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Larson & Edwards - Calculus 11/e HS (Homework)

James Finch

Math - High School, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 6 / 23

Due : Friday, August 16, 2019 20:30 EDT

Last Saved : n/a Saving...  ()

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Question
Points
1 2 3 4 5 6 7 8
–/7 –/4 –/3 –/2 –/1 –/1 –/1 –/1
Total
6/23 (26.1%)

  • Instructions

    Calculus, 11th edition (AP edition), by Larson and Edwards and published by Cengage Learning, carefully integrates engaging and challenging content with technology products for successful teaching and learning, ensuring that the exercise sets are rigorous and relevant. Multi-step, real-life exercises reinforce problem-solving skills and mastery of concepts by giving students the opportunity to apply the concepts in real-life situations. The Larson Calculus program has been widely praised by a generation of students and professors for its solid and effective pedagogy that addresses the needs of a broad range of teaching and learning styles.

    Question 1 showcases the ability to enter table values to summarize limit numerical answers.

    Question 2 highlights prompts describing the needed answer format and includes a pull down for word answers, such as “removable,” for continuity.

    Question 3 is a Just-In-Time review question with emphasis on trigonometry grading.

    Question 4 features answers presented as a comma-separated list emphasizing the Fundamental Theorem of Calculus.

    Question 5 exhibits differential equation grading.

    Question 6 highlights the use of calcPad for integration problems and integral grading.

    Question 7 includes a Master It with step-by-step tutorial for Taylor polynomials.

    Question 8 illustrates vector grading. 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.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

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. –/7 points LarCalc11HS 1.2.016. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7
–/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/7
 
Create a table of values for the function and use the result to estimate the limit. Use a graphing utility to graph the function to confirm your result. (Round your answers to four decimal places. If an answer does not exist, enter DNE.)
lim x→4 
[x/(x + 1)] − (4/5)
x − 4
x 3.9 3.99 3.999 4.001 4.01 4.1
f(x) (No Response) seenKey

0.0408

 		(No Response) seenKey

0.0401

 		(No Response) seenKey

0.0400

 		(No Response) seenKey

0.0400

 		(No Response) seenKey

0.0399

 		(No Response) seenKey

0.0392

lim x→4 
[x/(x + 1)] − (4/5)
x − 4
 ≈ (No Response) seenKey

0.0400


Solution or Explanation
x 3.9 3.99 3.999 4 4.001 4.01 4.1
f(x) 0.0408 0.0401 0.0400 ? 0.0400 0.0399 0.0392

lim x→4 
x/(x + 1) − 4/5
x − 4
 ≈ 0.0400
    
Actual limit is
1
25
.

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2. –/4 points LarCalc11HS 1.4.041. My Notes
Question Part
Points
Submissions Used
1 2 3 4
–/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100
Total
–/4
 
Consider the following.
f(x) = 5
–x2 + 1
Find the x-values at which f is not continuous. Which of the discontinuities are removable? (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks.)
x = (No Response) -1 ; (No Response) seenKey

nonremovable
x = (No Response) 1 ; (No Response) seenKey

nonremovable

Solution or Explanation
f(x) = 
5
1 − x2
 = 
5
(1 − x)(1 + x)
 has nonremovable discontinuities at x = ±1 because
 
lim
x→1
f(x) and 
lim
x→−1
 f(x)
do not exist.

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3. –/3 points LarCalc11HS 2.5.JIT.006. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/100 0/100 0/100
Total
–/3
 
Verify the identity.
 
1
1 − sin2 y
 = tan2 y + 1

Use a Pythagorean Identity in the denominator, and then use a Reciprocal Identity.
1
1 − sin2 y
 = 
1
(No Response) webMathematica generated answer key
 = (No Response) webMathematica generated answer key

Use a Pythagorean Identity.
               = (No Response) webMathematica generated answer key

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4. –/2 points LarCalc11HS 4.4.112. My Notes
Question Part
Points
Submissions Used
1 2
–/1 –/1
0/100 0/100
Total
–/2
 
Find the function f(x) and all values of c such that
x
c
f(t) dt = x2 + x − 56.

f(x) = (No Response) webMathematica generated answer key
c = (No Response) webMathematica generated answer key (Entered as a comma-separated list.)


Solution or Explanation
 
x f(t) dt
c
 = x2 + x − 56
Let f(t) = 2t + 1. Then
x f(t) dt
c
 = 
x (2t + 1)
c
dt = 
t2 + t
x
c
 = x2 + x − c2 − c		 = x2 + x − 56
−c2 − c = −56
c2 + c − 56 = 0
(c + 8)(c − 7) = 0 right double arrow implies c = 7, −8.
So, f(x) = 2x + 1, and c = 7 or c = −8.

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5. –/1 points LarCalc11HS 6.3.011. My Notes
Question Part
Points
Submissions Used
1
–/1
0/100
Total
–/1
 
Find the general solution of the differential equation. (Enter your solution as an equation.)
(2 + x)y' = 5y
(No Response) webMathematica generated answer key

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6. –/1 points LarCalc11HS 8.1.023. My Notes
Question Part
Points
Submissions Used
1
–/1
0/100
Total
–/1
 
Find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
 
x2
x − 9
 dx
(No Response) webMathematica generated answer key

Solution or Explanation
x2
x − 9
 dx
 = 
(x + 9) dx
 + 
81
x − 9
 dx
 = 
1
2
x2 + 9x + 81ln(|x − 9|) + C

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7. –/1 points LarCalc11HS 9.7.032.MI. My Notes
Question Part
Points
Submissions Used
1
–/1
0/100
Total
–/1
 
Find the nth Taylor polynomial centered at c.
f(x) = x2cos x,    n = 2,    c = π
P2(x) = (No Response) -pi^2-2pi(x-pi)+ (pi^2-2)/2 (x-pi)^2

Solution or Explanation
f(x) = x2 cos(x)
f '(x) = cos(x) − x2 sin(x)
f ''(x) = 2 cos(x) − 4x sin(x) − x2 cos(x)
    		    
f(π) = −π2
f '(π) = −2π
f ''(π) = −2 + π2
P2(x) = −π2 − 2π(x − π) + 
π2 − 2
2
(x − π)2

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8. –/1 points LarCalc11HS 11.4.016. My Notes
Question Part
Points
Submissions Used
1
–/1
0/100
Total
–/1
 
Find a unit vector that is orthogonal to both u and v.
u = 
−8, −6, 4
v = 
11, −12, −1
(No Response) webMathematica generated answer key

Solution or Explanation
u = 
−8, −6, 4
 
v = 
11, −12, −1
 
u × v = 
ijk
−8−64
11  −12   −1
 = 54i + 36j + 162k
u × v
u × v
 = 
1
18
94
54, 36, 162
 
 = 
3
94
,
2
94
,
9
94

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