C4 – Classifying Discontinuities
A. Determine if the function is continuous or not. If it is not continuous, identify the value(s) of x at which any
discontinuity occurs and classify the discontinuity
1. f  x   x2  8x  20
2. g  x   x2  1
3. h x  3 x  5
4. r  x  
x 2  25
x 5
x4
7. w  x  
x4
6. q  x  
2x
x3
x2  4
x2  4
x 2  x  12
8. t  x  
x 2  16
5. v  x  
x 3
x2  3
2x
12. b  x  
log 2 x
9. f  x  
x2  2 x
x3  4 x
x2  9
11. n  x   4
x  27 x
10. m  x  
x2  2x  1
2x 1  8
x7
15. h  x  
x7
14. g  x   x  2
13. c  x  
17. z  x  
19. j  x  
x7
x7
2x  5
16. f  x  
18. i  x  
x7
x7
x 2  9 x  18
x 3
2 x2  5x
Determine if the function is continuous or not. If it is not continuous, determine the value(s) of x at which any
discontinuity occurs and classify the discontinuity.
2, x   , 0 
20. u  x   
 x  2, x  0,  
2, x   , 0 
21. c  x   
 x  2, x   0,  
2, x   , 0 
22. w  x   
 x  1, x  0,  
2, x   , 0 

23. k  x    x  10
 x  5 , x   0,  

 x 2  1, x   , 2

24. e  x    2 x  1, x   2, 2 
1
 x  4, x   2,  
2
B Find the value of the constant p so that the function is continuous for all real numbers.

2 px  7, x   , 3
1. f  x    2

2 x  5 x  2, x   3,  
2, x   ,1

2. Find the value of the constants p and q so that the function g  x    px 2  qx, x  1, 4  is continuous for all real
2 x  20, x   4,  

numbers.
C. Define a function’s value so that a discontinuity can be removed. (Hint: use the hole location principle of C3)
2 x  22
1. Define f 11 so that the function f  x   2
will be continuous at x  11 .
x  121
x 2  16
will be continuous at x  4 .
2x  8
 x 
3. Define h  0  so that the function h  x   sin 
 will be continuous at x  0 .
 2x 
x 9
4. Define i  9  so that the function i  x  
will be continuous at x  9 .
x 3
2. Define g  4  so that the function g  x  
D. Sketch the graph of a function given particular discontinuities.
1. Sketch the graph of a function that is continuous everywhere except at x  2 where the function has an infinite
discontinuity.
2. Sketch the graph of a function that is continuous everywhere except at x  2 where the function has a jump
discontinuity and at x  3 where the function has a removable discontinuity.
C4 – Classifying Discontinuities Answer Key
A. Determine if the function is continuous or not. If it is not continuous, identify the value(s) of x at which any
discontinuity occurs and classify the discontinuity
1. continuous
2. continuous
3. continuous
4. infinite discontinuity at x = -3
5. removable discontinuity at x = 5
6. continuous
7. removable discontinuity at x = -4
8. infinite dis. @ x = -4 and removable dis. @ x = 4
9. infinite dis. @ x = -2 and removable dis. @ x = 0, 2
10.
infinite dis. @ x = 3
x 3
removable dis @ x =  3
m x 
x 3 x 3

11.
n x 
( x  3)( x  3)
x( x  3)( x 2  3x  9)
infinite dis. @ x = 0
removable dis. @ x = -3
infinite dis. @ x = 4
( x  1) 2
c
x



x

1
13.
2 8
Solve 2 x 1  8
15. jump discontinuity at x = -7
17. infinite discontinuity @ x = -7
19. infinite discontinuity @ x = 0
jump discontinuity @ x = 5/2
21. removable discontinuity @ x = 0
23. infinite discontinuity @ x = 5

12. b  x  

infinite dis. @ x = 1
2x
log 2 x
14. continuous
16. removable discontinuity @ x = -7
18. jump discontinuity at x = 3
20. continuous
22. jump discontinuity at x = 0
24. jump discontinuity at x = -3
B 1. 2 px  7 and 2x2  5x  2 are continuous because all polynomial functions are continuous (Continuity
Principles). Thus for the split function to be continuous, we must be sure that at the value of x where the function is
split, the y values are the same. In other words we must be sure that lim f  x   lim f  x  . Thus
x 3
x 3
2 p  3  7  2  3  5  3  2 . Solving the resulting equation 6 p  7  31 , yields p  4 . Thus for f  x  to
2
be continuous, p  4 .
.
2. lim g  x   lim g  x 


x 1
x 1
lim g  x   2
lim g  x   lim g  x 
x  4
x 4
lim g  x   2 x  20  2(4)  20  24
16 p  4q  24
p  2  q
16( 2  q )  4q  24
x 1
x  4
32  16q  4q  24
x 1
x  4
p (1) 2  q (1)  2
p  q  2
p(4) 2  q(4)  24
16 p  4q  24
12q  56
lim g  x   px 2  qx  2
lim g  x   px 2  qx  24
q
14
3
 14  20
p  2  

3
 3 
2( x  11)
2
1
 lim

( x  11)( x  11) x11 x  11 11
f 11 
x 2  16
( x  4)( x  4)
x4
 lim
 lim
8
x 4 2 x  8
x 4
x

4
2( x  4)
2
x
 
3. limsin 
  limsin    1
x 0
 2 x  x0  2 
g  4  8
C1. lim
x 11
2. lim
h  0  1
1
11
4. lim
x 9
x 9
 lim
x  3 x9

x 3

x 3
x 3
  lim
x 9
x 3  6
i 9  6
D. 1. The graph must have a vertical asymptote at x = 2.
2. The graph must have a hole at x = -3 and a vertical break at x = 2.