﻿ Continuity of Composite and Piece-wise Functions Worksheet | Problems & Solutions # Continuity of Composite and Piece-wise Functions Worksheet

Continuity of Composite and Piece-wise Functions Worksheet
• Page 1
1.
For the function $f$($x$) = 2$x$ + $c$ for $x$ ≤ 1,
= $x$2 + 3 for $x$ > 1,
find the value of $c$ which makes the function continuous. a. $\frac{1}{2}$ 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.

2.
Choose the correct statement for the function,
$f$($x$) = $x$ - 5 for $x$ ≤ 1
= 4$x$2 - 9 for 1 < $x$ < 2
= 2$x$ + 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.

3.
For the function $f$($x$) = cos ($c$$x$) for $x$ < 1,
= $x$2 - 1 for $x$ ≥ 1,
find the value(s) of $c$ that makes the function continuous. a. (2$k$ + 1)$\frac{\pi }{2}$, where $k$ is an integer b. $k$π, where $k$ is an integer c. $\frac{\pi }{2}$ 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.

4.
For the function $f$($x$) = | 2$x$ - $c$ | for $x$ < - 3,
= 2 - $x$2 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.]

5.
Find the value of $k$ for which the function,
$f$($x$) = $e$3$x$ + $k$ for $x$ ≥ 0,
= 2$x$ + 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.

6.
If $f$ is given by $f$($x$) = 6 for $x$ < 1,
= (2$x$ - 5)($x$ + 3) for $x$ ≥ 1,
then which of the following statements is true? a. $\underset{x\to 1-}{\mathrm{lim}}$ $f$($x$) = $f$(1) b. $f$($x$) is continuous at $x$ = 1 c. $\underset{x\to 1+}{\mathrm{lim}}$$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.

7.
Which of the following statements is true for the function,
$f$($x$) = $x$ for $x$ < 0,
= $\frac{1}{{x}^{2}+4}$ for $x$ ≥ 0? a. $\underset{x\to 0+}{\mathrm{lim}}$ $f$($x$) = 0 b. $\underset{x\to 0-}{\mathrm{lim}}$ $f$($x$) ≠ 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.

8.
Find the values of $a$ and $b$ for the function to be continuous.
$f$($x$) = 2 for $x$ ≤ 1
= - $\mathrm{ax}$2 + $b$ 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.

9.
For the function $f$($x$) = $\sqrt{\left(2x+c\right)}$ for $x$ < 2,
= $x$2 - 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.

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.