AT-140-J1353 21EW1 Precalculus PDF

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Assignment Worksheet

Online Homework System

8/27/21 - 12:38:51 AM EDT

Name:

Class:

Class #:

Section #:

Instructor: Delaine Cochran

Assignment: 1-5 Module One Problem Set

MAT-140-J1353 21EW1 Precalculus

Assignment Instructions:

Do you need help with this problem or this material? Remember to use the Academic Support resources available on your homepage in Brightspace or or in the learning modules.

Question 1: (2 points)

Solve the compound inequality:

6 ≤ 2 x − 8 < 10.

Solution: Above relation can be written as:

 6 ≤ 2x – 8 and 2x − 8 < 10.  6 is lesser and equal to 2x-8  2x-8 is lesser than 10.  Solving 6 ≤ 2x – 8 for value of x:  Adding +8 to both side:  6+8 ≤ 2x-8+8  14 ≤ 2x  Divide by “2” on both side 14 2𝑥 ≤  2 2  7≤x Or  x≥7

    

Solving 2x – 8 < 10 for value of x: Adding +8 to both side: 2x-8+8 < 10+8 2x < 18 Divide by “2” on both side 2𝑥 18  < 2 2  x9

Solution in interval Notation: [7, 9) Enter the exact answer in interval notation.

[7, 9) To enter

∞, type infinity. To enter ∪, type U.

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Question 2: (4 points) Solve the inequality involving absolute value.

|x − 3| + 6 ≥ 11 Solution:

   

adding “-6” to both sides of above relation |x-3|+6-6 ≥ 11- 6 |x-3| ≥ 5 Above relation can be solve as:  Solving x-3 ≥ + 5 for value of x:  Adding “3” to both side:  x-3+3 ≥ + 5+3  x ≥8

   

Solving x-3 ≤ -5 for value of x: Adding “3” to both side: x-3+3 ≤ - 5+3 x ≤ -2

Solution in interval Notation: (-∞, -2] U [8, ∞) Enter the exact answer in interval notation. (-∞, -2] U [8, ∞) To enter

∞, type infinity. To enter ∪, type U.

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Question 3: (2 points) Solve the inequality involving absolute value.

| Solution:

𝑥−4 5

| -2× 5

 𝑥 − 4 < 10  Adding “4” to both side:

 𝑥 − 4 > -10  Adding “4” to both side:







𝑥 − 4+4 < 10+4 𝑥 < 14



𝑥 − 4+4 > -10+4 𝑥 >-6

Solution in interval Notation: (-6, 14) Enter the exact answer in interval notation. (-6, 14)

To enter

∞, type infinity. To enter ∪, type U.

Note: There is a sample student explanation given in the feedback to this question.

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Question 4: (2 points) Describe all the x-values at a distance of 12 or less from the number 5. Solution:

Here, we want all numbers x whose distance from 5 is or less than 12:

|x − 5| ≤ 12  Above relation can be solve as:  Solving x-5 ≥ -12 for value of x:  Adding “5” to both side:  x-5+5 ≥ -12+5  x ≥ -7

   

Solving x-5 ≤ 12 for value of x: Adding “5” to both side: x-5+5 ≤ 12+5 x ≤ 17

[-7, 17] Solution in interval Notation: Enter the exact answer in interval notation. [-7, 17]

Enter your answer in interval notation.

To enter

∞, type infinity. To enter ∪, type U.

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Find the domain of the rational function.

f ( x) =

x −6 x+4

Solution:

 To find domain we will set the denominator of f(x) equal to 0 and solve for x. This value(s) of x will be where our domain does not exist.  𝑥+4 =0

 𝑥 = −4  We just solved where our function is not defined. If we plug x = - 4 into our function we get a 0 in the denominator. We can write our domain in either of the following ways:  Domain is (-∞, - 4) U (- 4, ∞) Enter your answer in interval notation. (-∞, -4) U (- 4, ∞)

To enter

∞, type infinity. To enter ∪, type U.

Note: There is a sample student explanation given in the feedback to this question.

Question 6: (4 points) Find the domain, vertical asymptotes, and horizontal asymptotes of the function.

f ( x) =

x+3 x2 −9

Solution: Doma Domain: in:  To find domain we will set the denominator of f(x) equal to 0 and solve for x. This value(s) of x will be where our domain does not exist.  𝑥2 − 9 = 0 



𝑥2 = 9 𝑥 = ±3

 We just solved where our function is not defined. If we plug 𝑥 = ±3 into our function we get a 0 in the denominator. We can write our domain in either of the following ways:  (-∞, -3) U (-3,3) U (3, ∞) Vertical asymptotes:



𝑓(𝑥) =

𝑥+3

𝑥 2 −9

 We can simplify denominator as: (x-3)(x+3) 

𝑓(𝑥) = (

𝑥+3

𝑥−3)(𝑥+3)

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1

 𝑓(𝑥) = 𝑥−3  To find the vertical asymptote(s) of a rational function, simply set the denominator equal to 0 and solve for x.  𝑥−3 =0



𝑥=3 𝑥=3

 Vertical asymptote is Horizontal asymptotes:

To find the horizontal asymptote we calculate

.

From the equation:



𝑓(𝑥) =

1

𝑥−3

 This tells us that y = 0 (which is the x-axis) is a horizontal asymptote.  Horizontal asymptote is y = 0  To enter ∞, type infinity. To enter ∪ , type U. Domain:

(-∞, -3) U (-3,3) U (3, ∞)

The fields below accept a list of numbers or formulas separated by semicolons (e.g. not matter.

2; 4; 6 or x + 1; x − 1 ). The order of the lists do

Vertical asymptotes:

x =3 Horizontal asymptotes:

y=0

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Question 7: (4 points) Find the x- and y -intercepts for the function.

f (x)

= x+10 x2 +16

Solution: X-in X-intercepts tercepts tercepts::  To find the x-intercept, we must substitute in 0 for y or f(x):  

𝑦 = 𝑓(𝑥) = 𝑥+10

0= 2 𝑥 +16 0 = 𝑥 + 10 𝑥 = −10

𝑥+10

=0

𝑥 2 +16

   There is a x-intercept at (-10,0). (Notice that 0 is the y coordinate because on the x-axis, y = 0.)

Y-inte Y-intercepts rcepts rcepts::  To find the y-intercept, we must substitute in 0 for each x:   

0+10

𝑦 = 𝑓(𝑥) = 0+16 𝑦 = 𝑓(𝑥) =

𝑦 = 𝑓(𝑥) =

10

16 5 8

 There is a y-intercept at (0, 58) (Notice that 0 is the x coordinate because on the y-axis, x = 0.) Enter your answers as points, (a, b). The x-intercept is (-10,0). 5

The y -intercept is (0, 8) Show your work and explain, in your own words, how you arrived at your answers. There are sample student explanations in the feedback to questions 3, 5, 9, and 14 that show the level of detail that is expected in your explanations.

Question 8: (4 points) Find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the function. Use that information to sketch a graph.

f (x ) =

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x2 −x−30 x2 −25

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Solution: Simplify above equation first: 

𝑓(𝑥) =

𝑥 2 −𝑥−30 𝑥 2 −25

𝑓(𝑥) =

𝑥 2 −6𝑥+5𝑥−30

 Solving using middle term breaking:    

𝑓(𝑥) =

𝑓(𝑥) =

𝑥 2 −52 𝑥(𝑥−6)+5(𝑥−6) (𝑥−5)(𝑥+5) (𝑥−6)(𝑥+5)

(𝑥−5)(𝑥+5) (𝑥−6)

𝑓(𝑥) = (𝑥−5)

X-in X-intercepts tercepts or horizon horizontal tal intercep intercepts: ts:  To find the x-intercept, we must substitute in 0 for y or f(x):

𝑦 = 𝑓(𝑥) =



𝑥−6

0= 𝑥−5 0 =𝑥−6 𝑥=6

 

𝑥−6

=0

𝑥−5

  There is an x-intercept at (6,0). (Notice that 0 is the y coordinate because on the x-axis, y = 0.)

Y-inte Y-intercepts rcepts rcepts::  To find the y-intercept, we must substitute in 0 for each x:    

𝑦 = 𝑓(𝑥) =

𝑦 = 𝑓(𝑥) =

𝑦 = 𝑓(𝑥) = 𝑦=

6

𝑥+6

𝑥−5 0−6

0−5 −6 −5

5

 There is a y-intercept at (0, 65) (Notice that 0 is the x coordinate because on the y-axis, x = 0.)

Vertical Asymptotes: Solution:



𝑓(𝑥) =

(𝑥−6)

(𝑥−5)

 The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:  𝑥−5 =0



𝑥=5

Horizontal Asymptotes:



(𝑥−6)

𝑓(𝑥) = (𝑥−5)

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.  Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

 To find the horizontal asymptote, we note that the degree of the numerator is one and the degree of the denominator is also one.  Y = Horizontal asymptote =

𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 𝑜𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

= 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 𝑜𝑓 𝑑𝑒𝑛𝑜𝑚𝑒𝑟𝑎𝑡𝑜𝑟

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1

1

=1

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Enter the intercepts as points, (a, b). The x-intercept is (6, 0). 6

The y –intercept is (0, 5) The field below accepts a list of numbers or formulas separated by semicolons (e.g. not matter.

2; 4; 6 or x + 1; x − 1 ). The order of the list does

Vertical asymptotes:

x=6 Horizontal or slant asymptote:

y= 1 Select the correct sketch of the function.

Since from the x and y intercept and from horizontal and vertical asymptote the correct graph of given function is (e)

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(a)

(b)

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(c)

(d)

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(e)

Question 9: (4 points) Write an equation for a rational function with the given characteristics.

Vertical asymptotes at x

= −2 and x = 5, x -intercepts at (−3, 0) and (1, 0), horizontal asymptote at y = −5

Solution: Given  x -intercepts at (−3, 0) and (1, 0),  As we know functions 

𝑦 = 𝑓(𝑥) =

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

 As we know in case of x-intercept y=0 

0=

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

We are given x = −3 and x = 1 or x + 3 = 0 and x − 1 = 0 and we can write it (x+3)(x-1)=numerator    

0=

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

0 ∗ 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

0 = 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 0 = (𝑥 + 3)(𝑥 − 1) = 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

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So, the function can be written as: 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑓(𝑥) = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

Or

(𝑥 + 3)(𝑥 − 1) 𝑓(𝑥) = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑥 (𝑥 − 1) + 3(𝑥 − 1) 𝑓(𝑥) = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑥 2 − 𝑥 + 3𝑥 − 3 𝑓(𝑥) = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑥 2 + 2𝑥 − 3 𝑓(𝑥) = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

Given

 Vertical asymptotes at x = −2 and x = 5 The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:  So we can find denominator of require equation using values of x=-2 and x=5    

(𝑥 + 2)(𝑥 − 5) = 0 = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑥 (𝑥 − 5) + 2(𝑥 − 5) = 0 𝑥 2 − 5𝑥 + 2𝑥 − 10 = 0 𝑥 2 − 3𝑥 − 10 = 0 = 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

Now, the function can be written as:

𝑓(𝑥) =

𝑥 2 + 2𝑥 − 3

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑥 2 + 2𝑥 − 3 𝑓(𝑥) = 2 𝑥 − 3𝑥 − 10

Given

 horizontal asymptote at y = −5

Now, the function can be written as:

Or

(−5){𝑥 2 + 2𝑥 − 3} 𝑓(𝑥) = 𝑥 2 − 3𝑥 − 10 (−5𝑥 2 − 10𝑥 + 15) 𝑓(𝑥) = (𝑥 2 − 3𝑥 − 10)

Enclose numerators and denominators in parentheses. For example, (a − b)/ (1 + n). Include a multiplication sign between symbols. For example, a * x.

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(−5𝑥 2 − 10𝑥 + 15) 𝑓(𝑥) = (𝑥 2 − 3𝑥 − 10) Note: There is a sample student explanation to this question given in the feedback to this question.

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Simplify the expression.

Solution:

      

3

√

−64 27

We can write -64=(-4)(-4)(-4) and 27= (3)(3)(3) 3

√

(−4)(−4)(−4)

3

(−4)3

√

(3)(3)(3)

(3)3

Applying cube root to both numerator and denominator 3

√(−4)3 3

√(3)3 −4 3

Question 11: (2 points) Simplify the expression.

Solution:

 √108 𝑥 4 +√27 𝑥 4  We can write 108 𝑥 4 = (2)(2)(3)(3)(3) (𝑥 2 )(𝑥 2 )     

and 27 𝑥4 =(3)(3)(3) (𝑥 2 )(𝑥 2 ) √(2)(2)(3)(3)(3) (𝑥 2 )(𝑥 2 ) +√(3)(3)(3) (𝑥 2 )(𝑥 2 ) √(2)2 (3)2 (3) (𝑥 2 )2 +√(3)2 (3)(𝑥 2 )2 Applying square: √(2)2 ∗ √(3)2 ∗ √3 ∗ √(𝑥 2 )2 +√(3)2 ∗ √3 ∗ √(𝑥 2 )2 2 * 3 𝑥 2 √3 + 3𝑥 2 √3 6 𝑥 2 √3 + 3𝑥 2 √3

  Adding both terms

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9 𝑥 2 √3

After simplification we have:

9 √3 𝑥 2

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Use function composition to determine if f (x) and g (x) are inverse functions.

Proof: 



𝑦 = 𝑓(𝑥) = √𝑥 − 1 3

Solving for “x” 3 𝑦 = √𝑥 − 1

 Taking cube power on both sides   

3

𝑦 3 = √𝑥 − 1 𝑦 3 = (𝑥 − 1) 𝑦3 + 1 = 𝑥 3

or

 

𝑥 = 𝑦3 + 1 Finally replace y with 𝑓 −1(𝑥) 𝑓 −1(𝑥) = 𝑥 3 + 1

(a) Yes, they are inverse functions.

(b) No, they are not inverse functions.

Show your work and explain, in your own words, how you arrived at your answers. There are sample student explanations in the feedback to questions 3, 5, 9, and 14 that show the level of detail that is expected in your explanations.

Question 13: (2 points) The circumference C of a circle is a function of its radius given by C (r)

= 2πr.

a. Express the radius of a circle as a function of its circumference. Call this function r (C).

       

By the definition of Pie π (that is, PI) = the ratio of the circumference of a circle over the diameter of the circle circumference of a circle π= diameter As we have C(r)=2 πr and diameter=2r, above equation can be written:

π=

C(r) 2r

Rearranging for r: C(r) r = 2π

So, radius of a circle as a function of its circumference can be written as :

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

r or f(C) = r(C) =

Enter the exact answer.

𝑟(𝐶) =

(C(r)) (2π)

(𝐶(𝑟)) (2𝜋)

Enclose numerators and denominators in parentheses. For example, (a − b)/ (1 + n).

r (C ) =

(𝐶(𝑟)) (2𝜋)

b. Find r (36 π) and interpret its meaning.

Solution:   

𝑟(𝐶) =

𝑟(36𝜋)

(𝐶(𝑟))

(2𝜋) (36𝜋) = (2𝜋)

𝑟(36𝜋) = 18

 Circle having circumference 36 r (36π ) = 18

(a) The radius of a circle with a circumference of

𝜋 has radius 18.

36 π is r (36π ) .

(b) The radius of a circle with a circumference of r (36π ) is

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36π .

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Find the inverse of the function on the given domain.

f (x ) = (x − 20)2, [20, ∞)

Solution:   

𝑓 −1 (𝑥) =? Given 𝑦 = 𝑓(𝑥) = (𝑥 − 20)2

Solving for “x” 𝑦 = (𝑥 − 20)2

 Taking square root on both sides   

√𝑦 = √(𝑥 − 20)2 √𝑦 = (𝑥 − 20) √𝑦 + 20 = 𝑥

or

 

𝑥 = √𝑦 + 20 Finally replace y with 𝑓 −1(𝑥) 𝑓 −1(𝑥) = √𝑥 + 20

f −1 (x) =

√𝑥 + 20

Note: There is a sample student explanation given in the feedback to this question.

Question 15: (2 points) Find the inverse of the function.

f (x) = 5 − 3x3

Hint: The cube root is the same as an exponent of 1/3, so for

Solution:  

3 19 x , you could type in (19*x)^(1/3). Remember your parentheses! √

𝑓 −1 (𝑥) =? Given 𝑦 = 𝑓(𝑥) = 5 − 3𝑥 3

Solving for “x”

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  

𝑦 = 5 − 3𝑥 3 3𝑥 3 = 5 − 𝑦

= 5−x 3

𝑥3

 Taking cube root on both sides  



f

5−y √𝑥 3 = √ 3 3

3

5−y 𝑥=√ 3

3 Finally replace y with 𝑓 −1(𝑥)

5−x 𝑓 −1(𝑥) = √ 3

−1 (x) =

3

√5−x

3

3

Question 16: (2 points) Find the inverse of the function.

f ( x) =

x +2 x+3

Solution:

𝑓 −1 (𝑥) =? Given



𝑦 = 𝑓(𝑥) =



Solving for “x”



𝑦=

x+2 x+3

Dividing



𝑦 = 1−

1 x+3 1  1−y 1





1−y

x+2

x+3 1 x+3

x+2 x+3

=1−

1

x+3

= 1−y =x+3...