Section 2.3b
Intermediate Value Property – describes a function that never
takes on two values without taking on all the values in between.
The Intermediate Value Theorem for Continuous Functions
A function y = f (x) that is continuous on a closed interval [a, b]
takes on every value between f (a) and f (b). In other words, if
y0 is between f (a) and f (b), then y0 = f (c) for some c in [a, b].
y  f  x
y
f b
y0
f a
0
a
c
b
x
Exploration 1: Removing a Discontinuity
x  7x  6
x  7x  6

f  x 
x2  9
 x  3 x  3
3
Let
3
1. Factor the denominator. What is the domain of f ?
Domain:
 , 3  3,3 3, 
2. Investigate the graph of f around x = 3 to see that f has a
removable discontinuity at x = 3.
It appears that the limit of f as x  3 exists and is a
little more than 3.
3. How should f be redefined at x = 3 to remove the
discontinuity? Use zoom-in and tables as necessary.
f  3 should be defined as 10 3
Exploration 1: Removing a Discontinuity
x  7x  6
x  7x  6

f  x 
x2  9
 x  3 x  3
3
Let
3
4. Show that (x – 3) is a factor of the numerator of f, and
remove all common factors. Now compute the limit as
x  3 of the reduced form for f.
x  3 x  1 x  2   x  1 x  2 

f  x 

for x  3
x3
 x  3 x  3
Thus,
x  1 x  2   3  1 3  2  20 10




lim
x 3
x3
33
6
3
Exploration 1: Removing a Discontinuity
x  7x  6
x  7x  6

f  x 
x2  9
 x  3 x  3
3
3
Let
5. Show that the extended function
 x  7x  6
,x 3

2
g  x   x  9
10 3, x  3

3
is continuous at x = 3. The function g is the continuous
extension of the original function f to include x = 3.
10
lim g  x  
 g  3 , so g is continuous at x = 3
x 3
3
Guided Practice
Give a formula for the extended function that is continuous
at the indicated point.
x 1
f  x  2 , x  1
x 1
3
x  1  x  1  x  x  1
x  1, f  x   2

x 1
 x  1 x  1
2
3
For
x  x 1

x 1
2
Guided Practice
Give a formula for the extended function that is continuous
at the indicated point.
x 1
f  x  2 , x  1
x 1
3
x  x 1 1 11 3
lim f  x   lim


x 1
x 1
x 1
11
2
2
2
The extended function:
 f ( x), x  1
g  x  
 3 2, x  1
Guided Practice
Sketch a possible graph for a function f that has the properties
stated below.
f  3
exists, but
lim f  x 
x 3
One possible answer:
does not.
y
y  f  x
3
x
Guided Practice
Sketch a possible graph for a function f that has the properties
stated below.
f
x

f

2




,
f  2  exists, xlim

2
but lim f  x 
x 2
One possible answer:
does not exist.
y
y  f  x
–2
x
Guided Practice
Sketch a possible graph for a function f that has the properties
stated below.
f  4
exists,
lim f  x  exists, but f is not continuous at x = 4.
x 4
One possible answer:
y
y  f  x
4
x
Guided Practice
Sketch a possible graph for a function f that has the properties
stated below.
f  x  is continuous for all x except x = 1, where f has a
non-removable discontinuity.
One possible answer:
y
y  f  x
x=1
x
Guided Practice
Find a value for a so that the given function is continuous.
 x  1, x  3
f  x  
2ax, x  3
2
We require that
What might the graph look like?
lim 2ax  lim  x  1
2
x 3
Substitute:
x 3
2
2a  3  3 1
6a  8
4
a
3
Does this answer make sense graphically?