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Continuity of Composite and Piece-wise Functions Worksheet

Continuity of Composite and Piece-wise Functions Worksheet
  • Page 1
 1.  
For the function f(x) = 2x + c for x ≤ 1,
                               = x2 + 3 for x > 1,
find the value of c which makes the function continuous.
a.
1 2
b.
1
c.
6
d.
2
e.
- 2


Solution:

For the function to be continuous, limx1- f(x) = limx1+ f(x)

limx1-(2x + c) = limx1+(x2 + 3)

2(1) + c = (1)2 + 3

2 + c = 4

c = 2
[Solve for c.]

The function f(x) is continuous at c = 2.


Correct answer : (4)
 2.  
Choose the correct statement for the function,
f(x) = x - 5 for x ≤ 1
     = 4x2 - 9 for 1 < x < 2
     = 2x + 1 for x ≥ 2
a.
f(x) is continuous at x = 2 but not continuous at x = 1
b.
f(x) is continuous on the interval [1, 2]
c.
f(x) is continuous at x = 1 and at x = 2
d.
f(x) is neither continuous at x = 1 nor at x = 2
e.
f(x) is continuous at x = 1 but not continuous at x = 2


Solution:

limx1- f(x) = limx1- (x - 5) = - 4
[Use f(x) = x - 5 for x ≤ 1.]

limx1+ f(x) = limx1+ (4x2 - 9) = - 5
[Use f(x) = 4x2 - 9 for x > 1.]

Since limx1- f(x) ≠ limx1+ f(x), limx1 f(x) does not exist, so f is not continuous at x = 1.

limx2- f(x) = limx2- (4x2 - 9) = 7
[Use f(x) = 4x2 - 9 for x < 2.]

limx2+ f(x) = limx2+ (2x + 1) = 5
[Use f(x) = 2x + 1 for x ≥ 2.]

Since limx2- f(x) ≠ limx2+ f(x), limx2 f(x) does not exist, so f is not continuous at x = 2.

So, f(x) is neither continuous at x = 1 nor at x = 2.


Correct answer : (4)
 3.  
For the function f(x) = cos (cx) for x < 1,
                               = x2 - 1 for x ≥ 1,
find the value(s) of c that makes the function continuous.
a.
(2k + 1)π2, where k is an integer
b.
kπ, where k is an integer
c.
π2 only
d.
π only
e.
0 only


Solution:

For the function to be continuous, limx1- f(x) = limx1+ f(x)

limx1- cos (cx) = limx1+ (x2 - 1)

cos ((1)c) = (1)2 - 1

cos (c) = 0

c = (2k + 1)π2 , where k is an integer.
[The general solution of cos x = 0 is x = (2k + 1)π2, where k is an integer.]

The function f(x) is continuous, when c = (2k + 1)π2, where k is an integer.


Correct answer : (1)
 4.  
For the function f(x) = | 2x - c | for x < - 3,
                               = 2 - x2 for x ≥ - 3,
find the value of c which makes the function continuous.
a.
does not exist
b.
- 7
c.
13
d.
1
e.
- 13


Solution:

limx- 3- f(x) = limx- 3- | 2x - c | = | 2(- 3) - c | = | - 6 - c |
[Evaluate limx- 3- f(x) by using the definition of f(x) for x < - 3.]

limx- 3+ f(x) = limx- 3+ (2 - x2) = (2 - (- 3)2) = - 7
[Evaluate limx- 3+ f(x) by using the definition of f(x) for x ≥ - 3.]

For the function f(x) to be continuous, limx- 3- f(x) = limx- 3+ f(x)

| - 6 - c | = - 7

The value of c does not exist.
[|- 6 - c| > 0.]


Correct answer : (1)
 5.  
Find the value of k for which the function,
               f(x) = e3x + k for x ≥ 0,
                    = 2x + 1 for x < 0
is continuous.
a.
1
b.
e
c.
- 1
d.
does not exist


Solution:

limx0- f(x) = limx0- (2x + 1) = 2(0) + 1 = 1
[Evaluate limx0- f(x) by using the definition of f(x) for x < 0.]

limx0+ f(x) = limx0+ e3x + k = e3(0) + k = ek
[Evaluate limx0+ f(x) by using the definition of f(x) for x ≥ 0.]

For the function to be continuous, limx0- f(x) = limx0+ f(x)

1 = ek

k = 0
[Solve for k.]

So, the function f(x) is continuous at k = 0.


Correct answer : (5)
 6.  
If f is given by f(x) = 6 for x < 1,
                             = (2x - 5)(x + 3) for x ≥ 1,
then which of the following statements is true?
a.
limx1- f(x) = f(1)
b.
f(x) is continuous at x = 1
c.
limx1+f(x) ≠ f(1)
d.
f(x) is not continuous at x = 1
e.
f(1) is not defined


Solution:

limx1- f(x) = limx1- 6 = 6
[Evaluate limx1- f(x) by using the definition of f(x) for x < 1.]

limx1+ f(x) = limx1+ (2x - 5)(x + 3) = (2(1) - 5)(1 + 3) = - 12
[Evaluate limx1+ f(x) by using the definition of f(x) for x ≥ 1.]

f(1) = (2x - 5)(x + 3) = (2(1) - 5)(1 + 3) = - 12
[Find f(1).]

Since limx1- f(x) ≠ limx1+ f(x), f(x) is not continuous at x = 1.


Correct answer : (4)
 7.  
Which of the following statements is true for the function,
        f(x) = x for x < 0,
             = 1x2+4 for x ≥ 0?
a.
limx0+ f(x) = 0
b.
limx0- f(x) ≠ 0
c.
f(0) is not defined
d.
f(x) is continuous at x = 0
e.
f(x) is not continuous at x = 0


Solution:

limx0- f(x) = limx0- x = 0
[Evaluate limx0- f(x) by using the definition of f(x) for x < 0.]

limx0+ f(x) = limx0+ 1x2+4 = 1(0)2+4 = 1 / 4
[Evaluate limx0+ f(x) by using the definition of f(x) for x ≥ 0.]

Since limx0- f(x) ≠ limx0+ f(x), f(x) is not continuous at x = 0.


Correct answer : (5)
 8.  
Find the values of a and b for the function to be continuous.
f(x) = 2 for x ≤ 1
      = - ax2 + b for 1 < x < 2
      = 8     for x ≥ 2
a.
2, 8
b.
1, 2
c.
0, 1
d.
2, 0
e.
- 2, 0


Solution:

limx1- f(x) = limx1- 2 = 2
[Use f(x) = 2 for x ≤ 1.]

limx1+ f(x) = limx1+ - ax2 + b = - a + b
[Use f(x) = ax2 + b for 1 < x < 2.]

For f(x) to be continuous at x = 1, limx1- f(x) = limx1+ f(x)

2 = - a + b ------------(1)

limx2- f(x) = limx2- - ax2 + b = - 4a + b
[Use f(x) = ax2 + b for 1 < x < 2.]

limx2+ f(x) = limx2+ 8 = 8
[Use f(x) = 8 for x ≥ 2.]

For f(x) to be continuous at x = 2, limx2- f(x) = limx2+ f(x)

8 = - 4a + b ------------(2)

a = - 2, b = 0
[Solve the equations (1) and (2).]

Therefore, the function is continuous when a = - 2 and b = 0.


Correct answer : (5)
 9.  
For the function f(x) = (2x+c) for x < 2,
                               = x2 - 4 for x ≥ 2,
find the value of c which makes the function continuous.
a.
1
b.
2
c.
- 4
d.
4


Solution:

limx2- f(x) = limx2- (2x+c) = 4+c
[Evaluate limx2- f(x) by using the definition of f(x) for x < 2.]

limx2+ f(x) = limx2+ (x2 - 4) = 4 - 4 = 0
[Evaluate limx2+ f(x) by using the definition of f(x) for x ≥ 2.]

For the function f(x) to be continuous, limx2- f(x) = limx2+ f(x)

4+c = 0

c = - 4
[Solve for c.]

The function f(x) is continuous at c = - 4.


Correct answer : (4)
 10.  
Which of the following is correct for the function f(x) from the graph?

a.
f(x) is not continuous at x = - 6
b.
f(x) is not continuous at x = 4
c.
f(x) is not continuous at x = - 2
d.
f(x) is not continuous at x = 6
e.
f(x) is not continuous at x = 0


Solution:

limx4- f(x) = 2, limx4+f(x) = 0
[From graph.]

f(4) = 0

Since, limx4- f(x) ≠ limx4+ f(x), f(x) is discontinuous at x = 4.

Therefore, f(x) is not continuous at x = 4.


Correct answer : (2)

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