3.2 - 
 
 MATH
1081
 Monday,
January
31
 
...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 
 
 MATH
1081
 Monday,
January
31
 
 Chapter
3
–
Section
2
 CONTINUITY
 Homework
#3
(due
2/7):


 Section
3.2
#6,
20,
26
 
 
 
 
 
 
 
 
 Section
3.3
#8,
30,
34
 
 Clicker
Check­In:

Choose
any
letter
to
check
in
now. Now,
we
saw
in
section
3.1
that
for
many
(although
certainly
 not
all)
functions
at
most
choices
of
a,

 
 
 
 
 
 
 
 
 lim f ( x ) = f ( a).
 
 
 
 That
is,
the
value
of
the
limit
coincides
with
the
value
of
the
 function.

So,
the
point
at
 x = a
is
where
we
would
“expect”
it
to
 be.
 x →a For
us,
this
is
precisely
the
criterion
for
continuity
of
a
function
 at
a
point.
 
 
 A
very
informal
notion
of
continuity
at
a
point
is
that
there
is
 no
hole
or
gap
or
jump
or
asymptote.

The
curve
can
be
drawn
 through
that
point
without
lifting
the
pencil.
 
 
 This
is
what
this
limit
equation
also
says…as
 x 
gets
close
to
 a ,
 the
values
of
the
function
are
getting
close
to
a
limiting
value
 AND
that
happens
to
be
exactly
equal
to
the
value
that
the
 function
takes
on
at
 x = a.
 The
book
describes
this
with
a
3‐part
definition.
 
 A
function
f
is
continuous
at
 x = a
if
the
following
3
conditions
 are
all
satisfied:
 
 1.

 f ( a) 
is
defined.


 (So
the
function
has
a
value
at
 x = a.)
 
 2.

 lim f ( x ) 
exists.
 
 (So
the
function
is
tending
to
one
specific
value.)
 
 3.

 lim f ( x ) = f ( a).
 
 x →a x →a (The
value
of
the
function
and
the
limit
are
the
same.)
 Example:

Determine
if
the
function
is
continuous
at
the

 indicated
point.

If
not,
state
which
of
the



 criteria
fail.
 
 1. f ( x ) = 9 x 2 + 5 
 at

 x = 1
 
 x2 − 4 2. f ( x ) = 

 x−2 at

 x = 2
 at

 x = 3
 
 Note,
if
a
function
is
not
continuous
at
a
point,
it
is
said
to
be
 discontinuous
there. x+9 3. f ( x ) = 

 x−3 
 
 • Polynomial
functions
are
continuous
for
all
 x .
 • Rational
functions
are
continuous
for
all
 x 
in
the
domain.
 • Exponential
functions
are
continuous
for
all
 x .
 • Logarithmic
functions
are
continuous
for
all
 x 
in
the
 domain.
 • Radical
(or
Root)
functions
are
continuous
for
all
 x 
in
the
 domain.
 Now,
as
we
observed
last
time,
 lim x 
doesn’t
exist.

So,
can
 the
function
 f ( x ) = x 
continuous
at
 x = 0 ?


 
 
 In
the
sense
that
there
is
no
hole,
gap,
jump,
vertical
asymptote
 at
 x = 0 
it
satisfies
the
intuitive
notion
of
continuous.
 
 However,
in
the
definition
that
 lim f ( x ) = f ( 0 ),
we
have
an
 issue,
since
the
limit
does
not
exist.
 
 x→ 0 x→ 0 So,
to
deal
with
this
and
other
like
situations,
we
can
define
a
 function
as
continuous
on
a
closed
interval
[ a, b ]
if
 
 1. It
is
continuous
at
all
points
in
the
open
interval
( a, b ).
 2. It
is
continuous
from
the
right
at
 x = a .

That
is,
 lim+ f ( x ) = f ( a ) .
 3. It
is
continuous
from
the
left
at
 x = b .

That
is,
 lim− f ( x ) = f ( b ) .
 
 Extending
this
in
the
obvious
way,
we
can
say
that
 f ( x ) = x 
is
 continuous
on
its
domain
[ 0, ∞ ).
 
 x→b x→ a A
particular
kind
of
function
that
requires
a
bit
more
analysis
is
 one
that
is
defined
not
as
a
single
algebraic
expression,
but
 maybe
two
or
more
pieces.

These
are
referred
to
as
 (appropriately
enough)
piecewise­defined
functions.
 
 For
example,
 
 is
such
a
function.

To
the
left
side
of
 x = 2 ,
the
function
 behaves
like
the
function
 f1 ( x ) = x + 4 .

To
the
right
side
of
 x = 2 ,
the
function
behaves
like
 f2 ( x ) = x 2 + 1.
 ⎧ x + 4 for f ( x) = ⎨ 2 ⎩ x + 1 for x<2 
 x≥2 Both
 f1 ( x ) = x + 4 
and
 f2 ( x ) = x 2 + 1
are
polynomials
which
are
 continuous
everywhere.

So
the
only
real
questionable
point
of
 ⎧ x + 4 for continuity
for
 f ( x ) = ⎨ 2 ⎩ x + 1 for x→ 2 x<2 
is
at
 x = 2 .
 x≥2 
 To
evaluate
 lim f ( x ) ,
we
MUST
analyze
each
one‐sided
limit
 separately,
since
the
function
is
defined
by
a
different
 expression
to
the
left
of
 x = 2 
than
to
the
right
of
 x = 2 .
 
 What
are
these
one‐sided
limits?


 Is
 f ( x ) 
continuous
at
 x = 2 ?
 Example:

Determine
all
values
of
 x 
where
the
function
is
 
 
 
 
discontinuous.

Which
of
the
3
criteria
for

 
continuity
fails
in
each
case?
 
 ⎧2 if x < −1 ⎪ 
 
 
 1.

 f ( x ) = ⎨ 3 + x if −1 < x ≤ 3 
 ⎪ 4 − x if x>3 ⎩ 
 ⎧ x 2 + 1 if x < 0 ⎪ 
 
 
 2.

 g ( x ) = ⎨ −1 if x = 0 
 ⎪ ex if x > 0 ⎩ 
 Clicker
Check­out:
Choose
any
letter
to
check
out
now.
 
 Next
time
–
Section
3.3

 
 ...
View Full Document

This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

Ask a homework question - tutors are online