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Cengage Testing Assignments, Fall 2019
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Stewart Calculus Early Transcendentals 9/e (Homework)

James Finch

Cengage Testing Assignments, Fall 2019

Instructor: Sarah Anders

Current Score : 29 / 29

Due : Tuesday, December 30, 2025 11:00 EST

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

Ask Your Teacher Extension Requests Print Assignment

Question

Points

1	2	3	4	5	6	7	8	9	10	11	12
2/2	1/1	4/4	1/1	2/2	8/8	4/4	1/1	1/1	3/3	1/1	1/1

Total

29/29 (100.0%)

Instructions
WebAssign provides a wide range of exercises that enable you to:
- Build problem-solving skills (#1–4: Read It, Watch It, Master It, and Problems Plus exercises)
- Develop conceptual understanding (#5–7: Explore It, Proof Problems, and Expanded Problems)
- Address readiness gaps (#8–12: Scaffolded exercises, Just-In-Time, Quick Prep, Diagnostics, and Calculus Readiness Bootcamp)
Please read the information in the box at the top of each exercise to learn more about the features, content, and/or grading. We encourage you to try correct and incorrect answers in any answer blanks.

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. 2/2 points | Previous Answers SCalcET9 2.2.001. My Notes

This exercise will build problem-solving skills.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

Explain the meaning of the following equation.

lim x → 5 f(x) = 7

The values of f(x) can be made as close to 5 as we like by taking x sufficiently close to 7.

If |x₁ − 5| < |x₂ − 5|, then |f(x₁) − 7|≤ |f(x₂) − 7|.

f(x) = 7 for all values of x.

If |x₁ − 5| < |x₂ − 5|, then |f(x₁) − 7| < |f(x₂) − 7|.

The values of f(x) can be made as close to 7 as we like by taking x sufficiently close to 5.

Is it possible for this statement to be true and yet

f(5) = 8?

Explain.

Yes, the graph could have a hole at (5, 7) and be defined such that f(5) = 8.

Yes, the graph could have a vertical asymptote at x = 5 and be defined such that f(5) = 8.

No, if f(5) = 8, then lim x→5 f(x) = 8.

No, if lim x→5 f(x) = 7, then f(5) = 7.

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2. 1/1 points | Previous Answers SCalcET9 2.5.047.MI. My Notes

This exercise will build problem-solving skills.
Master It tutorials (MIs) are an optional student help tool available within select questions for just-in-time support. Students can use the tutorial to guide them through the problem-solving process step-by-step using different numbers.

Let

f(x) =

	cx² + 6x		if x < 5
	x³ − cx		if x ≥ 5.

For what value of the constant c is the function f continuous on (−∞, ∞)?

c =

19/6

Solution or Explanation

We have the following.

f(x) =

cx² + 6x	if x < 5
x³ − cx	if x ≥ 5

f is continuous on

(−∞, 5)

and

(5, ∞).

Now

lim x → 5⁻ f(x) = lim x → 5⁻ (cx² + 6x) = 25c + 30 and

lim x → 5⁺ f(x) = lim x → 5⁺ (x³ − cx) = 125 − 5c.

So f is continuous ⇔

25c + 30 = 125 − 5c ⇔ 30c = 95 ⇔ c =

Thus, for f to be continuous on

(−∞, ∞), c =

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3. 4/4 points | Previous Answers SCalcET9 5.5.003.MI.SA. My Notes

This exercise will build problem-solving skills.
Master It tutorials–Stand Alone (MI.SA) is an embedded, step-by-step tutorial used to help students understand each step required to solve the problem, before inputting their final answer.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Evaluate the integral by making the given substitution.

x²

x³ + 16

dx, u = x³ + 16

Step 1

We know that if

u = f(x),

then

du = f ′(x) dx.

Therefore, if

u = x³ + 16,

then

du =

$$3x2

$webMathematica generated answer key$ dx.

Step 2

u = x³ + 16

is substituted into

x²

x³ + 16

then we have

	x² (u)^1⁄2 dx

	u^1⁄2 x² dx

We must also convert

x² dx

into an expression involving u.

We know that

du = 3x² dx,

x² dx =

$$13

$webMathematica generated answer key$ du.

Step 3

Now, if

u = x³ + 16,

then

x²

x³ + 16

dx =

u^1⁄2

u^1⁄2 du.

This evaluates as follows. (Enter your answer in terms of u.)

	u^1⁄2 du

$$2u(32)9

$webMathematica generated answer key$ + C

Step 4

Since

u = x³ + 16,

then converting back to an expression in x we get the following.

u^3⁄2 + C =

$$29(x3+16)(32)+C

$webMathematica generated answer key$

You have now completed the Master It.

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4. 1/1 points | Previous Answers SCalcET9 3.PP.008. My Notes

This exercise will build problem-solving skills.
Problems Plus exercises (PPs) feature examples of how to tackle challenging calculus problems and encourage students to recognize which problem-solving principles are relevant. (Note: "Show My Work" has been enabled for this exercise.)
Read It links are available as a learning tool under every WebAssign question so students can quickly jump to the corresponding section of the eTextbook.

A car is traveling at night along a highway shaped like a parabola with its vertex at the origin (see the figure). The car starts at a point 50 m west and 50 m north of the origin and travels in an easterly direction. There is a statue located 50 m east and 25 m north of the origin. At what point on the highway will the car's headlights illuminate the statue? (Round your coordinates to one decimal place.)

The x y-coordinate plane is given. A parabola, a car, and a statue are shown.
The parabola enters the window in the second quadrant, goes down and right, changes direction at the origin, goes up and right, and exits the window in the first quadrant.
The car is located in the second quadrant driving along the parabola.
The statue is located in the first quadrant, to the right of the parabola.

(x, y) =

14.6,4.2

$webMathematica generated answer key$

Solution or Explanation

We find the equation of the parabola by substituting the point

(−50, 50),

at which the car is situated, into the general equation

y = ax².

50 = a(−50)² right double arrow implies

a =

Now we find the equation of a tangent to the parabola at the point

(x₀, y₀).

We can show that

y ′ = a(2x) =

(2x) =

so an equation of the tangent is

y − y₀ =

x₀(x − x₀).

Since the point

(x₀, y₀)

is on the parabola, we must have

y₀ =

x₀²,

so our equation of the tangent can be simplified to

y =

x₀² +

x₀(x − x₀).

We want the statue to be located on the tangent line, so we substitute its coordinates

(50, 25)

into this equation.

x₀² +

x₀(50 − x₀) right double arrow implies

x₀² − 100x₀ + 1,250 = 0

x₀ =

100 ±

(−100)² − 4(1,250)

x₀ ≈ 14.6 and 85.4

But

x₀ < 50,

so the car's headlights illuminate the statue when it is located at the point

≈ (14.6, 4.3),

that is, about 14.6 m east and 4.3 m north of the origin.

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5. 2/2 points | Previous Answers SCalcET9 6.2.EI.001. My Notes

This exercise will develop conceptual understanding.
Explore It exercises (EIs) engage students with interactive learning modules that include video and explorations. The module is also available to use for studying in the eTextbook.

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

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6. 8/8 points | Previous Answers SCalcET9 2.5.054. My Notes

This exercise will develop conceptual understanding.
Automatically Graded Proof Problems develop an understanding of the entire process of writing proofs as students get practice and immediate feedback.

Suppose f is continuous on [1, 5] and the only solutions of the equation

f(x) = 6

are

x = 1

and

x = 4.

f(2) = 8,

show that

f(3) > 6.

Proof by contradiction: Select an appropriate statement to start the proof.

Suppose that f(3) ≥ 6. Suppose that f(3) = 8. correct

Suppose that f(3) ≤ 6. Suppose that f(3) > 8.

The only solutions of

f(x) = 6

are

x = 1

and

x =

therefore

f(3) ≠ 6.

By the

intermediate value theorem

applied to the continuous function f on the closed interval [2, 3], the fact that

f(2) = 8

and

f(3) < 6

implies that there is a number c in (2, 3) such that

f(c) = 6.

This

contradicts

the fact that the only solutions of the equation

f(x) = 6

are

x = 1

and

x =

Hence, our supposition that

f(3) ≤ 6

was

incorrect

. Therefore,

f(3)

Solution or Explanation

Suppose that f(3) ≤ 6. The only solutions of

f(x) = 6

are

x = 1

and

x = 4,

therefore

f(3) ≠ 6.

By the intermediate value theorem applied to the continuous function f on the closed interval [2, 3], the fact that

f(2) = 8 > 6

and

f(3) < 6

implies that there is a number c in (2, 3) such that

f(c) = 6.

This contradicts the fact that the only solutions of the equation

f(x) = 6

are

x = 1

and

x = 4.

Hence, our supposition that

f(3) ≤ 6

was incorrect. Therefore,

f(3) > 6.

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7. 4/4 points | Previous Answers SCalcET9 2.6.057.EP. My Notes

This exercise will develop conceptual understanding.
Expanded Problems (EPs) enhance student understanding by going beyond a basic exercise and asking students to solve each step of the problem in addition to their final answer.

Consider a rational function f that satisfies the following conditions.

lim x→±∞ f(x) = 0,	lim x→0 f(x) = −∞,	f(7) = 0,

lim x→8⁻ f(x) = ∞,	lim x→8⁺ f(x) = −∞,

Find the factors of the denominator of f corresponding to the vertical asymptotes. (Enter your answers as a comma-separated list. Enter at least one expression for each vertical asymptote.)

x,x−8

$webMathematica generated answer key$

Find the factors of the numerator of f corresponding to the x-intercepts. (Enter your answers as a comma-separated list. Enter at lest one expression for each x-intercept.)

x−7

$webMathematica generated answer key$

Which of the following must be true?

The degrees of the numerator and the denominator are equal. correct

The degree of the denominator is greater than the degree of the numerator. The degree of the numerator is greater than the degree of the denominator.

Find an equation for f.

f(x) =

7−xx2(x−8)

$webMathematica generated answer key$

Solution or Explanation

Let's look for a rational function.

lim x→± ∞ f(x) = 0
⇒ degree of numerator < degree of denominator
lim x→0 f(x) = −∞
there is a factor of
x²
in the denominator (not just
x,
since that would produce a sign change at
x = 0),
and the function is negative near
x = 0.
lim x→8⁻ f(x) = ∞
and
lim x→8⁺ f(x) = −∞
vertical asymptote at
x = 8;
there is a factor of
(x − 8)
in the denominator.
f(7) = 0

7
is an x-intercept; there is at least one factor of
(x − 7)
in the numerator.

Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us

f(x) =

7 − x

x²(x − 8)

as one possibility.

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8. 1/1 points | Previous Answers SCalcET9 5.5.003.MI. My Notes

This exercise will address readiness gaps.
Scaffolded exercises in some problem sets reinforce key skills and prepare students for the more rigorous exercises.

Evaluate the integral by making the given substitution. (Use C for the constant of integration.)

x²

x³ + 31

dx, u = x³ + 31

29(x3+31)(32)+C

$webMathematica generated answer key$

Solution or Explanation

Let

u = x³ + 31.

Then

du = 3x² dx

and

x² dx =

du,

x²

x³ + 31

u^3⁄2

+ C

u^3⁄2 + C =

(x³ + 31)^3⁄2 + C.

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9. 1/1 points | Previous Answers SCalcET9 2.JIT.1.001.MI. My Notes

This exercise will address readiness gaps.
Just In Time exercises (JITs) provide timely support by reviewing prerequisites within the context of new concepts throughout the course.

Find the slope of the line through P and Q.

P(0, 0), Q(8, 4)

1/2

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10. 3/3 points | Previous Answers SCalcET9 QP.1.011.MI. My Notes

This exercise will address readiness gaps.
Quick Prep exercises (QPs) address readiness gaps by reviewing prerequisite concepts and skills. They can be used early in the course or whenever needed.

Evaluate the function at each specified value of the independent variable and simplify. (If an answer is undefined, enter UNDEFINED.)

q(t) =

6t² + 7

t²

(a)

q(2)

7.75

$webMathematica generated answer key$

(b)

q(0)

UNDEFINED

$webMathematica generated answer key$

(c)

q(−x)

6+7x2

$webMathematica generated answer key$

Solution or Explanation

We have the following.

q(t) =

6t² + 7

t²

(a)

q(2) =

6(2)² + 7

(2)²

24 + 7

(b)

q(0) =

6(0)² + 7

(0)²

Division by zero is undefined.

(c)

q(−x) =

6(−x)² + 7

(−x)²

6x² + 7

x²

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11. 1/1 points | Previous Answers SCalcET9 DT.A.001a. My Notes

This exercise will address readiness gaps.
Four diagnostic tests in algebra, analytic geometry, functions, and trigonometry appear at the beginning of the text and provide sources of help.

Evaluate the expression without using a calculator.

(−4)²

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

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12. 1/1 points | Previous Answers WACalcBootcamp1 1.1.001. My Notes

This exercise will address readiness gaps.
The Calculus Readiness Bootcamp contains a formative student assessment on prerequisite topics needed for success and targeted learning modules for areas where students struggled.

Simplify expressions by using the distributive property and then combining like terms.

Simplify the expression.

−7 + 3(2a − 8) − 9a

−31−3a

$webMathematica generated answer key$

Solution or Explanation

Part 1 of 3

Rewrite subtraction as addition of the opposite.

−7 + 3(2a − 8) − 9a = −7 + 3(2a + (-8)) + (-9)a
Part 2 of 3

Use the distributive property to multiply each term inside the parenthesis by 3. Simplify using multiplication properties of real numbers.

−7 + 3(2a + (−8)) + (−9)a = −7 + 3(2a) + (3)(−8) + (−9)a

= −7 + 6a + (-24) + (−9)a
Part 3 of 3

Use properties of addition to group like terms. Then simplify by combining like terms.

−7 + 6a + (−24) + (−9)a = (−7 + (−24)) + (6a + (−9)a)

= (-31) + (6 + (−9))a

= -31 + (−3)a

= -31 − 3a

Use the results to simplify the expression
−7 + 3(2a − 8) − 9a.

${\color{red}-3 a-31}$

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Home My Assignments Extension Request

1	2	3	4	5	6	7	8
1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1
6/100	3/100	4/100	1/100	2/100	1/100	2/100	2/100

−7 + 3(2a + (−8)) + (−9)a	=	−7 + 3(2a) + (3)(−8) + (−9)a

	=	−7 + 6a + (-24) + (−9)a

−7 + 6a + (−24) + (−9)a	=	(−7 + (−24)) + (6a + (−9)a)

	=	(-31) + (6 + (−9))a

	=	-31 + (−3)a

	=	-31 − 3a

1	2
1/1	1/1
36/100	46/100

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Welcome, demo@demo

Stewart Calculus Early Transcendentals 9/e (Homework)

Current Score : 29 / 29

Due : Tuesday, December 30, 2025 11:00 EST

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

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