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Lesson 5 Continuity
Definition A function f is said to be continuity at the number x = a if all of the
following three conditions are satisfied:
1.
2.
3.
f is defined on some neighborhood of x = a (NOTE: This is not a deleted
neighborhood of x = a. Thus, f is defined at x = a.)
lim f ( x) exists
xa
lim f ( x)  f ( a )
xa
Notation: If f is not continuous at x = a, then we say that f is discontinuous at
x = a.
Examples Determine if the following functions are continuous at x = a for the
given value of a.
1.
f ( x)  2 x 2  x  5 ; a = 3
1.
f is defined on some neighborhood of x = 3, namely (   ,  ) .
2.
lim (2 x 2  x  5)  18  3  5  16 . Thus, lim f ( x) exists.
x3
x3
3.
f (3)  18  3  5  16 Thus, lim f ( x)  f (3)
x3
Answer: f is continuous at x = 3
2.
x2  2
g ( x) 
;a=0
2x  1
1.


1
2
g is defined on some neighborhood of x = 0, namely    ,  .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
2.
x2  2 0  2
lim

  2 . Thus, lim g ( x) exists.
x  0 2x  1
x0
0 1
3.
g ( 0) 
02
  2 Thus, lim g ( x)  g (0)
x0
0 1
Answer: g is continuous at x = 0
3.
 x 2  3x , x  2
h( x )  
; a=2
 x4 ,x2
1.
2.
h is defined on some neighborhood of x = 2, namely (   ,  ) .
lim h( x) = lim  ( x  4)  2  4   2
x2
x2 
x  2   x  2  h( x)  x  4
lim  h( x) = lim  ( x 2  3x)  4  6   2
x2
x2
x  2   x  2  h( x)  x 2  3x .
h( x)   2 . Thus, lim h( x) exists.
Thus, xlim
2
x2
3.
h(2)  2  4   2 Thus, lim h( x)  h(2)
x2
Answer: h is continuous at x = 2
4.
g ( x) 
1.
1
;a=3
x3
g is not defined on any neighborhood of x   3 .
Answer: g is discontinuous at x   3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
5.
  2x 2 , x  1
f ( x)  
; a=1
5  x , x  1
1.
2.
f is defined on some neighborhood of x = 1, namely (   ,  ) .
lim f ( x) = lim  (5  x)  5  1  4
x 1
x 1 
x  1  x  1  f ( x)  5  x
lim  f ( x) = lim  ( 2 x 2 )   2
x 1
x 1
x  1  x  1  f ( x)   2 x 2
f ( x) = DNE.
Thus, lim
x 1
Answer: f is discontinuous at x = 1
NOTE: The discontinuity at x = 1 is called a jump discontinuity because “the
graph of the function jumps” at the point where x = 1.
This graph was created using Maple.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
6.
 16  x 2
,x4

h( x )   x  4
;a=4
 6
,x4

1.
h is defined on some neighborhood of x = 4, namely (   ,  ) .
2.
(4  x)( 4  x)
16  x 2
lim
lim h( x) = xlim
= x4
=
4
x4
x4
x4
lim [ (4  x)]   (4  4)   8
x4
3.
h( x)  h(4)
By definition of the function h, h( 4)  6 . Thus, xlim
4
Answer: h is discontinuous at x = 4
NOTE: The discontinuity at x = 4 is called a removable discontinuity
because if the value of the function h at 4 was defined to be  8 , that is
h(4)   8 , then h would be continuous at x = 4.
Definition The function f is said to be continuous on the open interval (a, b) if f is
continuous at each number in the interval.
Definition The function f is said to be continuous on the closed interval [a, b] if f
f ( x)  f (a) and
is continuous on the open interval (a, b) and xlim
 a
lim f ( x)  f (b) .
x  b
f ( x)  f (a) , we say that f is continuous at x = a from the
Notation: When xlim
 a
f ( x)  f (b) , we say that f is continuous at x = b from the left.
right. When xlim
 b
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
The Graphical Aspect of Continuity on an Interval: If you can draw the graph
of a function on an interval without lifting your pencil (or pen), then the function is
continuous on that interval. (This property of continuous functions can be proved
in a mathematics course called Topology using the property of connectedness.)
Some familiar functions which are continuous:
1.
2.
3.
4.
Polynomials are continuous for all real numbers.
Rational functions are continuous wherever they are defined.
Root functions are continuous wherever they are defined.
The trigonometric functions are continuous wherever they are defined.
Theorem If the functions f and g are continuous at x = a and c is a constant, then
the following functions are continuous at x = a:
1.
f g
2.
4.
fg
5.
f g
cf
3.
f
g (a)  0
g , provided that
Examples Use the theorem above to determine the continuity of the following
functions.
1.
f ( x )  sin 2 x 
x  4
The function y  x  4 is continuous on its domain of definition, which
2
is the interval [ 4 ,  ) . The function y  sin x = ( sin x ) ( sin x ) is the
product of the function y  sin x with itself, which is continuous on its
domain of definition, which is the set of real numbers. Thus, the function
y  sin 2 x is continuous on the set of real numbers since it is the product of
two continuous functions. Thus, the function f given by
f ( x )  sin 2 x  x  4 is the difference of two continuous functions
for the set of real numbers given by [ 4 ,  ) . Thus, the function f is
continuous on the interval [ 4 ,  ) .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
2.
x 2  9x  5
g( x ) 
x  cos x
2
The polynomial function y  x  9 x  5 is continuous on its domain of
definition, which is the set of all real numbers. The function y  x  cos x
is the sum of the two functions y  x and y  cos x , which are continuous
on the set of real numbers. Since x  cos x  0 when cos x   x , then the
quotient function g is continuous on the set of real numbers such that
cos x   x .
  2 x2 , x  1
Example Find the sign of the piecewise function f ( x)  
. That is,
 5  x , x 1
find the values of x for when f (x ) is positive and find the values of x for when
f (x ) is negative.
NOTE: The domain of this function is the set of all real numbers. Thus, the
function is defined for all real numbers.
We can use a sketch of the graph of the piecewise function in order to solve this
problem. Since the value of the function f at x is f (x ) , then we can get these
values by using the y-coordinate of points on the graph.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
From the graph, we can see that the function f is positive on the interval (1 , 5 )
and that the function f is negative on the interval (   , 0 )  ( 0 , 1)  ( 5 ,  ) .
That is, f ( x)  0 on the interval (1 , 5 ) and f ( x )  0 on the interval
(   , 0 )  ( 0 , 1)  ( 5 ,  ) . This can be summarized by the following:



0

+

1
Sign of f (x )

5
NOTE: To find the sign of f (x ) , Step 1 of the three step process in Lesson 1
would have us doing the following two things: a) find when f ( x)  0 and b) find
when f (x) is undefined. We can see from the sketch of the graph of the function f
above, that f ( x)  0 when x  0 and x  5 . However, since the domain of f is
the set of all real numbers, f (x ) is defined everywhere.
So, how did the number 1 get on the number line above and more importantly, why
did the function change signs at this number? The function f is discontinuous at
x  1 . In general, expressions can change signs at discontinuities. Thus, in general
if you want to solve a non-linear inequality, the two “doors” that allow an
expression the opportunity to change signs are a) numbers where the expression is
equal to zero and b) numbers where the expression is discontinuous.
NOTE: The discontinuities of a rational expression (quotient of polynomials) is
exactly where the expression is undefined since a rational expression is continuous
on its domain of definition.
Theorem (The Intermediate Value Theorem) Let the function f be continuous on
the closed interval [a, b] and let N be any number between f (a ) and f (b ) , where
f (a)  f (b) . Then there exists a number c in the open interval (a, b) such that
f (c )  N .
The Intermediate Value Theorem (IVT) can be used to solve equations numerically.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Example
Using the Intermediate Value Theorem and Maple to find an
2
approximation to the solution of the equation x  x  1 in the interval (1, 2).
Calculations without using the graphs.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850