PREPARE FOR SUCCESS IN CALCULUS

Welcome ‘Canes!

If you are currently enrolled in MTH130, MTH140, MTH151, MTH161, or MTH171 for the

upcoming semester, we encourage you to spend some time on this packet before the

start of the term. We have carefully selected topics that are essential for success in this

course.

It is important that you work through this packet on your own, so that you get a clear

gauge of your preparation. For each main topic, you will find examples worked out with

explanations. After the examples, we have selected problems for you to try on your own.

The answers to these problems are found at the end, so you can check your work.

If you feel you need to review certain topics in more depth, we recommend these

websites – just search the related topics on the site:

https://www.khanacademy.org/math/algebra-home

YouTube - Patrick JMT

Here’s to a great start of the semester and much SUCCESS!!

CALCULUS Prerequisite Review

TOPICS

Simplifying Expressions

Your Turn

Solving Equations

Quadratic

Rational

Radical

Factoring

Absolute value

Quadratic in form

Your Turn

Systems of Linear Equations

Your Turn

Systems of Non-Linear Equations

Your Turn

Geometry

Functions

Transformations of basic graphs

Domain

Composition

Inverse

Trigonometry

Graphing

Your Turn (graphing)

Solving equations

Your Turn (equations)

Answers to Practice Problems

Simplifying expressions

Rational Expressions

•

2 2

2 2

25 2 15

20 7 12

x x x

x x x x

− − −

+ − + +

( )( )

( )( )

( )( )

( )( )

( )55 5 5 3

5 4 4 3

xx x x x

x x x x

++ − − + =

+ − + +

( )5x −

( )5x + ( )

( ) ( )4 3

4

x x

x

+ +

− ( )5x − ( )3x +

4

4

x

x

+=

−

• 3 4

2 2

x x

x x

− +−

+ −

( )( ) ( )( )

( )( ) ( )( ) ( )( )

2 23 2 4 2 5 6 6 8 11 2

2 2 2 2 2 2

x x x x x x x x x

x x x x x x

− − − + + − + − − − − −= = =

+ − + − + −

Complex Fractions (Mixed Quotients)

Recall that to divide rational expressions you must multiply by the reciprocal of the denominator. This only

works if you are dividing a single rational expression by another.

There are two methods for simplifying complex fractions.

Method 1:

=

Method 2:

=

22

64

32

yx

yx

+

−

2222

22

22

22

22

2

22

2 64

)32(

64

32

64

32

64

32

xy

xyxy

xy

yx

xy

xy

yx

xy

xy

xy

yx

x

yx

y

yx

x

xy

y

+

−=

+

−=

+

−

=

+

−

22

64

32

yx

yx

+

−

2222

22

2

22

2

22

2222

22

22

22

64

)32(

64

32

64

32

64

32

xy

xyxy

xy

yxxy

y

yx

x

yx

y

yx

x

yx

yx

yx

yx

yx

+

−=

+

−=

+

−

=

+

−

Examples. Simplify the expression. Write the expression as a single quotient in which only positive

exponents and/or radicals appear.

•

2 3

4 9

x y

y x

x y

−

−

( )

( ) ( )( )2 2

2 3

2 3 2 3 1

4 9 2 3 2 3 2 34 9

xyx y y x y x

xy y x y x y x y xy x

x y

−

− − = = =− + − +

−

• 2 2

1 1

25

5

x y

x y

− −

− −

−

− CAUTION:

( )( )

( )

2 2 2 2 2 2

2 2 2 2

1 25

5 525 5

5 51 5

y x y xx y x y y x y x

x y xy x y xy y x xy

x y

− + −− + = = =

− − −

Your Turn. Perform the indicated operation. Simplify if possible.

1.

2 2

2 2

2 4 4

4

c c c c

c c c

+ − +

− −

2.

2 22 4

5 3 6

z z z

z z

+ −

+ −

3. 2

15 2 5

3 3x x x x+ +

+ +

4. 2 3

4 1 12

1 1 1x x x x+ −

+ − + +

5.

16

1 4

x y

y x

y x

−

−

6.

2

12

31

4

x

x

x

+−

+−

111)(1 −−− ++=+

bababa

Solving equations

Quadratic equations

Factoring.

• ( )9 14y y− = −

( )( )

2 9 14 0 7 2 0

7 0 7

2 0 2

2, 7

y y y y

y y

y y

− + = → − − =

− = → =

− = → =

The square root property.

• 23 54 0x − =

2 18 18 3 2

3 2

x x= → = =

•

3 62 3 36

2

9 3,

2 2

x x−

+ = → =

−

Completing the square.

• 2 6 2 0x x+ + =

( )

2 222 26 6

6 2 6 9 2 9 3 7 3 7 3 72 2

3 7

x x x x x x x

+ + = − + → + + = − + → + = → + = → = −

−

The quadratic formula.

• ( )2 2 3x x− + = −

( ) ( )( )

( )

2

2 24 4 4 2 3 4 40 4 2 10 2 10

2 4 3 0 2 4 3 02 2 4 4 2

2 10

2

x x x x x− − − − − −

− − + = → + − = → = = = =

−

36)32( 2 =+x

Rational

• 1

3 1xx

= −

2 2 1 1 123 1 3 1 0

6

1 11

6

x x x x x

i

−= − → − + = → =

• 2

1 1 6

5 4 5 1x x x x= +

− − − +

( )( )( )( )

( )( )

( )

1 1 6

5 5 1 1

1 1 6 5 1 1 6 30 5 30 6

6

5 1 5 1 1, 5x x x x xx x x x

x x x x x x

= +

− − + +

→

− +

+ = + − + = + =

− + −

− → − − → =

Factoring

• 3 25 0p p− =

( )( ) 5 5 0 0, 5p p p p+ − = → =

• 4 3 23 4 12x x x x+ = +

( )

( ) ( )( ) ( )( )

4 3 2 3 2

2 2

3 4 12 0 3 4 12 0

3 4 3 0 3 4 0 0, 3, 2

x x x x x x x x

x x x x x x x x

+ − − = → + − − =

+ − + = → + − = → = −

• 3 8 0x + =

( )( )

3 3 2

2

2 0 2 2 4 0

2 0 2

2 4 16 2 12 2 2 32 4 1 3

2 2 2

2, 1 3

x x x x

x x

ix x x i

i

+ = → + − + =

+ = → = −

− − − + → = = = =

−

Quadratic in form

• 4 22 4 7x x+ =

( )( )4 2 2 2

2

2

2 7 4 0 2 1 4 0

1 22 1

2 22

4 0 2

22,

2

x x x x

i ix x

x x

i

− + = → + − =

+ → = − = =

− = → =

Square root

• 8 26 2t t− − + = −

( ) ( )

( ) ( )( )

( ) ( )

( ) ( )

2 2

2 2

8 2 26

8 4 4 26 26

2 22 4 26

11 2 26

22 121 4 26 18 17 0 17 1 0

1

1

8 7 26 17 5 3 2

8 1 26 1 3 5 2

2

t t

t t t

t t

t t

t t t t t t t

− = − + +

− = − + + +

− − = − +

+ = +

+ + = + → + + = → +

−

− − − + −

− − + − =

+ =

=

= − = −

−

−

−

Absolute Value

Recall the definition of absolute value: , 0

, 0

a aa

a a

=

−

• 2 5 3 12x− − =

2 5 15

5 152 5 15 5, 10

2

x

x x x

− =

− = → = → = −

Your Turn. Find all solutions to each equation.

Provide exact answers (no decimals). Simplify all answers as much as possible.

7. 12

8xx

−= +

8. 3 1

12 4 2x x

− =+

9. ( )2

4 7 81x − =

10. 22 5 3 0x x− + =

11. 4 229 100 0x x− + =

12. ( ) ( )2

2 5 6 4 2 5x x+ − = +

13. 4/3 2/34 13 9 0m m− + =

14. 3 16 10x x= −

15. ( )2

3 5 12 0x − + =

16. 2

12 63

2 2g g g+ =

− −

17. ( )( ) ( )2

9 1 1x x x+ − = +

18. 6 2 5x − + =

Systems of linear equations

Recall that the solution to a system of linear equations is an ordered pair ( ),x y , which corresponds to the

point of intersection of the graphs of the two lines.

Example: 3

4

2 3 1

5

x

x

y

y

− =

+ =

We can see from the graph that the two lines appear to intersect

at the point ( )2, 3− . It is not always easy to determine the point of

intersection simply by inspecting a graph. We find this solution by solving the

system algebraically using one of two methods: substitution or elimination.

Method 1: Substitution. Select one equation and one variable in that equation. Solve for the variable.

Then, substitute your expression into the other equation.

The variable y is easy to solve for in the first equation: 4 5 5 4x y y x+ = → = −

Substitute this expression into the second equation: ( )2 3 13 2 3 135 4x y x x−− = → − =

Solve this equation and we get 2x = .

We still need our y-coordinate. As we already have an equation that is solved for y, we plug in 2x = and

evaluate: ( ) 325 4 5 4y x y= − → = − = −

Therefore, the solution to our system is the point ( )2, 3− .

Method 2: Elimination (aka Addition). The goal here is to manipulate one or both equations such that the

coefficients on one of the variables is the same number, but opposite sign, in both equations. Then, we

add the equations, term by term. If done correctly, one of the variables should be eliminated.

For our example, we can see that we can make the coefficients of the variable x both equal to 4 by

multiplying the second equation by a factor of 2. However, remember that we want them to be opposite

signs, so we multiply by –2:

( )( ) ( )( )2 22 3 13x y =−− −

Now we have the following system and we add vertically, term by term:

3

4

2

4 6 26

5

7 1

y

x

x

y

y y

− + = −

= − →

=

= −

+

Once again, the solution to our system is found to be the point ( )2, 3− .

Your turn. Solve the systems algebraically using both methods. Then, graph the lines and inspect the

graphs to see if your answers seem reasonable.

19. 2 5

3 6

x y

x y

+ =− + =

20. 2 3 19

3 7 6

x y

x y

+ =

− = −

Systems of Non-Linear Equations

Examples. Find all points of intersection and graph the system of equations.

2 2 25

4 3 0

x y

x y

+ =

− =

22 9y x

x y

= − +

=

( )

( )

2

2

2

4

3

425

3

2525

9

3

43 4

3

43 4

3

y x

x x

x

x

y

y

=

+ =

=

=

= =

= − = −

( )2

2 9

2 9

3

3

x y

y y

y y

y

x

=

= − +

= − +

=

=

POI: ( ) ( )3, 4 , 3, 4− − POI: ( )3, 3

Intercepts: ( ) ( ) ( )5, 0 , 0, 0 , 0, 5 Intercepts: ( ) ( )9

, 0 , 0, 0 , 0, 92

Your Turn. Solve the systems algebraically. Then, graph the equations on the same coordinate system and

label all points of intersection.

21. 2 12 36

8

y x x

y x

= − +

= − +

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

-8 -7 -6 -5 -4 -3 -2 -1

1 2 3 4 5 6 7 8

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

x

y

22. 2 2

2

9

3

x y

y x

+ =

= −

23. 2 12 3

2

y x

x y

− = −

− = −

Geometry

Pythagorean Theorem

Find the length of the missing side in each right triangle.

24. Legs 24 and 48

25. Leg 5 and hypotenuse 3

Areas

Find the exact area of the shaded region in each figure.

26.

27.

12 cm

25 cm

2 cm

2.5 cm

r = 3 in

r = 5 in

28.

29.

r = 3 mm

15 mm

8 mm

2 ft 2 ft

20 ft

12 ft

Functions

Transformations

Sketch the graph of each of the following. Find domain, range, and intercepts.

30. ( ) ( )2

4 9f x x= − −

31. ( ) ( )3

1 8f x x= + −

32. ( )2

2 9x y= − −

33. ( ) 7 2f x x= − + −

34. ( ) 4f x x=

35. ( ) 2 5f x x= − − +

Domain

State the domain of each function.

36. ( )2

2

9

4 17 15

xf x

x x

−=

+ +

37. ( ) 216g x x= −

38. ( )3 5

6

xh x

x

+=

−

Composition

Def: Given two functions f and g, the composite function ( )( )f g x is defined by ( )( )f g x .

( )( )f g x is read as, “f composed with g at x.”

( )( )f g x is read as, “f of g of x.”

Composition uses two (or more) functions to create a new one.

The domain of ( )( )f g x is all x such that x is in the domain of g AND ( )g x is in the domain of f .

Example.

( ) ( )5 ; 1f x x g x x= + = +

( )( ) ( ) ( )1 5 1 6f g x f x x x= + = + + = +

Domain:

𝑥 ∈ domain𝑔 → 𝑥 ∈ ℝ 𝑔(𝑥) ∈ domain𝑓 → 5 + 𝑔(𝑥) ≥ 0 → 5 + (𝑥 + 1) ≥ 0 → 𝑥 + 6 ≥ 0 → 𝑥 ≥ −6 −6,∞)

Find ( )( )f g x and state the domain of the resulting function.

39. ( ) ( )30

;6 4

xf x g x

x x= =

+ +

40. ( ) ( ) 25; 4f x x g x x= − = −

Inverse

Def.: A function is said to be one-to-one if ( ) ( )1 2 1 2x x f x f x .

Simply put, different inputs (x-values) give you different outputs (y-values).

If ( )y f x= is a one-to-one function, then we define its inverse as ( )1x f y−= .

a) The domain of ( )y f x= is the range of ( )1x f y−= .

b) The range of ( )y f x= is the domain of ( )1x f y−= .

If we compose a function and its inverse, we have the following:

a) ( )( )1f f y y− = for all y in the domain of the inverse such that ( )1f y−is in the domain of f

b) ( )( )1f f x x− = for all x in the domain of the function such that ( )f x is in the domain of 1f −

Example. Show the given functions are inverses. State the domain and range of each function.

( )3

2 5

xf x

x=

+ and ( )

5

3 2

xg x

x=

−

( )( )( )

( )( )( )

53

5 15 153 2

53 2 10 5 3 2 152 5

3 2

35

3 15 152 5

32 5 3 2 5 6 153 2

2 5

x

x x xxf g x f x

xx x x

x

x

x x xxg f x g x

xx x x

x

− = = = = = − + − +

−

+ = = = = = + + − −

+

𝐷𝑓:ℝ\ {−5

2} = 𝑅𝑔

𝐷𝑔:ℝ\ {3

2} = 𝑅𝑓

Find the inverse function of f. State the domain of f and 1f −. For #62, sketch the graph of the function

and its inverse on the same coordinate plane.

41. ( ) ( )3

1 2f x x= + −

42. ( ) 4 5f x x= + −

Trigonometry

Graphing sine or cosine

Example. ( ) 3cos 24

g x x

= +

Amplitude: 3 3= Range: 3, 3− Period: 2

2

=

Phase Shift: ( ) 3cos 2 3cos 2 P.S.8 8

is4

xg x x

= + = →

+

−

Key Points for two periods:

Subinterval length = 44

Period =

Start at and keep adding

3 5 7 9 11 13 15, , , , , , , ,

8 8 8 8 8 8 8

2

8

88 4

8

x

x

= −

==

−

Graphing secant or cosecant

Example. 1

( ) 2csc3

g x x

= −

Range: ( ), 2 2,− − Period: 2

61/ 3

=

Phase Shift: ( )1 1

( ) 2csc 2cs i3c P.S. s3

33

g xx x

= − = →

−

Key Points for two periods:

Subinterval length = 6 3

4 24

Period ==

Start at and keep adding

6 9 12 15 18 21 24 27 30, ,

6

, , , , , ,2 2 2 2 2 2 2 2

33

22

2x

x

=

=

=

−

−

x

y

Graphing tangent or cotangent

( ) tan 46

h x x

= +

Period: 4

Phase Shift: ( ) tan 4 tan 4 P.S.

4 is

6 24 2h x xx

= + = →

−

+

Vertical asymptotes first period: 42 6 2 6 12

x x

− + → −

x-intercept first period: 6 12 , 02 24 24

x

− +

= = − → −

Vertical asymptote second period (moving to the right): 3412

x

= + =

x-intercept second period: 5 5

, 024 244 24

x

= − + = →

Your Turn

43. ( ) ( )4sin 2 3f x x = −

44. ( )1

5sec4

h x x

=

45. ( )cot 6 2y x = −

−

−

−

−

x

y

Solving equations

Find a general formula for all solutions to each equation and all solution in any specified interval.

• 22sin 3cos 3 − =

( ) ( ) ( )( )2 22 1 cos 3cos 3 0 2cos 3cos 1 0 2cos 1 cos 1 0

1cos and QII QIII cos 1 , 3 ,...r

2 3oR or

− − − = → − + + = → + + =

= − → = = − → =

General: 2 4

2 , 2 , 2 where 0, 1, 2, ....3 3

k k k k

= + + + =

• tan 2 04

x

− =

in 0, 2

tan 0

2 24 4 8 2

k

x k x k x k

= → =

− = → = + → = +

In 16

0, 2 0,8

=

: 4 5 9 13

, , ,8 8 8 8 8 8

x k

= + →

• sin 2 3 sin in , = −

( )2sin cos 3 sin 0 sin 2cos 3 0

3sin 0 cos and QI or QIV

2 6

11, 2 ,

or

26 6

Rk

k k k

− = → − =

= → = = → =

= + +

In , − : , 0, , ,6 6

= − −

Your Turn

46. 2 2sin cos sin 0 − − =

47. sin sin 2 = in ( )0, 4

48. 3 22sin sin 2sin 1 0x x x+ − − =

Answers to Your Turn Problems

1.

( )

( )( )

( )

( )

22 2 2

2 2 1 1

c c c c

c c c c c

+ − − =

+ − − −

2.

( ) ( )

( )( )

( )

( )( )

2 3 2 3 2 3

5 2 2 5 2 5

z z z z z z

z z z z z z

+ − − = = −

+ − + + − +

3.

( )

( ) ( )

( )

( )

15 2 3 5 7 37 21 7

3 3 3

x x xx

x x x x x x x

+ + + ++= = =

+ + +

4.

( ) ( )

( )( ) ( )( )( )( )

( )( )

2 2

22 2 2

4 1 1 1 12 4 7 14 3 7 4 7

11 1 1 1 1 1

x x x x xx x x

x xx x x x x x x x x

− + + + − − +− − −= = =

− ++ − + + − + + − +

5.

( )( )2 2

16

4 4164

4 41 4

x y

x y x yy x xy x yx y

xy x y x y

y x

− + −− = = − +

− − −

6 ( ) ( )( )

( )( )

( )2

2 2

2 2

21 2 1 2 2 2 22 2 42 2

3 3 4 2 1 2 1 1 11

4 4

x x xx x x xx xx x

x x x x x x x

x x

+ −+ − + − +− −− −= = = =

+ − − − − + − ++

− −

7. ( )( ) 2 2 1and 2012 8 8 2 0 6 2 0 6,x x x x xx x− = + → + + = → + + = → − −

8.

( ) ( )

2 2

1and 0,

4

3 2 1 2 2 1

3 9 48 3 576 3 4 2 3 3 0

8 8

2x x x x

x x x x x x x

x+ − = +

+ → + − = + → − − = → = =

−

9. 7 9 1

4 7 9 , 44 2

x x x

− = → = → = −

10. ( )( )3

2 3 1 0 1,2

x x x

− − = → =

11. ( )( ) 2 2 2 225 4 0 25 or 4 5, 2x x x x x− − = → = = → =

12.

( ) ( )

( )

2

22 2

2 5 4 2 5 6 0 2 5

4 6 0 4 4 6 4 2 10 2 10

3 102 5 2 10

2

x x u x

u u u u u u

x x

+ − + − = → = +

− − = → − + = + → − = → =

− + = → =

13.

( )( )2/3 2/3 2/3 2/3

1/3 1/3

94 9 1 0 or 1

4

3 27 27or 1 or 1 , 1

2 8 8

m m m m

m m m m m

− − = → = =

→ = = → = = → =

14.

( )( )

( ) ( )

?

?

2 2 89 16 10 9 10 16 0 9 8 2 0 o

3

2

r 29

8 8 8 643 16 10

9

3 16 10 2 6 368

9

9 9

no

y

x x x x x x x x

x

es

= −

−

− = − − →

=

− =

= → + − = → − + = → = = −

→

→

=

15. 5 2 3

3 5 123

x x

− = → =

16. ( ) ( )22 2and 0, 2 212 3 2 6 3 12 0 3 0 !1 22g g gg g g g g+ − = → − =+ = → − = → →

17. 2 2 5

8 9 2 1 6 103

x x x x x x

+ − = + + → = → =

18. 515 6 2 3

6

9 2 2 5

9 15

1

x x

x− + = +

− = → =

→= == +

19. 9 17

,7 7

20. ( )5, 3

21.

( )( )

( ) ( )

2

2

12 36 8

11 28 0 4 7 0 4, 7

8 4, 4 , 7, 1

x x x

x x x x x x

y x

− + = − +

− + = → − − = → = =

= − + →

22.

( )

( )( )

( ) ( )

2

2 2

3

3 9 6 0

3 2 0 3, 2

3 0, 3 , 5, 2

x y

y y y y

y y y y

x y

= +

+ + = → + − =

→ + − = → = − =

= + → −

23.

( )

( )( )

( ) ( )

2 2

2

12 3 2 3 18 0

6 3 0 6, 3

2 8, 6 , 1, 3

x y

y y y y

y y y y

x y

= −

− = − − → + − =

→ + − = → = − =

= − → − −

-4 -2 2 4

-4

-2

2

4

x

y

24. 2 224 48 2880 24 2+ = =

25. 2

23 5 4 2− = =

26. ( )( ) ( )( ) 225 12 20 8 140A cm cm cm cm cm= − =

27. ( ) ( )2 2 25 3 16A in in in = − =

28. ( )( ) ( )2 215 8 3 120 9A mm mm mm mm = − = −

29. ( )( ) ( )( ) 2120 12 16 12 144

2A ft ft ft ft ft= − =

30.

x-int: ( ) ( )

( ) ( )

2 24 9 0 4 9

4 03 7, 0 ; 1,4 3

x x

x x

− − = → − =

→ − = → = →

y-int: ( ) ( ) ( )2

70 0 4 9 1 9 0,6 7f = − − = − = →

Vertex: ( )4, 9V −

Additional Points (there are many options): ( ) ( )2, 5 , 6, 5− −

Domain:

Range: )9,−

31.

x-int: ( ) ( )3 3 01 8 0 1 8 1 2 1,x x x+ − = → + = → = − + →

y-int: ( ) ( ) ( )3

70 0 1 8 7 0,f = − = − → −+

Ref. Point: ( )1, 8− −

Additional Point : ( )2, 9− −

Domain:

Range:

-1 1 2 3 4 5 6 7 8 9

-10

-8

-6

-4

-2

2

4

6

8

10

x

y

-4 -2 2 4

-12

-10

-8

-6

-4

-2

x

y

32.

x-int: ( ) ( )2

00 2 9 4 59 ,5x x= − − − − → −→ = =

y-int: ( ) ( ) ( )2

10 2 9 2 3 0, 5 ; 0,y y= − → = → −−

Vertex: ( )9, 2V −

Additional Points: ( )5, 4−

Domain: )9,−

Range:

33.

Domain: )7 0 7,x+ → −

x-int: 7 2 0 7 2 NONx x E− + − = → + = − →

y-int: ( ) ( )20 0 7 2 0, 7f = − → −− −+

Starting point: ( )7, 2− −

Additional Points: ( ) ( )2, 5 , 9, 6− −

Range: ( , 2− −

34.

Domain: )4 0 0,x →

x-int: ( )0, 0

y-int: ( )0, 0

Starting point: ( )0, 0

Additional Points: ( ) ( )1 9

, 1 , 1, 2 , , 3 , 4, 44 4

Range: )0,

35.

-10 -8 -6 -4 -2 2

-4

-2

2

4

6

8

x

y

-10 -5 5 10

-10

-5

5

10

x

y

6

y

x-int: ( ) ( )2 7, 0 , 3, 02 5 0 2 5 5x x x− − + = → − = → → −− =

y-int: ( ) ( )0 2 5 2 5 0, 3f x= − − + = − + →

Vertex: ( )2, 5

Additional Points: ( )4, 3

Domain: ℝ

Range: ( , 5−

36. 4𝑥2 + 17𝑥 + 15 ≠ 0 → (4𝑥 + 5)(𝑥 + 3) ≠ 0 → ℝ\ {−3,−5

4}

37. ( )( ) 216 0 4 4 0 4, 4x x x− → − + → −

38. ( )5 5

3 5 0 6 0 6 , 6 6,3 3

x and x x and x

+ − → − → −

39.

𝑥 ∈ domain𝑔 → 𝑥 ≠ −4

𝑔(𝑥) ∈ domain𝑓 → 𝑔(𝑥) ≠ −6 →30

𝑥 + 4≠ −6 → 30 ≠ −6𝑥 − 24 → 𝑥 ≠ −9

ℝ\{−9,−4}

40.

𝑥 ∈ domain𝑔 → 𝑥 ∈ ℝ 𝑔(𝑥) ∈ domain𝑓 → 𝑔(𝑥) − 5 ≥ 0 → 𝑥2 − 4 − 5 ≥ 0 → 𝑥2 − 9 ≥ 0 −∞− 3 ∪ 3,∞)

41. ( ) ( )3

1 2f x x= + −

𝑥 = 1 + (𝑦 − 2)3 → (𝑦 − 2)3 = 𝑥 − 1 → 𝑓−1(𝑥) = √𝑥 − 13

+ 2 𝐷𝑓:ℝ = 𝑅𝑓−1 𝑅𝑓:ℝ = 𝐷𝑓−1

42. ( ) 4 5f x x= + −

)

)

( )

( ) ( )

1

1

2

21

: 4,

: 5,

4

5

5

5

4

4,

5

Df Rf

Rf Df

x y x

f x x x

y

−

−

−

− =

− =

= + − → + = +

→ = + − −

43. Amplitude: 4 4= Range: 4, 4−

Period: 2

2

= Phase Shift:

3

2

(start) Subinterval:

4 4

T = (add)

Key points first period: 4

6 7 8 9, , , ,

4 4

0

4

1

4

Key points second period (moving to the right): 4

1 11 12 13 14, ,

4

0, ,

4 4 4

−

−

−

−

x

44. Range: ( ), 5 5,− −

Period: 2

81/ 4

= Phase Shift: None (start at x = 0) Subinterval:

82

4 4

T = = (add)

Key points first period: 0, 2 , 4 8, 6 ,

Key points second period (moving to the right): , 10 , 12 , 14 ,8 16

Asymptotes: 2 , 6 , 10 , 14 2 4x or x k = = +

**Two extra key points were added to show a complete curve on both the right and left.

45. Period: 1

6 6

= Phase Shift:

2 1

6 3

=

Vertical asymptotes first period:

1 10 6 2

3 2x x − →

x-intercept first period:

1 1

5 53 2 , 02 12 12

x

+

= = →

Vertical asymptote second period (moving to the left):

1 1 1

3 6 6x = − =

−

−

−

x

y

1 12

1 6

1 4

1 3

5 12

1 2

0 x

y

x-intercept second period: 5 1 1 1

, 012 6 4 4

x

= − = →

46.

( ) ( )( )2 2 2sin 1 sin sin 0 2sin sin 1 0 2sin 1 sin 1 0

1sin and QIII or QIV sin 1

2 6 2orR R

− − − = → − − = → + − =

= − → = = → =

General: 7 11

2 , 2 , 26 6 2

k k k

= + + +

47.

( )

Vor

sin 1 2cos 0

1sin 0 0 cos and QI QI

2 3R R or

− =

= → = = → =

General: 5

, 2 , 23 3

k k k

= + +

In ( )0, 4 : 5 7 11

, 2 , 3 , , , ,3 3 3 3

48.

( ) ( ) ( )( )2 2sin 2sin 1 1 2sin 1 0 2sin 1 sin 1 0

1sin and QIII or QIV sin 1

2 6 2orR R

x x x x x

x x x x x

+ − + = → + − =

= − → = = → =

General: 7 11

2 , 2 ,6 6 2

x k k k

= + + +