Extra Practice - Continuity
1.
 x  7, x  2

Is the function f  x   9, x  2
continuous at the point x  2 ? Explain.
3x  3, x  2

2.
 4 x 2  2 x, x  3

Is the function f  x   10 x  1, x  3 continuous at the point x  3 ? Explain.
30, x  3

3.
Is the function f  x   
4.
Is the function f  x   sec x continuous everywhere? Explain.
5.
Is the function f  x   sec x continuous on the closed interval  
6.
Is the function f  x   sec x continuous on the open interval  
7.
For what values of k is the function f  x   
8.
6 x  12, x  3

For what values of k is the function f  x   k 2  5k , x  3 continuous at x  3 ?
6, x  3

9.
x 2  5 x  24
At what point is the removable discontinuity for the function f  x   2
?
x  x6
5 x  7, x  3
continuous at the point x  3 ? Explain.
7 x  1, x  3
  
,
? Explain.
 2 2 
  
,  ? Explain.
 2 2
kx  5, x  4
2
 x  x, x  4
continuous at x  4 ?
Answers:
1.
Yes, the function is continuous at x  2 .
Condition 1:
Condition 2:
f  2  9 , so f  2  exists. Therefore, condition 1 is satisfied.
lim f  x   9 and lim f  x   9 , so lim f  x   9 . Therefore, condition 2 is
x2
x 2
x2
satisfied.
Condition 3:
lim f  x   9  f  2  . Therefore, condition 3 is satisfied.
x 2
Since all three conditions are satisfied, f  x  is continuous at x  2 .
2.
No, it fails condition 3.
Condition 1:
Condition 2:
f  3  29 , so f  2  exists. Therefore, condition 1 is satisfied.
lim f  x   30 and lim f  x   30 , so lim f  x   30 . Therefore, condition
x 3
x3
x 3
2 is satisfied.
Condition 3:
lim f  x   f  3 . Therefore, condition 3 is not satisfied.
x 3
Since not all three conditions are satisfied, f  x  is not continuous at x  3 .
3.
No, it fails condition 1, since the function is not defined at f  3 .
4.
No, it is discontinuous at any odd integral multiple of
5.
No, it is discontinuous at the endpoints of the interval.
6.
Yes. Since sec x 

.
2
1

, sec x is undefined when cos x  0 . Since cos  0 and
cos x
2
 
  
cos     0 , sec x is defined everywhere on   ,  because the interval does not include
 2
 2 2
the endpoints.
7.
7
.
4
The function is continuous for k 
We need to find a value (of values) of k which satisfy the conditions of continuity.
Because f  4  42  4  12 , the first condition is satisfied.
For Condition 2 to be satisfied, lim f  x   4k  5 must equal lim f  x   12 . So, set
x 4
4k  5  12 and solve for k. If k 
x 4
7
, the limit will exist at x  4 and the other two conditions
4
will be fulfilled.
8.
The function is continuous for k  6 or k  1 .
We need to find a value (of values) of k which satisfy the conditions of continuity.
Condition 1: f  3  k 2  5k
Condition 2: lim f  x   6 and lim f  x   6 , so lim f  x   6 .
x3
x3
x 3
Condition 3: lim f  x   6  f  3 . So, k 2  5k  6  k  6 or k  1 .
x 3
9.


The removable discontinuity is at the point  3,
11 
.
5
x 2  5 x  24  x  8  x  3 x  8
f  x  2


, x  2 or x  3 .
x  x6
 x  2  x  3 x  2
The removable discontinuity occurs at the x-value of the canceled factor, which is x  3 .
In the reduced version, f  x  
x 8
3  8 11
 .
, we have f  3 
x2
3 2 5


Therefore, the removable discontinuity is at the point  3,
11 
.
5