Download 2.5 Continuity

Section 2.5 Notes Page 1
2.5 Continuity
Continuity: graph is connected with no breaks or holes.
Let f(x) be a graph. Continuity at x = c occurs if all three of the following are true:
1.) f ( x0 ) is defined. (There are no vertical asymptotes or holes at x  x 0 )
2.) lim f ( x) exists. (If it didn’t exist, then there must be a break in the graph or a vertical asymptote.)
x x0
3.) lim f ( x)  f ( x 0 ) (I can plug x0 in for x since there is not a hole).
x  x0
Let’s look at the below graph from the previous section. Can you tell which places the graph is discontinuous?
What value should be assigned to f(2) to make the function continuous at x = 2?
We see that the graph will be discontinuous at x = -2, 0, 2, and 4. .
The value of ½ should be assigned to f(2) to make it continuous.
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
1
f ( x)  (2  x) 5
For all these problems you want to look for any place where the equation will be undefined. This only occurs if
you are dividing by a negative number or taking the even root of a negative number. In this problem there are
no places where we are dividing by zero, and we have an odd root here since f(x) can be written as:
f ( x)  5 2  x . Therefore, it will be continuous at every value of x. So our answer is  ,  .


EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
f ( x)  6 x  35
We are only allowed to take the square root of positive numbers or zero. Therefore this is the only place where
the graph is continuous. So to solve this one, we will let everything inside the square root be greater than or
35
 35 
. In interval notation the answer is:  ,   .
equal to zero: 6x  35  0 . Solving this will give us x 
6
6

Section 2.5 Notes Page 2
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
1
f ( x)  2
x 1
In the problem above, since no x value will make the bottom zero then this graph will be continuous for all x.
So our answer is  ,   .
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
f ( x)  4 x  5  cos(2 x)
There is no place where there could be a division of zero or square root of a negative number. Therefore, this is
continuous at all values of x. Therefore our interval is:  ,   .
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
3
 7x
f ( x) 
x4
The only place where this is not going to be continuous is at x = 4. Therefore our interval is:
 , 4  (4, ) .
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
sin x
f ( x) 
5x  5
The only place where this is not going to be continuous is at x = 1. Therefore our interval is:
 , 1  (1, ) .
EXAMPLE: Fill in the statement: f(x) is continuous on  ,   except for:_______.
tan x
f ( x)  2
x 1
From the previous example we know that the bottom won’t ever be zero However we still need to consider the
sin x
top. We need to find where tangent could be undefined. Identities tell us that tan x 
so we need to see
cos x
 3 5
where cos x  0 . This happens at x  , , ,... . Therefore, f(x) is continuous on  ,   except for:
2 2 2
n
x
, where n is odd.
2
Section 2.5 Notes Page 3
EXAMPLE: Fill in the statement: f(x) is continuous on  ,   except for:_______.
f ( x)  4 csc(4 x)
4
. We need to see where sin(4 x)  0 . Suppose the 4x
sin( 4 x)
was just x. Then we know that would be zero at 0,  ,2 ,... . However we have 4x so basically
  3
4 x  0,  ,2 ,3 ,4 ... Therefore by solving for x we get x  0, , , ,  ... Therefore, f(x) is continuous on
4 2 4
 ,   except for n4 where n is any integer.
This one can be written with identities: f ( x) 
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
x
f ( x)  2
x 1
Here there will be two points of discontinuity, at 1 and -1 that refer to vertical
asymptotes as shown on the graph to the right. Therefore this is continuous on
(, 1)  (1, 1)  (1, ) .
EXAMPLE: Describe the set of x-values where the function is continuous, using interval notation:
x3
f ( x)  2
x  3x  10
x3
.
( x  2)( x  5)
Nothing cancels, so both x = -2 and x = 5 would be discontinuities.
Therefore, this will be continuous on  , 2   2, 5  5,   .
Notice our graph has two vertical asymptotes.
First factor the top and bottom. We will get: f ( x) 