such a case?
If wewish to try to provethat continuous functions always map
intervals onto intervals, it is worth noticing first that this result fails if
weareworking with only therational numbrs.
11 Let the function f :
2. This is
certainly a continuous function. Now restrict the domain to the set
[1, 2] Q. Thenumbrs 1 and 2 arein therang, but what
rational number between them is not?
Q Q
be defined by f ( x )
x
This result suggests that it will be necessary to use the completeness
of the real numbers in a proof that continuous functions map intervals
onto intervals.
Intermediate Value Theorem
A simpleform of theproblem which wearefacing is this: suppose
that A is an interval and that f : A R is a continuous function.
Supposefurthr that a and b arepoints in A and that f ( a ) 0 and
f ( b ) 0. Can webesurethat threis a point
c in theintrval ( a , b ) such
that f ( c ) 0?
12 ( Bolzano , 1817) Let f :[ a , b ] R bea continuous function with
f ( a )
0. Sketch the graph of such a function. Try
different sketches which illustrate the same conditions. Notice that
neither of the sets
0 and f ( b )
need be an interval.
We construct sequences in the domain of the function which either
reach a point x where f ( x ) 0 or converge to such a point by
locating smaller and smaller intervals on each of which the function
f changes sign.
Let a
x
f ( x )
0
or
x
f ( x )
0
). If f ( d ) 0 wehavefound the
point wewant; if f ( d ) 0, explain how to choose a
a , b
b and d
( a
b
and b
so that
a
a
f ( a
)
0, f ( b
)
0 and b
( b
). Extend this process to
an inductivedefinition of a
, b
where f ( a
) 0 f ( b
), and
a
a
b
b
, with b
a
( b
a
).
13 Let f :[ a , b ] R bea continuous function with f ( a ) 0 and
f ( b ) 0. With thenotation of qn 12, supposethat theprocss of
repeated bisection of the interval [ a , b ] has not located a point at
which thefunction is 0.
 
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