Name: _____________________________ Date: ________________ Period: ____

AP Calculus AB Summer Assignment

Congratulations on making it into AP Calculus! I am very pleased and excited to be teaching this course and

can’t wait to challenge you! AP Calculus AB is a fast-paced course that is taught at the college level. There is a

lot of material in the curriculum that must be covered before the AP exam in May. Therefore, we cannot

spend a lot of class time re-teaching prerequisite skills.

The purpose of this assignment is to have you practice the mathematical skills necessary in order for you to be

successful throughout the year and ultimately on the AP exam. I am including some reference sheets at the

end of the packet that may help you, but don’t rely on them too heavily. They are a reference, not a crutch! I

have also included some websites you may want to refer to if you need help with the topics presented in the

packet.

One piece of advice: Don’t fake your way through any of these problems because you will need to understand

everything in this very well. If not, it is possible that you will get calculus problems wrong in the future – not

because you do not understand the calculus concept, but because you do not understand the algebra or

trigonometry behind it.

Each problem should be done in the space provided. For the AP exam, you will not box, circle or underline

your final answer so let’s practice that now! Please do not circle or box your final answer. Just leave it as is.

Remember, NO WORK = NO CREDIT!! (Double exclamation points because it’s that important) I do not care

how good you are with mental math; all work must be provided. You may use additional sheets and please be

neat. Also, do not rely on a calculator. Half of your AP exam next year is taken without a calculator so use

paper and pencil techniques as much as possible.

It is a mistake to decide to do this right as school ends. Let it go until mid-summer. I want these techniques to

be relatively fresh in your mind in the fall, but of course do not wait until last minute. Spend some quality

time with this packet ☺

In addition to the questions provided, you must memorize the unit circle. You will not be allowed to use the

unit circle on any test or quiz in the class. We use trigonometry often (radians only) so be familiar with it!

You will also need a graphing calculator for this class. It is highly recommended that you purchase the TI-84

calculator and have it the first week of school. Half of the AP test is taken with a graphing calculator. It is too

time consuming to explain how to use a variety of calculators during class so only TI-84 will be taught. If you

cannot purchase one, another version of the calculator will be lent to you for the year. A scientific calculator is

not allowed during any parts of the exam; therefore, it will not be allowed during class.

If you have any questions please do not hesitate to contact me at kramos@ppmhcharterschool.org. Good luck

and see you in the fall!

Please refer to the following websites should you need any help:

http://purplemath.com/modules/index.htm

http://khanacademy.org

http://fireflylectures.com/free-algebra/

http://whyu.orghttp://whyu.org

Topic 1: Equations of Lines

Important Things to Remember About Lines

Slope-Intercept Form: 𝑦 = 𝑚𝑥 + 𝑏

** Where 𝑚 is the slope and 𝑏 is the y-intercept. **

Point-Slope Form: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)

** Where (𝑥1, 𝑦1) is the point going through the line. **

Vertical line: 𝑥 = 𝑐

** Where 𝑐 is the value of 𝑥-value that the line passes through and the slope is undefined. **

Horizontal line: 𝑦 = 𝑐

** Where 𝑐 is the value of 𝑦-value that the line passes through and the slope is zero. **

Parallel lines have equal slopes.

Perpendicular lines have opposite reciprocal slopes.

Example: Write an equation for each line.

(a) Containing (0,1) and with a slope of −2. (b) Containing points (3, −4) and (9,0).

Solution: (a) The slope, 𝑚, is given as −2. The line contains (0,1), so this point is the 𝑦-intercept, or 𝑏, which is 1.

Substituting these numbers into slope-intercept form gives you the equation 𝑦 = −2𝑥 + 1.

(b) First find the slope: 𝑚 =𝑦2−𝑦1

𝑥2−𝑥1=

−4−0

3−9=

2

3

• Next, substitute the coordinates of one of the given points (does not matter which point you choose to use) into the point-slope equation (since you have the points and the slope) → 𝑦 +

4 =2

3(𝑥 − 3) OR 𝑦 =

2

3(𝑥 − 9).

• Both are acceptable answers!

• For the purpose of this course, you do not need to go any further. You may leave your equation in point-slope form rather than distributing to get the equation in slope-intercept form.

Now you practice!

1. Determine the equation of a line passing through the point (5,-3) with an undefined slope.

2. Determine the equation of a line passing through the point (-4,2) with a slope of 0.

3. Use point-slope form to find a line passing through the point (2,8) and parallel to the line 𝑦 =5

6𝑥 − 1.

4. Find the equation of the line passing through the points (-3,6) and (1,2).

5. Use point-slope form to find a line perpendicular to 𝑦 = −2𝑥 + 9 and passing through the point (4,7).

Find the x and y intercepts for each equation.

** The 𝑥-intercepts are found by plugging in 𝑦 = 0 and solving for 𝑥. The 𝑦-intercepts are found by plugging

in 𝑥 = 0 and solving for 𝑦. **

6. 𝑦 = 2𝑥 − 5

7. 𝑦 = 𝑥2 + 𝑥 − 2

8. 𝑦 = 𝑥√16 − 𝑥2

9. 𝑦2 = 𝑥3 − 4𝑥

Topic 2: Points of Intersection

Remember that a system of equations is made up of equations of lines. You can solve a system by using the substitution or elimination method. Your answer is the POINT OF INTERSECTION. Therefore, your answer

must be expressed as an ordered pair.

Example:

Solve the system {3𝑥 + 5𝑦 = 112𝑥 + 3𝑦 = 7

using the

elimination method.

Example:

Solve the system {𝑥2 − 𝑦 = 3𝑥 − 𝑦 = 1

using the substitution

method.

Solution:

• First decide which variable you want to eliminate (does not matter which). For this example, I am choosing to eliminate the 𝑥.

• In order to eliminate the variable, their coefficients must be equal and different signs so that when I add the equations together, they will cancel out. To do this, I will multiply the top equation by −2 and the bottom equation by 3 (although you could have also multiplied the top equation by 2 and the bottom equation by −3; it does not matter.

• As a result, I end up with this system:

{−6𝑥 − 10𝑦 = −22

6𝑥 + 9𝑦 = 21

• Add the equations together and you get → −𝑦 = −1 → 𝑦 = 1.

• Now that you have the value for 𝑦, substitute it into either original equation to solve for 𝑥.

• Final answer: (2,1)

Solution:

• Solve one of the equations for one variable. In this case, it will be easier to solve for 𝑦 in either equation. I will choose to solve for 𝑦 in the second equation → 𝑦 = 𝑥 − 1.

• Plug in (substitute) what you got for 𝑦 in the first equation. → 𝑥2 − (𝑥 − 1) = 3.

• Distribute and set the equation equal to zero so you may factor. → 𝑥2 − 𝑥 + 1 = 3 → 𝑥2 − 𝑥 − 2 = 0 → (𝑥 − 2)(𝑥 + 1) = 0 → 𝑥 = 2, −1

• Plug in both 𝑥-values independently into either original equation to solve for 𝑦.

• When 𝑥 = 2 … → 2 − 𝑦 = 1 → 𝑦 = 1

• When 𝑥 = −1 … → −1 − 𝑦 = 1 → 𝑦 = −2

• Final answers: (2,1) and (−1, −2)

Find the point(s) of intersection of the graphs for the given equations.

10. {𝑥 + 𝑦 = 8

4𝑥 − 𝑦 = 7

11. {𝑥2 + 𝑦 = 6𝑥 + 𝑦 = 4

12. {2𝑥 + 7𝑦 = 4

3𝑥 + 5𝑦 = −5

13. {𝑥 = 3 − 𝑦2

𝑦 = 𝑥 − 1

Topic 3: Rational and Negative Exponents

Rules of Exponents

Generalization Explanation

Product Rule

𝑥𝑚 ∙ 𝑥𝑛 = 𝑥𝑚+𝑛 When multiplying like bases, add

the exponents

Quotient Rule

𝑥𝑚

𝑥𝑛= 𝑥𝑚−𝑛

When dividing like bases, subtract the exponents

Power Rule

(𝑥𝑚)𝑛 = 𝑥𝑚𝑛

When raising a term with an exponent with another exponent,

multiply exponents.

Expanded Power Rule

(𝑥𝑦)𝑚 = 𝑥𝑚𝑦𝑚

AND

(𝑎𝑥

𝑏𝑦)

𝑚

=𝑎𝑚𝑥𝑚

𝑏𝑚𝑦𝑚

When raising a product of terms to an exponent, raise each

individual term to the product (this is not the same as when you have a sum or difference of terms

raised to an exponent).

Negative Powers

𝑥−𝑝 =1

𝑥𝑝 OR 1

𝑥−𝑝 = 𝑥𝑝

Negative exponents mean that factors remain but they are in the

denominator if the factor was originally in the numerator and

vice versa.

Zero Powers

𝑥0 = 1

Any non-zero 𝑥-values raised to an exponent of zero is always

equal to 1

Rational Exponents

𝑥𝑛𝑑 = √𝑥𝑛𝑑

The denominator of a rational exponent is the root (index of the radical) and the numerator is the

power.

Simplify using only positive exponents.

14. 3𝑥−3

𝑥−2

15. (9𝑥2𝑦−4)−1

2

16. (16𝑥2𝑦)3

4

17. (𝑥2)

3𝑥

𝑥7

18. √4𝑥−16

√(𝑥−4)34

19. √𝑥 ∙ √𝑥3 ∙ 𝑥1

6

20. [(𝑥−1𝑦1

3) (𝑥−4

3𝑦2)]2

21. (𝑥

12𝑦−2

𝑦𝑥−

74

)

4

Topic 4: Factoring

When factoring a trinomial in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐, remember that you are looking for the two numbers that

multiply to give you 𝑐 and also add to give you 𝑏. If there 𝑎 ≠ 1, check to see if the coefficient can be factored out first before you continue to factor.

Example: 2𝑥2 − 4𝑥 − 30 = 2(𝑥2 − 2𝑥 − 15) = 2(𝑥 − 5)(𝑥 + 3)

Special Factorization Methods

General Example(s)

Perfect Square Trinomials

𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2 𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2

• 𝑥2 + 6𝑥 + 9 = (𝑥 + 3)2

• 𝑥2 − 8𝑥 + 16 = (𝑥 − 4)2

Difference of Squares

𝑎2 − 𝑏2 = (𝑎 + 𝑏)(𝑎 − 𝑏)

• 4𝑥2 − 9 = (2𝑥 + 3)(2𝑥 − 3)

AC Method

𝑎𝑥2 + 𝑏𝑥 + 𝑐 where 𝑎 ≠ 1 and 𝑎 cannot be factored.

• Find ac.

• Find factors of ac that add up to b.

• Write the original expression, but rewrite bx using your new factors.

• Factor by grouping.

• 2𝑥2 − 5𝑥 − 3

• 𝑎𝑐 = −6, 𝑏 = −6,1

• 2𝑥2 − 6𝑥 + 𝑥 − 3

• 2𝑥(𝑥 − 3) + 1(𝑥 − 3)

= (𝑥 − 3)(2𝑥 + 1)

Factoring by Grouping

Only use this method if the polynomial you are trying to

factor has four terms:

𝑎𝑐 + 𝑐𝑑 − 𝑎𝑏 − 𝑏𝑑 = 𝑐(𝑎 + 𝑑) − 𝑏(𝑎 + 𝑑)

= (𝑎 + 𝑑)(𝑐 − 𝑏)

• Factor the GCF of the first two factors and factor the GCF of the second two factors.

• What is in parenthesis will be one of your final factors.

• What is left outside the parenthesis is the other factor.

• 𝑥3 + 3𝑥2 − 2𝑥 − 6

• (𝑥3 + 3𝑥2) + (−2𝑥 − 6)

• 𝑥2(𝑥 + 3) − 2(𝑥 + 3)

= (𝑥 + 3)(𝑥2 − 2)

** Although there are many other factoring methods, these are the ones that we will mostly be using

throughout the course. Factoring is SUPER important! I will not have time to reteach factoring throughout the

year so make sure that this is a skill that you have mastered. **

Factor each of the following polynomials completely.

22. 𝑥2 − 20𝑥 + 36

23. 2𝑥3 − 2𝑥2 − 4𝑥

24. 𝑥3 − 5𝑥2 − 𝑥 + 5

25. 2𝑥2 + 𝑥 − 6

26. 9𝑥2 − 4𝑦4

27. 3𝑥2 + 3𝑥 − 60

28. 5𝑥2 + 29𝑥 + 20

29. 𝑥6 + 2𝑥4 − 16𝑥2 − 32

Topic 5: Solving Polynomials by Factoring

To solve a polynomial by factoring simply means to set the polynomial equal to zero first, factor and then set each individual factor equal to zero and solve for 𝑥.

Solve by factoring.

30. 4𝑥2 + 𝑥 − 36 = 𝑥2 + 4𝑥 31. 9𝑥2 + 2𝑥 − 4 = 2𝑥 − 3

Topic 6: The Quadratic Formula

The solutions to a quadratic equation written in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎 ≠ 0, can be found by using the quadratic formula:

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

** Where the values of 𝑥 represent the zeros or roots or 𝑥-intercepts of the quadratic equation. **

Example: Use the quadratic formula to solve: 4𝑥2 + 4𝑥 − 8 = 1

Solution:

• First, you have to set the quadratic equal to zero in order to use the quadratic formula: → 4𝑥2 + 4𝑥 − 9 = 0

• In this case, 𝑎 = 4, 𝑏 = 4, and 𝑐 = −9. Plug in these values into the quadratic formula.

• 𝑥 =−4±√42−4(4)(−9)

2(4)=

−4±√16+144

8=

−4±√160

8

• Simplify the square root, if possible.

• 𝑥 =−4±4√10

8→ Simplify your final answer

• Final answer: −1±√10

2

Use the quadratic formula to solve each equation.

32. 4𝑥2 + 3 = 8𝑥

33. 3𝑥2 + 𝑥 = 2𝑥2 + 6𝑥 − 4

Topic 7: Functions

In function notation, the symbol 𝑓(𝑥) is read 𝑓 of 𝑥 and interpreted as the function value of 𝑓 at 𝑥. In other words, 𝑦 = 𝑓(𝑥).

** To find a function value, simply plug in the value into every 𝑥 and simplify. **

Example: If 𝑔(𝑥) = 𝑥2 + 3𝑥, find 𝑔(−5).

Solution:

• 𝑔(−5) means to replace 𝑥 with the value −5 and evaluate for 𝑔(𝑥).

• 𝑔(−5) = (−5)2 + 3(−5) = 25 − 15 = 10

• Final answer: 𝑔(−5) = 10

Evaluate each function for the given function value.

Let 𝑓(𝑥) = 5 −2𝑥

3 and 𝑔(𝑥) =

1

2𝑥2 + 3𝑥

34. 𝑔 (1

2)

35. −3𝑓(−2)

36. 𝑓(6) + 2𝑔(1)

37. 𝑔(4) − 𝑓(5)

Topic 8: Operations with Functions

To write a sum, difference, product, or quotient of two functions, 𝑓 and 𝑔, write the sum difference, product, or quotient of the expressions that define 𝑓 and 𝑔, then simplify. The domain of the combined

function is restricted to the domains of each original function.

Example: Let 𝑓(𝑥) = 𝑥2 + 3𝑥 + 2 and 𝑔(𝑥) = 5𝑥 − 1. Write an expression for each function.

(a) (𝑓 + 𝑔)(𝑥) (b) (𝑓 − 𝑔)(𝑥) (c) (𝑓𝑔)(𝑥)

(d) (𝑓

𝑔) (𝑥)

** Keep in mind that the domain of both 𝑓 and 𝑔 is (−∞, ∞) since both functions are polynomial functions and are continuous everywhere and extend in both directions. **

Solution:

(a) (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = (𝑥2 + 3𝑥 + 2) + (5𝑥 − 1) = 𝑥2 − 2𝑥 + 3

• Domain: (−∞, ∞)

(b) (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) = (𝑥2 + 3𝑥 + 2) − (5𝑥 − 1) = 𝑥2 + 3𝑥 + 2 − 5𝑥 + 1 = 𝑥2 − 2𝑥 + 3

• Domain: (−∞, ∞)

(c) (𝑓𝑔)(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥) = (𝑥2 + 3𝑥 + 2)(5𝑥 − 1) = 5𝑥3 − 𝑥2 + 15𝑥2 − 3𝑥 + 10𝑥 − 2 = 5𝑥3 + 14𝑥2 + 7𝑥 − 2

• Domain: (−∞, ∞)

(d) (𝑓

𝑔) (𝑥) =

𝑓(𝑥)

𝑔(𝑥)=

𝑥2+3𝑥+2

5𝑥−1, where 𝑥 ≠

1

5 (because that is the value that would make the denominator

equal to zero.

• Domain: (−∞,1

5) ∪ (

1

5, ∞)

Find each new function and state the combined function’s domain.

Let 𝑓(𝑥) = 3𝑥2 + 2, 𝑔(𝑥) = 2𝑥 − 1, and ℎ(𝑥) = 𝑥2 + 5𝑥.

38. (𝑓 + 𝑔)(𝑥)

39. (𝑔ℎ)(𝑥)

40. (𝑓 − ℎ)(𝑥)

41. (𝑔

ℎ) (𝑥)

Topic 9: Composition of Functions

A composition of functions denoted as 𝑓(𝑔(𝑥)) or 𝑓[𝑔(𝑥)], is read as “𝑓 of 𝑔 of 𝑥,” and means to plug in

the inside function (in this case 𝑔(𝑥)) in for 𝑥 in the outside function (in this case 𝑓(𝑥)).

Example:

Given 𝑓(𝑥) = 2𝑥2 + 1 and 𝑔(𝑥) = 𝑥 − 4, find 𝑓[𝑔(1)]. ** To find a composition at a function value, you first need to find 𝑓[𝑔(𝑥)] and then you plug in the given 𝑥-value, in this case 𝑥 = 1. **

Solution:

• 𝑓[𝑔(𝑥)] = 𝑓(𝑥 − 4) = 2(𝑥 − 4)2 + 1 • 2(𝑥2 − 8𝑥 + 16) + 1 • 2𝑥2 − 16𝑥 + 32 + 1 • 𝑓[𝑔(𝑥)] = 2𝑥2 − 16𝑥 + 33

→Now you can plug in 𝑥 = 1.

• 𝑓[𝑔(1)] = 2(1)2 − 16(1) + 33 • Final answer: 𝑓[𝑔(1)] = 19

Find the following composition of functions at the indicated function value, if given.

Let 𝑓(𝑥) = 𝑥2 − 1, 𝑔(𝑥) = 3𝑥 − 𝑥2, and ℎ(𝑥) = 5 − 𝑥

42. ℎ[𝑓(𝑥)]

43. 𝑔[ℎ(2)]

44. 𝑓[𝑔(𝑥)]

45. 𝑓[ℎ(𝑔(1))]

Topic 10: Evaluating for Piecewise Functions

A piecewise function is exactly what it sounds like. It is a function that is defined in “pieces.” It is a function that combines pieces of different equations that are only defined at specified intervals. To evaluate for a function value for a piecewise function, you must first identity the interval where the indicated function

value lies and plug it into the equation that is defined in that interval.

Visual Representation

𝑓(𝑥) = {𝑥 + 3, 𝑥 < −1

𝑥2 , − 1 ≤ 𝑥 ≤ 23, 𝑥 > 2

To graph a linear piecewise function, make sure you remember how to graph lines by using the slope and y-intercept. Also, keep in mind the endpoints of each “piece.” If the < or > symbols are used, then that 𝑥-value is NOT included and would be graphed with an open circle. If the ≤ or ≥ symbols are used, the that 𝑥-value IS included and would be graphed with a solid circle.

Example:

Graph the piecewise function given below and find 𝑓(3).

𝑓(𝑥) = {1

3𝑥 + 2, 𝑥 ≤ 3

−𝑥 + 3, 3 < 𝑥 ≤ 4

Solution:

Although you could find 𝑓(3) graphically, if you were to find the function value algebraically, you would have to see where 𝑓(3) is defined. The value of 𝑥 = 3 is only included in the first equation and not in the second. Therefore, to find 𝑓(3), you would plug in 𝑥 = 3 into the first equation.

𝑓(3) =1

3(3) + 2 = 3

Graph each function and find the indicated function value algebraically.

46. 𝑓(𝑥) = {𝑥 + 3, 𝑥 < 0−2𝑥 + 5, 𝑥 ≥ 0

• 𝑓(3) = _____________

47. 𝑓(𝑥) = {1

2𝑥, − 4 ≤ 𝑥 ≤ 2

2𝑥 − 3, 𝑥 > 2

• 𝑓(2) = _____________

Topic 11: Transformations of Function Graphs

Transformations of function graphs occur when there are translations, reflections and dilations. The operations performed on a function indicate what transformations occurred. Addition and subtraction

mean that there was a translation. Multiplication and division mean that there was a dilation. A negative (not subtraction) means that the graph was reflected either over the 𝑥-axis or the 𝑦-axis.

** Anything “outside” affects the 𝑦-values and does the same as the operation indicates. Anything “inside” affects the 𝑥-values and does the opposite operation indicated. **

Transformation Notation

Verbal Description

𝑓(𝑥 + ℎ)

Horizontal shift left ℎ units

𝑓(𝑥 − ℎ)

Horizontal shift right ℎ units

𝑓(𝑥) + 𝑘

Vertical shift up 𝑘 units

𝑓(𝑥) − 𝑘

Vertical shift down 𝑘 units

𝑎𝑓(𝑥)

Vertical stretch with a scale factor of 𝑎

1

𝑎𝑓(𝑥)

Vertical shrink with a scale factor of 1

𝑎

𝑓(𝑏𝑥)

Horizontal shrink with a scale factor of 1

𝑏

𝑓 (1

𝑏𝑥)

Horizontal stretch with a scale factor of 𝑏

−𝑓(𝑥)

Reflection over the 𝑥-axis

𝑓(−𝑥)

Reflection over the 𝑦-axis

Identify the parent function for each of the following. Then, describe the transformations of the parent

functions indicated from the function’s equation.

48. 𝑓(𝑥) = −2|𝑥 + 1| − 4

• Parent function: __________________

49. 𝑓(𝑥) =1

3√−2𝑥 + 3

• Parent function: __________________

50. 𝑓(𝑥) = (1

4(𝑥 − 1))

2

• Parent function: __________________

51. 𝑓(𝑥) =3

−(𝑥+2)− 7

• Parent function: __________________

Topic 12: Domain and Range

Domain: The set of all the 𝑥-values for which a function is defined (input values). Range: The set of all the 𝑦-values (output values).

• All polynomial functions have a domain of (−∞, ∞) since their graphs go forever left and right encompassing all possible 𝑥-values.

** Please refer to the reference sheet at the end of this packet for graphs of all parent functions in order to be able to determine domain and range. **

Below are examples of when the domain is restricted based on the function’s graph and how to find the domain algebraically.

Function Type

Finding Domain Algebraically

Radical Functions (𝑓(𝑥) = √𝑥)

You can never have a negative inside of a square root (because then you would end up with the imaginary unit). Therefore, to

find the domain you set the radicand (inside of the radical) greater than or equal to zero and solve for 𝑥.

Rational Functions (𝑓(𝑥) =1

𝑥)

Rational functions are basically fractions (division). Since you can never divide by zero, to find the domain, you set the denominator to “not equal” zero and find the restriction.

Trigonometric Functions (𝑓(𝑥) = sin 𝑥 and cos 𝑥 only )

The graphs of sine and cosine are waves that oscillate forever in both directions. Therefore, the domains of both functions are

always (−∞, ∞).

Natural Log Functions (𝑓(𝑥) = ln 𝑥)

The graph of natural log has a vertical asymptote at 𝑥 = 0. Therefore, to find the domain, like the radical function, you set

what is “inside” of 𝑙𝑛 to be greater than (not equal) to zero.

State the domain and range of each of the following functions.

** To find the domain of the functions, follow the guidelines in the chart on the previous page. To determine

the range, you have to visualize the function’s behavior based on the vertical shifts. If the parent function

normally has a range of [0, ∞) but the function was shifted up 5 units, then the range would be [5, ∞). **

52. 𝑓(𝑥) = 𝑥2 − 3

• Domain: __________________________

• Range: ___________________________

53. 𝑓(𝑥) = 3 sin 𝑥

• Domain: __________________________

• Range: ___________________________

54. 𝑓(𝑥) = −√𝑥 + 3

• Domain: __________________________

• Range: ___________________________

55. 𝑓(𝑥) =2

𝑥−1

• Domain: __________________________

• Range: ___________________________

Topic 13: Inverses of Functions

Functions are considered inverses if and only if their 𝑥 and 𝑦 values have switched places. Graphically, this means that inverse functions are reflections along the line 𝑦 = 𝑥.

** Inverse function notation is denoted as 𝑓−1(𝑥). **

To find the inverse of a function, simply interchange the 𝑥 and 𝑦 and solve for 𝑦. You can also verify your answer by checking that 𝑓[𝑓−1(𝑥)] = 𝑥 and 𝑓−1[𝑓(𝑥)] = 𝑥.

Example:

Find the inverse of the function 𝑓(𝑥) = √𝑥 + 13

. Then, verify that your answer is the inverse by finding at least one of the compositions 𝑓[𝑓−1(𝑥)] = 𝑥 or 𝑓−1[𝑓(𝑥)] = 𝑥.

Solution:

• Rewrite 𝑓(𝑥) as 𝑦 → 𝑦 = √𝑥 + 13

• Interchange the 𝑥 and 𝑦 → 𝑥 = √𝑦 + 13

• Solve for 𝑦 → 𝑥3 = 𝑦 + 1 → 𝑥3 − 1 = 𝑦

• Rewrite using appropriate inverse function notation → 𝑓−1(𝑥) = 𝑥3 − 1

Verify your answer by finding one of the compositions indicated.

• 𝑓[𝑓−1(𝑥)] = √(𝑥3 − 1) + 13

= √𝑥3 + 1 − 13

√𝑥33= 𝑥

Since the composition equals to 𝑥, the inverse is

correct!

Find the inverse for each function. Verify that your answer is correct through composition.

56. 𝑓(𝑥) = 2𝑥 + 1

57. 𝑓(𝑥) =𝑥2

3

58. 𝑓(𝑥) =5

𝑥−2

59. 𝑓(𝑥) = √4 − 𝑥 + 1

Topic 14: Even and Odd Functions

Even functions are functions that are symmetric over the 𝑦-axis.

• To determine algebraically if a function is even, you must find out if 𝑓(−𝑥) = 𝑓(𝑥).

• This means that if you plug in −𝑥 and all of the terms remain the same sign, then the function is even.

Odd functions are functions that are symmetric over the origin.

• To determine algebraically if a function is odd, you must find out if 𝑓(−𝑥) = −𝑓(𝑥).

• This means that if you plug in −𝑥 and all of the terms change signs, then the function is odd.

** If you plug in −𝑥 into the function and only some terms change signs and others don’t, then the function is neither even nor odd. **

Example of an even function:

𝑓(𝑥) = 2𝑥2 + 3|𝑥|

→ 𝑓(−𝑥) = 2(−𝑥)2 + 3|−𝑥| = 2𝑥2 + 3|𝑥|

Since 𝑓(−𝑥) looks identical to 𝑓(𝑥), then 𝑓(𝑥) is an even function.

Example of an odd function:

𝑓(𝑥) = −5𝑥3 + 3𝑥

→ 𝑓(−𝑥) = −5(−𝑥)3 + 3(−𝑥) = 5𝑥3 − 3𝑥

Since all of the terms in 𝑓(−𝑥) are the opposite signs of 𝑓(𝑥), then 𝑓(𝑥) is an odd function.

Example of a function that is neither even nor odd:

𝑓(𝑥) = 2𝑥5 + 1

→ 𝑓(−𝑥) = 2(−𝑥)5 + 1 = −2𝑥5 + 1

Since only one term changed signs and the 1 did not, this means that 𝑓(𝑥) is neither even nor odd.

State whether each of the following functions are even, odd, or neither.

60.

61.

62. 𝑓(𝑥) = 2𝑥4 − 5𝑥2

63. 𝑓(𝑥) = 𝑥5 − 3𝑥3 + 𝑥

64. 𝑓(𝑥) = 2𝑥2 − 5𝑥 + 3

65. 𝑓(𝑥) = 2 cos 𝑥

Topic 15: Asymptotes of Rational Functions

Asymptotes are vertical or horizontal lines that the function will get infinitely close to but never touch.

Finding Asymptotes Algebraically

Example

To find vertical asymptotes, set the denominator

equal to zero to find the 𝑥-value for which the function is undefined (granted that the factor that

makes the denominator equal to zero is not a common factor of the numerator – this is a

removable discontinuity aka a hole).

Find the vertical asymptote of 𝑓(𝑥) =𝑥2−5𝑥−14

𝑥2−3𝑥−28.

• Factor both the numerator and denominator

• 𝑓(𝑥) =(𝑥−7)(𝑥+2)

(𝑥−7)(𝑥+4)→ Simplify

• 𝑓(𝑥) =𝑥+2

𝑥+4

• At 𝑥 = −4 there is a vertical asymptote where at 𝑥 = 7 there is a hole.

Horizontal asymptotes can be found depending on the following three cases:

(a) If the degree of the numerator < the degree of the denominator, then there is a horizontal asymptote at 𝑦 = 0.

(b) If the degree of the numerator = the degree of the denominator, then there is a horizontal asymptote that is equal to the quotient of the leading coefficients.

(c) If the degree of the numerator > the degree of the denominator, then there is NO horizontal asymptote.

Finding the horizontal asymptotes for all three cases:

(a) 𝑓(𝑥) =3𝑥

𝑥2+2→ HA @ 𝑦 = 0

(b) 𝑓(𝑥) =2𝑥2+3𝑥−1

4𝑥2+5→ HA @ 𝑦 =

2

4=

1

2

(c) 𝑓(𝑥) =5𝑥3−𝑥

𝑥2+2𝑥→ No horizontal asymptote

For each function, find the equations of the vertical and horizontal asymptotes as well as any holes that may

be present on the graph (if any).

66. 𝑓(𝑥) =𝑥+4

𝑥2+1

67. 𝑓(𝑥) =𝑥2+5𝑥−6

𝑥2−1

68. 𝑓(𝑥) =2𝑥3

𝑥2−4

69. 𝑓(𝑥) =2𝑥2−4𝑥+2

𝑥2−3𝑥−4

Topic 16: Solving Rational Equations

There are two ways of solving rational equations: 1. Cross multiplication – this method only works if you have one rational expression that equals to

another rational expression. 2. Multiply the least common denominator (LCD) to the entire equation – this method will always work

but is most used when you have addition or subtraction of rational functions on at least one side of the equation.

** Keep in mind that rational equations have excluded values which can lead to extraneous solutions. The excluded values are the values of 𝑥 that make the denominator equal to zero. **

Example of cross multiplication:

• 𝑥−5

𝑥+1=

𝑥−9

𝑥+5→ Excluded values: 𝑥 = −1, −5

• Cross multiply

• (𝑥 − 5)(𝑥 + 5) = (𝑥 − 9)(𝑥 + 1) → Distribute

• 𝑥2 − 25 = 𝑥2 − 8𝑥 − 9 → Solve for 𝑥.

• Although the resulting equation is a quadratic, if you subtract 𝑥2 from both sides, the term will cancel out.

• −16 = 8𝑥 → Divide by 8 to both sides.

• Final answer: 𝑥 = −2

** Since 𝑥 = −2 is not an excluded value, that is the real solution. **

Example of multiplying the LCD:

• 12

𝑥−1−

8

𝑥= 2 → Excluded values: 𝑥 = 1, 0

• LCD: 𝑥(𝑥 − 1)

• Multiply the LCD to the entire equation.

• 𝑥(𝑥 − 1) [12

𝑥−1−

8

𝑥= 2]

• 12𝑥 − 8(𝑥 − 1) = 2𝑥(𝑥 − 1) → Distribute

• 12𝑥 − 8𝑥 + 8 = 2𝑥2 − 2𝑥 → Since the resulting equation is a quadratic, set it to equal to zero and solve by factoring.

• 2𝑥2 − 6𝑥 − 8 = 2(𝑥2 − 3𝑥 − 4) = 2(𝑥 − 4)(𝑥 + 1) = 0

• Final answer: 𝑥 = 4, −1

** Since 𝑥 = 4, −1 are none of the excluded values, both are the real solutions of the equation. **

Solve for 𝑥 for each rational equation:

70. 𝑥−5

𝑥+1=

3

5

71. 𝑥+3

𝑥−4=

𝑥+4

𝑥−10

72. 1

𝑥−2+

1

𝑥2−7𝑥+10=

6

𝑥−2

73. 2𝑥+3

𝑥−1=

10

𝑥2−1+

2𝑥−3

𝑥+1

Topic 17: Logarithms and Exponential Functions

For 𝑥 > 0 and 𝑏 > 0, 𝑏 ≠ 1 … 𝑦 = log𝑏 𝑥 is equivalent to 𝑏𝑦 = 𝑥,

…where 𝑏 is the base and 𝑦 is the exponent.

** Basically, a logarithmic function is the inverse of an exponential function. **

*** The common logarithm, log10 𝑥, is usually written as log 𝑥. ***

The logarithmic function can have any base as long as the base is not zero or negative. However, the natural logarithmic function (ln 𝑥) has a base of 𝑒, where 𝑒 is the exponential base and is

equal to the irrational number 2.718281 …

Throughout the course, we will be working with logarithmic functions, more commonly the natural logarithmic function.

Below is a list of properties you should familiarize yourself with regarding the properties of log. These properties can be applied to both logarithmic and natural logarithmic ( ln 𝑥) functions.

Property

Definition

Example

Product

log𝑏 𝑚𝑛 = log𝑏 𝑚 + log𝑏 𝑛

• log3 9𝑥 = log3 9 + log3 𝑥 • ln 𝑥(𝑥 + 1) = ln 𝑥 + ln 𝑥 + 1

Quotient

log𝑏

𝑚

𝑛= log𝑏 𝑚 − log𝑏 𝑛

• log2 (4𝑥+1

𝑥+2)

= log2(4𝑥 + 1) − log2(𝑥 + 2)

• ln (8

𝑥) = ln 8 − ln 𝑥

Power

log𝑏 𝑚𝑝 = 𝑝 ∙ log𝑏 𝑚

• log2 8𝑥 = 𝑥 ∙ log2 8 • ln 𝑥2 = 2 ln 𝑥

Equality

log𝑏 𝑚 = log𝑏 𝑛 , then 𝑚 = 𝑛

• log3(3𝑥 − 4) = log3(5𝑥 + 2), then 3𝑥 − 4 = 5𝑥 + 2

• This property does not apply to the natural log function.

Logarithmic Exponent with Equal Bases

𝑏log𝑏 𝑥

• 4log4(𝑥−1) = 𝑥 − 1 • 𝑒ln 2𝑥 = 2𝑥

• This is because the base of ln 𝑥 is 𝑒 so they would cancel each other out.

Expand the following logarithmic functions using the properties of logarithms.

74. ln(𝑥𝑦)6 75. log[(𝑥 + 1)(𝑥2)] 76. ln (5𝑥+3

𝑦2 )

Topic 18: Natural Log and Exponential Functions with Base 𝒆

Since the natural logarithmic function has a base of 𝑒, this means that ln 𝑥 and 𝑒 are reciprocal functions and would cancel each other out if they are applied to one another.

When 𝑒 has an exponent of ln 𝑥:

𝑒ln 𝑥 = 𝑥

Example:

𝑒ln(2𝑥+4) = 2𝑥 + 4

When 𝑒 is “inside” of ln 𝑥:

ln 𝑒𝑥 = 𝑥

Example:

ln 𝑒𝑥2+1 = 𝑥2 + 1

To solve for 𝑥 when given an exponential equation with base 𝑒, apply the function 𝑙𝑛 to both sides of the equation to cancel out the 𝑒.

Example: Solve for 𝑥 given that 𝑒2𝑥 = 5

Solution:

• Apply 𝑙𝑛 to both sides → ln 𝑒2𝑥 = ln 5

• Simplify → 2𝑥 = ln 5

• Divide by 2 → 𝑥 =ln 5

2

** Unless this was a calculator question, you are not expected to know the ln 5, so you would leave your

final answer as 𝑥 =ln 5

2. **

To solve for 𝑥 when given a natural logarithmic equation, apply the function 𝑒 to both sides of the equation to cancel out the 𝑙𝑛.

Example: Solve for 𝑥 given that ln(4𝑥 + 1) = 3

Solution:

• Apply 𝑒 to both sides → 𝑒ln(4𝑥+1) = 𝑒3

• Simplify → 4𝑥 + 1 = 𝑒3

• Isolate the 𝑥 → 𝑥 =𝑒3−1

4

** Just like in the previous example, you would leave your answer as is. **

Solve for 𝑥 for each of the given equations.

77. ln 𝑒2𝑥 = 5𝑥 − 3

78. 5 ln 𝑒3𝑥 = 30

79. 8𝑒𝑥 = 51

80. ln(3𝑥 − 2) = 7

Topic 19: The Unit Circle

Remember that you are expected to know the Unit Circle! However, you only need to remember the first quadrant. The values of sine and cosine never change as you revolve around the Unit Circle, only the signs.

** I have provided a copy of the Unit Circle at the end of this packet. **

• All trigonometric function values are positive in

quadrant 1.

• Only sine and cosecant function values are

positive in quadrant 2.

• Only tangent and cotangent function values are

positive in quadrant 3.

• Only cosine and secant function values are

positive in quadrant 4.

On the Unit Circle, the 𝑥 = cos 𝜃 and 𝑦 = sin 𝜃.

• Tangent is found by dividing sine by cosine

→ tan 𝜃 =sin 𝜃

cos 𝜃.

• Secant is the reciprocal of cosine.

• Cosecant is the reciprocal of sine.

• Cotangent is the reciprocal of tangent.

Examples of finding a trigonometric function value:

(a) sec𝜋

3→ Since secant is the reciprocal of cosine,

we would first need to find cos𝜋

3 and then its

reciprocal.

• cos𝜋

3=

1

2→ Therefore sec

𝜋

3= 2

(b) tan𝜋

3=

sin𝜋

3

cos𝜋

3

=√3

21

2

• To divide the fractions, “copy, change, flip.”

• √3

2(

2

1) = √3

• Therefore, tan𝜋

3= √3

Example of finding the inverse of a trigonometric function value:

To evaluate for the inverse of a trigonometric function means to find the value of the angle, 𝜃, that would give the value inside the parenthesis.

(a) cos−1 (−√2

2) =

3𝜋

4,

5𝜋

4

• Both angles are the solutions because the

cosine value for both angles is equal to −√2

2.

(b) Evaluate sin (cos−1 (1

2)) on the interval

0 < 𝜃 <𝜋

2.

• The interval provided lets you know what quadrant of the Unit Circle that the angle is located in. According to the interval given, the angle is in quadrant 1.

• First, we would have to evaluate for the inverse cosine and then find the sine value at the angle that you found.

• sin (cos−1 (1

2)) = sin (

𝜋

3) =

√3

2

Evaluate the following trigonometric and inverse trigonometric function values.

81. sec11𝜋

6

82. cot 𝜋 83. sin5𝜋

4

84. sin−1(−1)

85. csc (sin−1 (√2

2) ) on the

interval 𝜋

2< 𝑥 < 𝜋

86. tan (sin−1 (−1

2)) on the

interval 𝜋 < 𝑥 <3𝜋

2

Topic 20: Solving Trigonometric Equations

To solve a trigonometric equation follows the same logical pathways as solving any other equation. The goal is to isolate the trigonometric function and then applying the inverse of the trigonometric function to solve for 𝑥. Depending on how the equation is presented, you may isolate the trigonometric function from

the beginning or (if given a squared trigonometric function) you may have to square root or factor.

Examples:

Solve the following trigonometric equations on the interval [0,2𝜋).

** Side note: Remember that intervals could be written in either set notation or interval notation. The interval given indicates that 𝑥 = 0 is included and could be a solution but 𝑥 = 2𝜋 is not included and therefore will not be a solution. The interval simply means that you will state all of the solutions in one full revolution around the Unit Circle. **

(a) 4 sin 𝑥 = 2 sin 𝑥 + √2

(b) 4 cos2 𝑥 + 1 = 4

(c) 2 sin2 𝑥 + 3 sin 𝑥 = −1

Solutions:

(a) Isolate sin 𝑥 by first subtracting 2 sin 𝑥 from both sides of the equation

→ 2 sin 𝑥 = √2

• Divide by 2 → sin 𝑥 =√2

2

• Apply the inverse of sine to isolate the 𝑥 → 𝑥 = sin−1 (√2

2)

• Final answer: 𝑥 =𝜋

4,

3𝜋

4

(b) Isolate cos2 𝑥 by subtracting the 1 and then dividing the 4

→ cos2 𝑥 =3

4

• Square root both sides to eliminate the exponent.

→ √cos2 𝑥 = √3

4=

√3

√4=

√3

2→ cos 𝑥 =

√3

2

• Apply the inverse of cosine to isolate the 𝑥 → 𝑥 = cos−1 (√3

2)

• Final answer: 𝑥 =𝜋

6,

11𝜋

6

(c) Since this equation is quadratic, first set it equal to zero so we can factor → 2 sin2 𝑥 + 3 sin 𝑥 + 1 = 0

• Just so that it is easier to see how this can be factored, I will be replacing sin 𝑥 with 𝑢 during the factoring process.

• If 𝑢 = sin 𝑥, then 2 sin2 𝑥 + 3 sin 𝑥 + 1 = 2𝑢2 + 3𝑢 + 1

• We can factor the quadratic using the 𝑎𝑐 method. In this case, 𝑎𝑐 = 2 and 𝑏 = 2,1. Therefore…

• 2𝑢2 + 2𝑢 + 𝑢 + 1 = 2𝑢(𝑢 + 1) + 1(𝑢 + 1) = (2𝑢 + 1)(𝑢 + 1) • Now replace 𝑢 back with sin 𝑥.

• (2 sin 𝑥 + 1)(sin 𝑥 + 1) = 0 • Set each individual factor equal to zero and isolate sin 𝑥.

• 2 sin 𝑥 + 1 = 0 → sin 𝑥 = −1

2 and sin 𝑥 + 1 = 0 → sin 𝑥 = −1

• Apply the inverse of sine to solve for 𝑥.

• 𝑥 = sin−1 (−1

2) =

7𝜋

6,

11𝜋

6 and 𝑥 = sin−1(−1) =

3𝜋

2

• Final answer: 𝑥 =7𝜋

6,

11𝜋

6,

3𝜋

2

Solve the following trigonometric equations on the interval [0,2𝜋).

87. 4 sin2 𝑥 − 1 = 0

88. cos2 𝑥 = cos 𝑥

89. 2 cos 𝑥 + √3 = 0

90. 1 − sin2 𝑥 = 2 + 2 sin 𝑥

** That’s all folks! I hope you found this packet a helpful review. If you have any questions and would like

to speak to me, feel free to message me via Remind. See you in the fall! **