WebAssign is not supported for this browser version. Some features or content might not work. System requirements

WebAssign

Welcome, demo@demo

(sign out)

Saturday, March 29, 2025 03:38 EDT

Home My Assignments Grades Communication Calendar My eBooks

My Cengage Home Notifications Help My Account

Math - College, section 1, Fall 2019
My Assignments

Stewart - Calculus 9/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 36 / 39

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

Ask Your Teacher Print Assignment

Question

Points

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
0/1	0/1	1/1	3/3	2/2	4/5	5/5	9/9	3/3	1/1	3/3	1/1	2/2	1/1	1/1

Total

36/39 (92.3%)

Instructions

Calculus, 9th edition by James Stewart, Daniel Clegg, and Saleem Watson, published by Cengage Learning, is widely renowned for its mathematical precision and accuracy, clarity of exposition, and outstanding examples and problem sets. The WebAssign enhancement to this textbook engages students with immediate feedback, rich tutorial content, video examples, interactive questions, and a media-rich eBook.

Questions 1, 2, 4, 6, 9, 13, and 15 have Watch Its.

Questions 1 and 2 have Master Its

Question 3 includes Enhanced Feedback designed to help guide students to the correct answer.

Question 4 is an Expanded Problem, which goes beyond the basic exercise and asks the student to show steps of their work or reasoning.

Question 5 is an Explore It question, which is an interactive resource focusing on the real world applications of Calculus. Explore Its allow Calculus students to work with animations and video explanations to deepen their understanding of key concepts by helping them visualize the concepts they are learning.

Questions 6 and 7 are questions that require the student to complete a proof.

Question 8 is an application question focused on an application from engineering.

Question 9 is a Just In Time exercise to remediate algebra skills required for the section.

Question 10 is from the challenging set of Problems Plus to stretch students' problem solving skills.

Question 11 is an application question focused on an application from biology.

Question 12 uses differential equation grading to test the validity of the answer. Accepts any correct form of the answer and runs student's responses through a series of tests to ensure the assumptions and requirements of the question are met.

Question 13 illustrates vector grading.

Question 14 is from one of the four diagnostic tests covering Basic Algebra, Analytic Geometry, Functions, and Geometry.

Question 15 is a question from the Quick Prep module on trigonometric identities. These modules can be used early in the course, or whenever the review is needed. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

We have also turned on the display of the solutions after the first submission so you can view them. Normally, instructors elect to display these solutions only after the due date.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 0/1 points | Previous Answers SCalc9 1.8.047.MI. My Notes

Let

f(x) =

	cx² + 2x		if x < 3
	x³ − cx		if x ≥ 3

Find the value of the constant c such that the function f continuous on (−∞, ∞)?

c =

7/4

Solution or Explanation

f(x) =

cx² + 2x	if x < 3
x³ − cx	if x ≥ 3

f is continuous on

(−∞, 3) and (3, ∞).

Now

lim x → 3⁻ f(x) = lim x → 3⁻ (cx² + 2x) = 9c + 6 and lim x → 3⁺ f(x) = lim x → 3⁺ (x³ − cx) = 27 − 3c.

So f is continuous ⇔

9c + 6 = 27 − 3c ⇔ 12c = 21 ⇔ c =

Thus, for f to be continuous on

(−∞, ∞), c =

Enhanced Feedback

Please try again. Choose c so that the function is continuous at the point where once piece of the function ends and the other begins.

Need Help?

Viewing Saved Work Revert to Last Response

2. 0/1 points | Previous Answers SCalc9 4.5.003.MI. My Notes

Evaluate the indefinite integral by using the indicated substitution. (Use C for the constant of integration.)

x²

x³ + 19

dx, u = x³ + 19

(19)·2π(x3+19)(32)+C

Need Help?

Viewing Saved Work Revert to Last Response

3. 1/1 points | Previous Answers SCalc9 1.6.046. My Notes

Find the limit, if it exists. (If an answer does not exist, enter DNE.)

lim x→−3

3 − |x|

3 + x

$webMathematica generated answer key$

Solution or Explanation

Since

|x| = −x

for

x < 0,

we have

lim x→−3

3 − |x|

3 + x

= lim x→−3

3 − (−x)

3 + x

= lim x→−3

3 + x

= lim x→−3 1 = 1.

Viewing Saved Work Revert to Last Response

4. 3/3 points | Previous Answers SCalc9 1.5.027.EP. My Notes

Consider the following.

lim x→7⁺

x + 6

x − 7

If x is chosen to be sufficiently close to 7 with x greater than 7, then f(x)

0 and f(x) can be made

arbitrarily large positive

Determine the infinite limit.

∞ −∞

Solution or Explanation

lim x→7⁺

x + 6

x − 7

= ∞

since the numerator is positive and the denominator approaches 0 from the positive side as

x → 7⁺.

Need Help?

Viewing Saved Work Revert to Last Response

5. 2/2 points | Previous Answers SCalc9 5.2.EI.001. My Notes

Review the Explore It, then use it to complete the exercise below.

Click here to access the Explore It in a new window.

(a)

For which of the three regions (given on the Explore & Test page of the Explore It) can the disk method alone be used to find the volume of the solid generated by revolving the region around the x-axis? (Select all that apply.)

Region 1: y = −

(12² − x²)

, y = 0

Region 2: y = (x − 3)² when x ≥ 3, x = 0, y = 4

Region 3: y = x² − 2x + 2, x = 0, x = 3, and y = 0

(b)

Region 1: y = −

(12² − x²)

, y = 0

Region 2: y = (x − 3)² when x ≥ 3, x = 0, y = 4

Region 3: y = x² − 2x + 2, x = 0, x = 3, and y = 0

Viewing Saved Work Revert to Last Response

6. 4/5 points | Previous Answers SCalc9 1.7.015. My Notes

Prove the statement using the ε, δ definition of a limit.

lim x → 3

x − 3

= 4

Given ε > 0, we need δ

> 0

such that if

0 < |x − 3| < δ,

then

3 +

− 4

< ε

But

3 +

− 4

< ε ⇔

x − 1

< ε ⇔

|x − 3| < ε ⇔ |x − 3| <

3ε

So if we choose

δ =

3ε

then

0 < |x − 3| < δ

3 +

− 4

< ε.

Thus,

lim x → 3

3 +

= 4

by the definition of a limit.

Illustrate with a diagram.

The x y coordinate plane is given. The line starts at a point on the positive y-axis, goes down and right, passes through the point (3, 4), and exits the window in the first quadrant. The point (3 − δ, 4 − ε) occurs just below the line. The point (3 + δ, 4 + ε) occurs just above the line.

The x y coordinate plane is given. The line starts at a point on the positive y-axis, goes up and right, passes through the point (3 − δ, 4 − ε), passes through the point (3, 4), passes through the point (3 + δ, 4 + ε), and exits the window in the first quadrant.

The x y coordinate plane is given. The line starts at a point on the positive y-axis, goes up and right, passes through the point (3 − ε, 4 − δ), passes through the point (3, 4), passes through the point (3 + ε, 4 + δ), and exits the window in the first quadrant.

Solution or Explanation

Given ε > 0, we need δ > 0 such that if

0 < |x − 3| < δ,

then

x − 3

− 4

< ε.

But

x − 3

− 4

< ε ⇔

x − 2

< ε ⇔

|x − 3| < ε ⇔ |x − 3| < 3ε.

So if we choose

δ = 3ε,

then

0 < |x − 3| < δ right double arrow implies

x − 3

− 4

< ε.

Thus,

lim x → 3

x − 3

= 4

by the definition of a limit.

Need Help?

Viewing Saved Work Revert to Last Response

7. 5/5 points | Previous Answers SCalc9 7.1.057. My Notes

Use integration by parts to prove the reduction formula.

(ln(x))ⁿ dx = x(ln(x))ⁿ − n

(ln(x))^{n − 1} dx

Let u = (ln(x))ⁿ, then dv =

dx

$webMathematica generated answer key$ . Then

du =

nln(x)n−1x

$webMathematica generated answer key$

and v =

x

$webMathematica generated answer key$ . By Equation 2, which states that

u dv = uv −

v du,

the integration by parts gives the following.

(ln(x))ⁿ dx

x

$webMathematica generated answer key$

(ln(x))ⁿ −

nln(x)n−1x

$webMathematica generated answer key$

x(ln(x))ⁿ − n

(ln(x))^{n − 1} dx

Viewing Saved Work Revert to Last Response

8. 9/9 points | Previous Answers SCalc9 1.7.012. My Notes

9. 3/3 points | Previous Answers SCalc9 1.6.JIT.003. My Notes

Find

f(a), f(a + h),

and the difference quotient

f(a + h) − f(a)

where h ≠ 0.

f(x) = 3 − 6x + 2x²

f(a)

3−6a+2a2

$webMathematica generated answer key$

f(a + h)

3−6(a+h)+2(a+h)2

$webMathematica generated answer key$

f(a + h) − f(a)

4a+2h−6

$webMathematica generated answer key$

Need Help?

Viewing Saved Work Revert to Last Response

10. 1/1 points | Previous Answers SCalc9 2.PP.012. My Notes

f(x) =

x⁴⁸ + x⁴⁷ + 2

1 + x

calculate

f ⁽⁴⁸⁾(3).

Express your answer using factorial notation:

n! = 1 · 2 · 3 ·

· (n − 1) · n.

48!297

$webMathematica generated answer key$

Solution or Explanation

f(x) =

x⁴⁸ + x⁴⁷ + 2

1 + x

x⁴⁷(x + 1) + 2

x + 1

x⁴⁷(x + 1)

x + 1

= x⁴⁷ + 2(x + 1)⁻¹,

f ⁽⁴⁸⁾(x) = (x⁴⁷)⁽⁴⁸⁾ + 2[(x + 1)⁻¹]⁽⁴⁸⁾.

The forty-eighth derivative of any forty-seventh degree polynomial is 0, so

(x⁴⁷)⁽⁴⁸⁾ = 0.

Thus,

f ⁽⁴⁸⁾(x) = 2[(−1)(−2)(−3)

(−48)(x + 1)⁻⁴⁹] = 2(48!)(x + 1)⁻⁴⁹

and

f ⁽⁴⁸⁾(3) = 2(48!)(4)⁻⁴⁹

(48!)2⁻⁹⁷.

Viewing Saved Work Revert to Last Response

11. 3/3 points | Previous Answers SCalc9 11.2.072. My Notes

A patient is injected with a drug every 12 hours. Immediately before each injection the concentration of the drug has been reduced by 90% and the new dose increases the concentration by 2.7 mg/L.

(a)

What is the concentration after three doses?

2.997

mg/L

(b)

If C_n is the concentration after the nth dose, find a formula for C_n as a function of n.

C_n =

3(1−0.1n)

$webMathematica generated answer key$

(c)

What is the limiting value of the concentration?

mg/L

Solution or Explanation

(a)

The concentration of the drug in the body after the first injection is 2.7 mg/L. After the second injection, there is 2.7 mg/L plus 10% (90% reduction) of the concentration from the first injection, that is,

[2.7 + 2.7(0.10)] = 2.97 mg/L.

After the third injection, the concentration is

[2.7 + 2.97(0.10)] = 2.997 mg/L.

(b)

The drug concentration is

0.1C_n

(90% reduction) just before the

nth + 1

injection, after which the concentration increases by 2.7 mg/L. Hence

C_{n + 1} = 0.1C_n + 2.7.

From

a_n = rⁿa + c

1 − rⁿ

1 − r

the solution to

C_{n + 1} = 0.1C_n + 2.7,

C₀ = 0 mg/L

C_n = (0.1)ⁿ(0) + 2.7

1 − 0.1ⁿ

1 − 0.1

2.7

0.9

(1 − 0.1ⁿ) = 3(1 − 0.1ⁿ).

(c)

The limiting value of the concentration is

lim n→∞ C_n = lim n→∞ 3(1 − 0.1ⁿ) = 3

lim n→∞ 1 − lim n→∞ 0.1ⁿ

= 3(1 − 0) = 3.

Viewing Saved Work Revert to Last Response

12. 1/1 points | Previous Answers SCalc9 9.3.010. My Notes

Solve the differential equation.

+ 2e^{t + z} = 0

z=−ln(2et−c)

$webMathematica generated answer key$

Solution or Explanation

+ 2e^{t + z} = 0 ⇒

−2e^te^z ⇒

e^−z dz

−2

e^t dt ⇒

−e^−z

−2e^t + C

e^−z

2e^t − C

e^z

2e^t − C

e^z

2e^t − C

= −ln(2e^t − C)

Viewing Saved Work Revert to Last Response

13. 2/2 points | Previous Answers SCalc9 12.4.019. My Notes

Find two unit vectors orthogonal to both

7, 4, 1

and

−1, 1, 0

(smaller i-value)

−1√123i−1√123j+11√123k

$webMathematica generated answer key$ (larger i-value)

1√123i+1√123j−11√123k

$webMathematica generated answer key$

Solution or Explanation

A theorem states that the vector a ✕ b is orthogonal to both a and b. By this theorem, the cross product of two vectors is orthogonal to both vectors. So we calculate

7, 4, 1

−1, 1, 0

i	j	k
7	4	1
−1	1	0

4	1
1	0

i −

7	1
−1	0

j +

7	4
−1	1

k = −i − j + 11k.

So two unit vectors orthogonal to both are

−1, −1, 11

1 + 1 + 121

= ±

−1, −1, 11

123

that is,

−

123

, −

123

and

123

, −

123

Need Help?

Viewing Saved Work Revert to Last Response

14. 1/1 points | Previous Answers SCalc9 DT.1.005b. My Notes

Simplify the rational expression.

4x² − 3x − 1

x² − 9

x + 3

4x + 1

(x−1)(x−3)

$webMathematica generated answer key$

If you have had difficulty with this problem, you may wish to consult the Review of Algebra on the website StewartCalculus.com.

Viewing Saved Work Revert to Last Response

15. 1/1 points | Previous Answers SCalc9 QP.20.009. My Notes

Write the expression in terms of sine only.

5(sin(2x) − cos(2x))

5√2sin(2x+7π4)

$webMathematica generated answer key$

Read more about Topic 20: Trigonometric Identities

Need Help?

Viewing Saved Work Revert to Last Response

Home My Assignments

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
0/1	0/1	1/1	3/3	2/2	4/5	5/5	9/9	3/3	1/1	3/3	1/1	2/2	1/1	1/1

1	2	3	4	5	6	7	8	9
1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1
1/50	1/50	1/50	1/50	1/50	1/50	1/50	2/50	2/50

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
0/1	0/1	1/1	3/3	2/2	4/5	5/5	9/9	3/3	1/1	3/3	1/1	2/2	1/1	1/1

WebAssign

Welcome, demo@demo

Stewart - Calculus 9/e (Homework)

Current Score : 36 / 39

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
0/1	0/1	1/1	3/3	2/2	4/5	5/5	9/9	3/3	1/1	3/3	1/1	2/2	1/1	1/1