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Math - College, section 1, Fall 2019
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Stewart-Essential Calculus: ET 2/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

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Question

Points

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
3/3	5/6	2/2	2/2	1/1	1/1	27/31	0/8	–/9	–/1	1/1	0/1	–/2	–/1	–/3

Total

42/72 (58.3%)

Instructions

This demo is for the new textbook Stewart "Essential Calculus: Early Transcendentals", 2nd edition. Create your course assignments by selecting questions from our bank of end-of-section exercises, as well as enhanced interactive examples with videos.

While doing their homework, students can link to the relevant interactive examples from the book and work through them again and again for additional practice before answering the question.

Students can view and hear additional instruction through the 2-5 minute Watch It links. Students will also find helpful links to online excerpts from their textbook, online-live help, tutorials, and videos.

- relevant textbook pages

- videos of worked examples

- tutorials

This course includes Cengage's new QuickPrep review, a new resource designed to address the varying levels of student preparedness. QuickPrep (QP) reviews 25 key precalculus topics to help your students with their readiness for calculus. Assign the entire module or specific questions from the module early in the course, or whenever you think the review is most needed throughout the course. See question #15 below as an example of a QP problem.

If additional review is needed beyond QuickPrep, assign the new Just-in-Time (JIT) modules offered for each section. Each module consists of approximately 10 carefully selected problems reviewing the prerequisite skills used in that section. Assign the entire module or specific questions from the module. See question #1 below as an example of a JIT problem.

Stewart's Tools for Enriching Calculus (TEC) are also included in this course. The TECs function as a powerful presentation tool for instructors, and as a tutorial environment in which students can explore and review selected topics. Now you will be able to assign WebAssign questions connected to the TECs—see question #10 below. Also, click here for a full demonstration of Stewart's Tools for Enriching Calculus.

Exercise #2 includes an example of Show My Work, a new tool that is available now. With the new Show My Work feature, students will be able to upload images, files, or photos of their detailed work or they can type it directly into WebAssign.

See questions #5, #6, #8, and #13 as examples of Cengage’s new Enhanced Feedback. These exercises deliver additional tips on how to work the problem when a student enters an incorrect response. Enhanced Feedback is included for the most highly assigned questions in the course.

Click here for a list of all of the questions coded in WebAssign. 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.

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. 3/3 points | Previous Answers SEssCalcET2 4.5.JIT.006.MI. My Notes

A pair of points is graphed.

(0, 8), (8, 14)

(a) Plot the points in a coordinate plane.

0,0

-10

-8

-6

-4

-2

-20

-18

-16

-14

-12

-10

-8

-6

-4

-2

WebAssign Graphing Tool

Graph LayersToggle Open/Closed

Point 1
Remove This Layer

P1 (, )
Point 2
Remove This Layer

P1 (, )

Submission Data

(b) Find the distance between them.

(x, y) = (

)

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2. 5/6 points | Previous Answers SEssCalcET2 4.1.029.MI. My Notes

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

g(y) =

y − 3

y² − 3y + 9

y =	68767868767869

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Show My Work
(Required)
What steps or reasoning did you use? Your work counts towards your score.

My dog has fleas.

My dog has fleas.
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3. 2/2 points | Previous Answers SEssCalcET2 4.1.039.MI. My Notes

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = 6x³ − 9x² − 216x + 5,

[−4, 5]

absolute minimum	−619
absolute maximum	410

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4. 2/2 points | Previous Answers SEssCalcET2 4.1.043.MI. My Notes

Find the absolute minimum and absolute maximum values of f on the given interval.

f(t) = t

9 − t²

, [−1, 3]

absolute minimum	−√8
absolute maximum	4.5

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5. 1/1 points | Previous Answers SEssCalcET2 4.2.003. My Notes

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.)

f(x) =

−

x, [0, 9]

c =

2.25

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6. 1/1 points | Previous Answers SEssCalcET2 4.2.023. My Notes

If f(1) = 15 and f '(x) ≥ 2 for 1 ≤ x ≤ 3, how small can f(3) possibly be?

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7. 27/31 points | Previous Answers SEssCalcET2 4.3.001.MI.SA. My Notes

Question Part

Points

Submissions Used

1/1

0/1

1/1

–/1

1/1

0/1

1/1

1/50

0/50

1/50

5/50

1/50

2/50

1/50

Total

27/31

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.

Consider the equation below.

f(x) = 2x³ + 3x² − 36x

Find the interval on which f is increasing. Find the interval on which f is decreasing.

Part 1 of 6

For

f(x) = 2x³ + 3x² − 36x,

we have

f '(x) =

$$6x2+6x−36

This factors into

x + 3

(x − 2).

Part 2 of 6

f '(x)

is negative, then

f(x)

is decreasing

decreasing

. If

f '(x)

is positive, then

f(x)

is increasing

increasing

Part 3 of 6

x < −3,

then

6(x + 3)

is negative

negative

and

(x − 2)

is negative

negative

. Therefore, their product,

f '(x),

is positive

positive

, and

f(x)

is increasing

increasing

Part 4 of 6

−3 < x < 2,

then

6(x + 3)

is positive

positive

and

(x − 2)

is negative

negative

. Therefore, their product,

f '(x),

is negative

negative

, and

f(x)

is decreasing

decreasing

Part 5 of 6

x > 2,

then

6(x + 3)

is positive

positive

and

(x − 2)

is positive

positive

. Therefore, their product,

f '(x),

is positive

positive

, and

f(x)

is increasing

increasing

Part 6 of 6

Thus, the interval on which f is increasing is as follows. (Enter your answer in interval notation.)

$$(−∞,−3)∪(2,∞)

$webMathematica generated answer key$

The interval on which f is decreasing is as follows. (Enter your answer in interval notation.)

$$(−3,2)

$webMathematica generated answer key$

Find the local minimum and maximum values of f.

Part 1 of 2

The function

f(x)

changes from increasing to decreasing at

x = −3.

Therefore,

f(−3) = 0

is a maximum

a maximum

The function

f(x)

changes from decreasing to increasing at

x = 2.

Therefore,

f(2) =

is .

Find the inflection point. Find the interval on which f is concave up. Find the interval on which f is concave down.

Part 1 of 3

We have

f '(x) = 6x² + 6x − 36,

f ''(x) =

$$12x+6

which equals 0 when

(x, y) =

$$−3, −12, 2

$webMathematica generated answer key$

Part 2 of 3

When

x < −

, f ''(x) = 12x + 6

is negative

negative

, and

f(x)

is concave down

down

Part 3 of 3

When

x > −

, f ''(x) = 12x + 6

is positive

positive

, and

f(x)

is concave up

. Since the concavity changes

changes

x = −

an inflection point.

You have now completed the Master It.

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8. 0/8 points | Previous Answers SEssCalcET2 4.3.033. My Notes

Consider the function below.

C(x) = x^1/3(x + 4)

(a) Find the interval of increase. (Enter your answer using interval notation.)

(0,3]

Find the interval of decrease. (Enter your answer using interval notation.)

(b) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

(x, y)

(smaller x-value)

(x, y)

(larger x-value)

Find the intervals where the graph is concave upward. (Enter your answer using interval notation.)

Find the interval where the graph is concave downward. (Enter your answer using interval notation.)

Enhanced Feedback

Please try again. Remember, for a function f, if

f '(x) > 0

f '(x) < 0

on an interval, then f is increasing or decreasing, respectively, on that interval. If f ' changes from positive to negative or negative to positive at a critical number c, then f has a local maximum or minimum, respectively, at c.

If

f ''(x) > 0

f ''(x) < 0

on an interval, then f is concave upward or concave downward, respectively, on that interval. An inflection point is a point P on the curve

y = f(x)

at which f changes from concave upward to concave downward or concave downward to concave upward.

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9. –/9 points SEssCalcET2 4.3.AE.001. My Notes

Video Example

EXAMPLE 1 Find where the function

f(x) = 3x⁴ − 20x³ − 216x² + 1

is increasing and where it is decreasing.

SOLUTION

f '(x) = 12x³ − 60x² − 432x = 12x

x −

x +

To use the I/D Test, we have to know where

f '(x) > 0

and where

f '(x) < 0.

This depends on the signs of the three factors of

f '(x),

namely, 12x,

x − ,

and

x + .

We divide the real line into intervals whose endpoints are the critical numbers (smallest), 0 and (largest) and arrange our work in a chart. A plus sign indicates that the given expression is positive, and a negative sign indicates that it is negative. The last column of the chart gives the conclusion based on the I/D Test. For instance,

f '(x) < 0

for

0 < x < 9,

so f is on

(0, 9).

(It would also be true to say that f is decreasing on the closed interval

[0, 9].)

Interval	12x	x − 9	x + 4	f '(x)	f
x < −4	−	−	−	−	decreasing on (−∞, −4)
−4 < x < 0	−	−	+	+	on (−4, 0)
0 < x < 9	+	−	+	−	decreasing on (0, 9)
x > 9	+	+	+	+	on (9, ∞)

The graph of f shown in the figure confirms the information in the chart.

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10. –/1 points SEssCalcET2 4.3.TEC.003. My Notes

Concept

This module initially shows you graphs of a function's first and second derivatives but not the graph of the function itself. By analyzing these derivative graphs, you can record key information about f: intervals where f is increasing or decreasing, locations of local maximum or minimum values, intervals where f is concave up or down and locations of inflection points.

With this information, you can the sketch an approximate graph of f on paper. When you are confident with your drawing, the module will reveal the graph of f and you can check the accuracy of your sketch.
Instructions

Select one of the four available functions from the pull-down menu at the upper right.

The graph of f ' is visible in the upper diagram. Click on the graph to mark any x-value where you believe f (not f ' ) has a local maximum value. You will see a marker appear on the first horizontal bar labeled "f maxima." (You can click on the bar itself to set values as well.)

When you have recorded all local maximum points, click the button for the next category, "f minima." Similarly, click on the graph or bar at all locations where the graph of f should have a local minimum point.

Next, click the button for the third category, "f increasing", and click on the graph or bar anywhere in an interval where you believe f (again, not f ' ) should be increasing. The interval, bounded by local maximum and minimum locations, will appear in the horizontal bar.

After marking all intervals where f is increasing, click the fourth button and click on the graph or bar where f is decreasing.

If you would like to change any of your selections, click the appropriate Clear button to the left of the bar and all markers in that category as well as the categories below it will be deleted.

When you are finished recording the information for these four categories, use the graph of f '' at the lower portion of the screen to record locations of inflection points and intervals of concavity of f in a similar manner as above.

When you are finished click Done. The graphs are replaced by two number lines that summarize the information you recorded. Use this information to sketch a graph of f on paper. When you are satisfied with your sketch, click Done again. (A timer prevents you from proceeding before making a drawing.) You will see four graphs on your screen. Choose the graph that most closely resembles your drawing. Although your graph may differ slightly in shape or size, the key features should match one of the four choices.

If you make an incorrect choice, you will be returned to the original screen to revise your selections and try again. If you choose correctly, you will see the graph of f. You can enable the graphs of f ' and f '' by clicking the check boxes on the right so that you can compare the features of the three curves. Moving the mouse over the graph will trace across the curves and display the x-value.
Simulation

Click here to access the TEC simulation.
Exercise

Select Function 1 from the pull-down menu. Follow the detailed module instructions to produce a hand-drawn sketch of f and check your drawing against the computer-generated graphs of f. If your sketch is inaccurate, revise the information gained from the graphs of f ' and f '' and try your sketch again. Keep in mind that the shape and size of your graph may differ slightly from the graph given by the module, but the locations of local maximum and minimum values, inflection points and intervals where f is increasing, decreasing, concave up and concave down should be accurate.

If you repeat the steps above with Function 1 but you assume that
f(0) = 3,
what will change in your sketch of f? What will not change?

The graph will be stretched vertically by a factor of 3. Other than that nothing else in the graph will be changed. The graph will be shifted left or right in order to adjust to the x-intercept of 3. This will affect the locations of local maximum and minimum values, inflection points, and intervals where f is increasing, decreasing, concave up and concave down. The graph will reflect about the line y = 3. This will affect the locations of local maximum and minimum values, inflection points, and intervals where f is increasing, decreasing, concave up and concave down. The graph will be stretched horizontally by a factor of 3. This will affect the locations of local maximum and minimum values, inflection points, and intervals where f is increasing, decreasing, concave up and concave down. The graph will be shifted up or down in order to adjust to the y-intercept of 3. Other than that nothing else in the graph will be changed.

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11. 1/1 points | Previous Answers SEssCalcET2 4.3.VE.001. My Notes

Watch the video below then answer the question.

f '(3) = 0 and f '(x) < 0 for 0 < x < 3 and f '(x) > 0 for 3 < x < 4

then f has a local minimum at x = 3.

True False

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12. 0/1 points | Previous Answers SEssCalcET2 4.3.045.MI. My Notes

Suppose the derivative of a function f is

f '(x) = (x + 2)⁴(x − 4)³(x − 6)⁶.

On what interval is f increasing? (Enter your answer in interval notation.)

(6,∞)

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13. –/2 points SEssCalcET2 4.5.027. My Notes

A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

(a) How much wire should be used for the square in order to maximize the total area?

m

(b) How much wire should be used for the square in order to minimize the total area?

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14. –/1 points SEssCalcET2 4.7.047.MI. My Notes

A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. What is the height of the cliff? (Use 32 ft/s² for the acceleration due to gravity.)
ft

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15. –/3 points SEssCalcET2 QP.14.N.001. My Notes

Topic 14: Root Functions

0. When will I need this in Calculus?

Functions involving roots appear in examples and applications in calculus. It's important to be able to convert functions that contain root signs to exponent form, to be able to identify the domain of root functions, and to visualize the graphs of some basic root functions.
1. Functions with terms of the form x^1/n, n an integer

For example, x^1/3, a number raised to a fractional exponent, is a number that when cubed gives the answer x. But $\sqrt[3]{x}$ is also a number that when cubed gives the answer x. This explanation gives us the general rule: $\sqrt[n]{x} = x^{1/n}.$ We will use both the root form and the fractional exponent form of root functions.

The most basic root function is the square root function, $y = f(x) = \sqrt{x}.$ (Recall that $\sqrt{a}$ is defined as the nonnegative value b satisfying b² = a, so a must be greater than or equal to 0 in order for $\sqrt{a}$ to be defined.) A function such as f, in which the independent variable appears under a radical sign, is sometimes called an irrational function. (See Topic 13 for a discussion of rational functions.)

In general, the domain of a function of the form f(x) = x^1/n for n an integer, n ≥ 0, depends on whether n is even or odd. If n is an even integer, the graph of f(x) = x^1/n behaves similarly to the graph of $y = \sqrt{x}:$ the domain is the interval [0, ∞), because we cannot take an even root of a negative number; the range is also the interval [0, ∞). For n an odd integer, we can take the n^th root of a negative number, because a negative number raised to an odd power is negative. For example, $\sqrt[3]{-8} = -2,$ because (−2)³ = −8. Thus, the domain of f(x) = x^1/n where n is an odd integer, is the interval (-∞, ∞); the range is also the interval (-∞, ∞).
Example 1: The square root function

Let $f(x) = \sqrt{x}$ and $g(x) = -\sqrt{x}.$
a. Find the domain of each function.

b. Find the range of each function.

c. Sketch the graph of each function.

Solution:

a. In order for $y = \sqrt{x}$ to be a real number, x must be greater than or equal to 0. (Recall that the square root of a negative number is an imaginary number. For example, by definition $\sqrt{-1} = i$ and $\sqrt{-5} = \sqrt{5}\sqrt{-1} = \sqrt{5}i.)$ Similarly, the domain of g is the same interval: [0, ∞).

b. Because the symbol $\sqrt{x}$ tells us to take the positive square root of x, $\sqrt{x}$ cannot be negative. Also, $\sqrt{0} = 0,$ so the range of f is the interval [0, ∞). The negative sign preceding the symbol $\sqrt{x}$ tells us to take the negative square root of x, so values of y = g(x) are never positive; the range of g is the interval (−∞, 0].

c. Here are the graphs of $f(x) = \sqrt{x}$ and $g(x) = -\sqrt{x}.$ Note that the graph of y = g(x) is the reflection about the x-axis of the graph of y = f(x).
Example 2: Functions with fractional exponents and root functions

Find the domain and sketch the graph of each function.
a.     $f(x) = x^{1/3}$

b.     $g(x) = 4x^{1/6}$

c.     $h(x) = \sqrt{x} - 4$
Solution:

1. Here $f(x) = x^{1/3} = \sqrt[3]{x},$ so the function f is defined for all real numbers; the domain is the interval (−∞, ∞). If x > 0, then $\sqrt[3]{x}$ is also positive. If x < 0, then $\sqrt[3]{x}$ is negative. Here is the graph of $y = f(x) = \sqrt[3]{x}.$

2. The function g(x) = 4x^1/6 involves taking an even root of x. Because no number raised to the 6^th power will give us a negative number, the domain of g is the interval [0, ∞). The graph of g has a shape similar to the graph of $y = \sqrt{x}.$ However, the graph of g rises more steeply for small values of x. Here is the graph of g(x) = 4x^1/6.

3. The function $h(x) = \sqrt{x} - 4$ is defined when x ≥ 0, so its domain consists of values of x in the interval [0, ∞). The graph of y = h(x) is like the graph of $f(x) = \sqrt{x},$ but each y-value is moved 4 units down. Here is the graph of $h(x) = \sqrt{x} - 4.$
2. Functions with terms of the form x^m/n, m and integers n integers, n > 0

Here are three important rules for exponents that are useful when working with functions that contain roots and exponents. In all cases, we assume m and n are integers and n > 0.
(1)     $x^{-m/n} = \frac{1}{x^{m/n}}$

(2)     $x^{1/n} = \sqrt[n]{x}$

(3)     $x^{m/n} = (\sqrt[n]{x})^m = (x^{1/n})^m$

(4)     $(x^m)^n = x^{nm}$
Example 3: Equating functions

In each part, determine if the pair of functions f and g is equal or not.
a.     $f(x) = \sqrt{x^2 - 4} \text{ and } g(x) = x - 2.$

b.     $f(x) = \sqrt{x^3} \text{ and } g(x) = x^{2/3}$

c.     $f(x) = \frac{1}{\sqrt[3]{x^4}} \text{ and } g(x) = x^{-4/3}$
Solution:

a. The functions $f(x) = \sqrt{x^2 - 4}$ and g(x) = x − 2 are not equal. If we square g(x) = x − 2, we get (x − 2)² = x² − 4x + 4. Because $\sqrt{x^2 - 4}$ represents an expression whose square is x² − 4, and (x − 2)² ≠ x² − 4, the two functions are not equal.

b. Using the properties of exponents, $f(x) = \sqrt{x^3} = (x^3 )^{1/2} = x^{3/2}.$ This is not the same as g(x) = x^2/3, so the two functions are not equal.

c. We can use the properties of exponents to rewrite $f: f(x) = \frac{1}{\sqrt[3]{x^4}} = \frac{1}{(x^4)^{1/3}} = \frac{1}{x^{4/3}} = x^{-4/3}.$ This is the same as the expression for g(x), so the two functions are equal.
Example 4: Rewriting functions with radicals

If possible, write each function as a sum of terms of the form a · x^p. If it is not possible, explain why not.
a.     $f(x) = \frac{1}{3\sqrt{x}} + \frac{4}{5\sqrt[3]{x}} - 2\sqrt{x}$

b.     $g(x) = \frac{1}{\sqrt[3]{x^2 + 2}} + 3$

c.     $h(x) = \frac{1}{\sqrt{x} + 9}$
Solution:

a. We use the properties of exponents to rewrite each term of the function f(x). The first term can be rewritten as: $\frac{1}{3\sqrt{x}} = \frac{1}{3x^{1/2}} = \frac{1}{3}x^{-1/2}.$ Similarly, the second term can be written as $\frac{4}{5\sqrt[3]{x}} = \frac{4}{5}x^{-1/3}.$ The final term is: $-2\sqrt{x} = -2x^{1/2}.$ Putting the three terms together, we have the function written as a sum of terms of the required form: $f(x) = \frac{1}{3\sqrt{x}} + \frac{4}{5\sqrt[3]{x}} - 2\sqrt{x} = \frac{1}{3}x^{-1/2} + \frac{4}{5}x^{-1/3} - 2x^{1/2}.$

b. The first term of g(x) is: $\frac{1}{\sqrt[3]{x^2 + 2}}.$ Because the denominator involves the cube root of a sum, we cannot separate the two terms of the sum. We can write $\frac{1}{\sqrt[3]{x^2 + 2}} = (x^2 + 2)^{-1/3}$ so $g(x) = \frac{1}{\sqrt[3]{x^2 + 2}} + 3 = (x^2 + 2)^{-1/3} + 3,$ but this is not as a sum of terms of the form a · x^p. It is not possible to write g in that form.

c. The function h(x) consists of a single term. We can write $h(x) = \frac{1}{\sqrt{x}+9} = \frac{1}{x^{1/2}+9}$ but we cannot simplify h further.
3. Functions of the form [p(x)]^m/n, m and n integers, n > 0

One of the important features of a function of the form [p(x)]^m/n is its domain. We recall that in order to take an even root of a number or an expression, the number or expression must be greater than or equal to 0. We can take an odd root of both positive and negative numbers. We use these properties to find the domains of irrational functions. (See Topic 11 for more details on solving polynomial inequalities.)
Example 5: The domain of functions with radicals

Find the domain of each function.
a.     $f(x) = \sqrt{9 - x^2}$

b.     $f(x) = \sqrt{x^2 + 3}$

c.     $f(x) = \sqrt[3]{2x^5 - 5}$

d.     $f(x) = \frac{1}{\sqrt[4]{2x^3 + 7x}}$
Solution:

a. The function $f(x) = \sqrt{9 - x^2}$ involves taking a square root of the expression 9 − x². In order for this to be defined, 9 − x² ≥ 0. We can solve this inequality by factoring the quadratic polynomial 9 − x² into 2 linear factors: 9 − x² = (3 − x)(3 + x). The product of two terms is greater than or equal to 0 when both terms are greater than or equal to 0 or when both terms are less than or equal to 0. So x must satisfy (3 − x) ≥ 0 and (3 + x) ≥ 0 or x must satisfy (3 − x) ≤ 0 and (3 + x) ≤ 0. In the first case, we need 3 ≥ x and x ≥ −3. This gives us the interval −3 ≤ x ≤ 3. In the second case, we need x ≥ 3 and x ≤ −3. There are no x-values that are both greater than or equal to 3 and less than or equal to −3. So the domain of f is the interval [−3, 3]. See Topic 11 for more information on solving quadratic inequalities.

b. The domain of $f(x) = \sqrt{x^2 + 3}$ consists of all x-values satisfying x² + 3 ≥ 0. Because x² ≥ 0 for all x, x² + 3 ≥ 3 for all real numbers x. Therefore the domain of f is the interval (-∞, ∞).

c. Because we can take the cube root of a positive or negative number, there are no restrictions on the domain of this function. Therefore, the domain of $f(x) = \sqrt[3]{2x^5 - 5}$ is the interval (-∞, ∞).

d. The function $f(x) = \frac{1}{\sqrt[4]{2x^3 + 7x}}$ involves an even root of an expression. In order for the 4^th root of 2x³ + 7x to be a real number, we need 2x³ + 7x ≥ 0. However, the 4^th root is in the denominator, so the denominator cannot be 0. Therefore, we need 2x³ + 7x > 0. We factor the expression 2x³ + 7x = x(2x² + 7). Because (2x² + 7) > 0 for all real values of x, 2x³ + 7x = x(2x² + 7) > 0 when the first factor is greater than 0; that is, when x > 0. The domain of $f(x) = \frac{1}{\sqrt[4]{2x^3+7x}}$ is the interval (0, ∞).
Example 6: Root functions

Find the domain and sketch the graph of each function.
a. $f(x) = \sqrt{2 + x}$

b. $g(x) = \sqrt{9 - x^2}$
Solution:

a. Because f involves a square root of an expression, the expression under the square root must be greater than or equal to 0. The domain of f is the set of x-values satisfying 2 + x ≥ 0; that is, x ≥ −2. We observe that the graph of y = f(x) is the graph of $y = \sqrt{x}$ translated two units to the left. This is consistent with the domain of [−2, ∞). Here is the graph of $y = \sqrt{2 + x}.$

b. As we found in part (a) of Example 5 of this Topic, the domain of the function $g(x) = \sqrt{9 - x^2}$ is the interval [−3, 3]. We rewrite this function as $y = \sqrt{9 - x^2},$ and note that y ≥ 0. Suppose we square both sides of this equation to get y² = 9 − x². Rearranging terms, we can rewrite this as x² + y² = 9. This is the equation of a circle of radius 3 centered at the origin. When we squared both sides, we introduced negative values of y, but if we restrict y to be greater than or equal to 0, we see that the graph of $g(x) = \sqrt{9 - x^2}$ is the top half of the circle. Here is the graph.

4. Solving equations involving root functions

When working with equations involving radicals, we want to convert the problem to one we already know how to solve. For solving equations involving square roots, we will square both sides to transform the equation into a polynomial equation. When we square both sides, however, we may introduce solutions to the squared equation that are not solutions to the original problem. These are called extraneous solutions. We need to always check the solutions we get by substituting all solutions back into the original equation; we then eliminate any extraneous solutions.

How are extraneous solutions introduced?

Suppose we want to solve the equation $\sqrt{x + 5} = x - 1.$ If we square both sides, we get $(\sqrt{x + 5})^2 = (x - 1)^2.$ Expanding both sides, yields x + 5 = x² − 2x + 1. Collecting like terms gives us a quadratic polynomial equation: 0 = x² − 3x - 4. Factoring, we see that 0 = (x − 4)(x + 1) and the solutions of this equation are x = 4 and x = −1. We substitute x = 4 back into the original equation, and we see that $\sqrt{4 + 5} = 3 = 4 - 1,$ so 4 is a solution of the original equation. Substituting x = −1 into the original equation, we see that $\sqrt{-1 + 5} \neq -1 - 1,$ so −1 is not a solution to the original equation. By squaring both sides of the original equation, we introduced an extraneous solution that is not a solution to the original equation. This happens because if we start with an equation a = b and square both sides, we get a² = b². But we cannot "undo" the squaring operation and necessarily assume that if a² = b², then a = b. From the equation a² = b², we get two solutions: a = b and a = −b.

Example 7: Solving equations with radicals

Solve each of the following equations
a.     $\sqrt{x + 5} - 1 = \sqrt{x}$

b.     $\sqrt{6 - 2x} + 3 = x$

c.     $\sqrt{2x + 5} + 2\sqrt{x + 6} = 5$
Solution:

a. We start by squaring both sides of the given equation to get: $(\sqrt{x + 5} - 1)^2 = (\sqrt{x})^2.$ Expanding, we get $x + 5 - 2\sqrt{x + 5} + 1 = x.$ Collecting like terms, we get $6 = 2\sqrt{x + 5}.$ We still have a root in our equation, so we need to again square both sides to get: 36 = 4(x + 5). We solve this linear equation to get 36 = 4x + 20 or x = 4. If we substitute x = 4 into the original equation, we see that $\sqrt{9} - 1 = \sqrt{4},$ so x = 4 is the solution.

b. We first rewrite the equation with the root by itself on the left side of the equation. This will simplify the computations. We then square both sides of the rewritten equation: $(\sqrt{6-2x})^2 = (x - 3)^2.$ Expanding, we get 6 - 2x = x² − 6x + 9. Collecting like terms we get 0 = x² − 4x + 3. We factor this quadratic to find the equation's solutions: 0 = (x − 1)(x − 3), so x = 1 and x = 3 are the solutions we obtain. We need to check these solutions in the original equation. We see that x = 1 does not satisfy the original equation $(\sqrt{6 - 2} + 3 \neq 1)$ but x = 3 does satisfy it $(\sqrt{0} + 3 = 3),$ so x = 3 is the only solution.

c. We first subtract $2\sqrt{x + 6}$ from both sides of the given equation to get: $\sqrt{2x + 5} = 5 - 2\sqrt{x + 6}.$ We square both sides of this equation to get: $2x + 5 = 25 - 20\sqrt{x + 6} + 4(x + 6).$ Rearranging terms, we get $20\sqrt{x + 6} = 2x + 44.$ We can simplify the computations a bit if we first divide both sides of this equation by 2 to get: $10\sqrt{x + 6} = x + 22.$ We now square both sides again to get: 100(x + 6) = x² + 44x + 484. We collect like terms to obtain a quadratic equation: 0 = x² − 56x − 116. We factor this quadratic: 0 = (x − 58)(x + 2). The two solutions to this quadratic equation are: x = 58 and x = −2. If we substitute x = 58 into the original equation, the left side of the equation yields 11 + 16, which is not equal to 5, so x = 58 is not a solution. By substituting, we see that x = −2 satisfies the original equation, so it is the only solution.
5. Concept Questions

1.     $\sqrt[n]{x} = x^{1/n}$

True False

2. The domain of a function of the form f(x) = x^1/n for n an integer, n ≠ 0, is [0, ∞) if n is even and (-∞, ∞) if n is odd.

True False

3. There is no need to check the solutions when solving equations involving root functions.

True False

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0. When will I need this in Calculus?

1. Functions with terms of the form x^1/n, n an integer

Example 1: The square root function

Example 2: Functions with fractional exponents and root functions

2. Functions with terms of the form x^m/n, m and integers n integers, n > 0

Example 3: Equating functions

Example 4: Rewriting functions with radicals

3. Functions of the form [p(x)]^m/n, m and n integers, n > 0

Example 5: The domain of functions with radicals

Example 6: Root functions

4. Solving equations involving root functions

Example 7: Solving equations with radicals

5. Concept Questions

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Stewart-Essential Calculus: ET 2/e (Homework)

Current Score : 42 / 72

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

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

Instructions

Assignment Submission

Assignment Scoring

WebAssign Graphing Tool

Graph LayersToggle Open/Closed

Show My Work

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Concept

Instructions

Simulation

Exercise

0. When will I need this in Calculus?

1. Functions with terms of the form x1/n, n an integer

Example 1: The square root function

Example 2: Functions with fractional exponents and root functions

2. Functions with terms of the form xm/n, m and integers n integers, n > 0

Example 3: Equating functions

Example 4: Rewriting functions with radicals

3. Functions of the form [p(x)]m/n, m and n integers, n > 0

Example 5: The domain of functions with radicals

Example 6: Root functions

4. Solving equations involving root functions

Example 7: Solving equations with radicals

5. Concept Questions

1. Functions with terms of the form x^1/n, n an integer

2. Functions with terms of the form x^m/n, m and integers n integers, n > 0

3. Functions of the form [p(x)]^m/n, m and n integers, n > 0