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
canโt 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 donโt 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: Donโt 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 โ 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 letโs 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 itโs 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 โบ
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 [email protected]. 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: ๐ฆ = ๐๐ฅ + ๐
** Where ๐ is the slope and ๐ is the y-intercept. **
Point-Slope Form: ๐ฆ โ ๐ฆ1 = ๐(๐ฅ โ ๐ฅ1)
** Where (๐ฅ1, ๐ฆ1) is the point going through the line. **
Vertical line: ๐ฅ = ๐
** Where ๐ is the value of ๐ฅ-value that the line passes through and the slope is undefined. **
Horizontal line: ๐ฆ = ๐
** Where ๐ is the value of ๐ฆ-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 โ2. (b) Containing points (3, โ4) and (9,0).
Solution: (a) The slope, ๐, is given as โ2. The line contains (0,1), so this point is the ๐ฆ-intercept, or ๐, which is 1.
Substituting these numbers into slope-intercept form gives you the equation ๐ฆ = โ2๐ฅ + 1.
(b) First find the slope: ๐ =๐ฆ2โ๐ฆ1
๐ฅ2โ๐ฅ1=
โ4โ0
3โ9=
2
3
โข 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) โ ๐ฆ +
4 =2
3(๐ฅ โ 3) OR ๐ฆ =
2
3(๐ฅ โ 9).
โข Both are acceptable answers!
โข 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 ๐ฆ =5
6๐ฅ โ 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 ๐ฆ = โ2๐ฅ + 9 and passing through the point (4,7).
Find the x and y intercepts for each equation.
** The ๐ฅ-intercepts are found by plugging in ๐ฆ = 0 and solving for ๐ฅ. The ๐ฆ-intercepts are found by plugging
in ๐ฅ = 0 and solving for ๐ฆ. **
6. ๐ฆ = 2๐ฅ โ 5
7. ๐ฆ = ๐ฅ2 + ๐ฅ โ 2
8. ๐ฆ = ๐ฅโ16 โ ๐ฅ2
9. ๐ฆ2 = ๐ฅ3 โ 4๐ฅ
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 {3๐ฅ + 5๐ฆ = 112๐ฅ + 3๐ฆ = 7
using the
elimination method.
Example:
Solve the system {๐ฅ2 โ ๐ฆ = 3๐ฅ โ ๐ฆ = 1
using the substitution
method.
Solution:
โข First decide which variable you want to eliminate (does not matter which). For this example, I am choosing to eliminate the ๐ฅ.
โข 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 โ2 and the bottom equation by 3 (although you could have also multiplied the top equation by 2 and the bottom equation by โ3; it does not matter.
โข As a result, I end up with this system:
{โ6๐ฅ โ 10๐ฆ = โ22
6๐ฅ + 9๐ฆ = 21
โข Add the equations together and you get โ โ๐ฆ = โ1 โ ๐ฆ = 1.
โข Now that you have the value for ๐ฆ, substitute it into either original equation to solve for ๐ฅ.
โข Final answer: (2,1)
Solution:
โข Solve one of the equations for one variable. In this case, it will be easier to solve for ๐ฆ in either equation. I will choose to solve for ๐ฆ in the second equation โ ๐ฆ = ๐ฅ โ 1.
โข Plug in (substitute) what you got for ๐ฆ in the first equation. โ ๐ฅ2 โ (๐ฅ โ 1) = 3.
โข Distribute and set the equation equal to zero so you may factor. โ ๐ฅ2 โ ๐ฅ + 1 = 3 โ ๐ฅ2 โ ๐ฅ โ 2 = 0 โ (๐ฅ โ 2)(๐ฅ + 1) = 0 โ ๐ฅ = 2, โ1
โข Plug in both ๐ฅ-values independently into either original equation to solve for ๐ฆ.
โข When ๐ฅ = 2 โฆ โ 2 โ ๐ฆ = 1 โ ๐ฆ = 1
โข When ๐ฅ = โ1 โฆ โ โ1 โ ๐ฆ = 1 โ ๐ฆ = โ2
โข Final answers: (2,1) and (โ1, โ2)
Find the point(s) of intersection of the graphs for the given equations.
10. {๐ฅ + ๐ฆ = 8
4๐ฅ โ ๐ฆ = 7
11. {๐ฅ2 + ๐ฆ = 6๐ฅ + ๐ฆ = 4
12. {2๐ฅ + 7๐ฆ = 4
3๐ฅ + 5๐ฆ = โ5
13. {๐ฅ = 3 โ ๐ฆ2
๐ฆ = ๐ฅ โ 1
Topic 3: Rational and Negative Exponents
Rules of Exponents
Generalization Explanation
Product Rule
๐ฅ๐ โ ๐ฅ๐ = ๐ฅ๐+๐ When multiplying like bases, add
the exponents
Quotient Rule
๐ฅ๐
๐ฅ๐= ๐ฅ๐โ๐
When dividing like bases, subtract the exponents
Power Rule
(๐ฅ๐)๐ = ๐ฅ๐๐
When raising a term with an exponent with another exponent,
multiply exponents.
Expanded Power Rule
(๐ฅ๐ฆ)๐ = ๐ฅ๐๐ฆ๐
AND
(๐๐ฅ
๐๐ฆ)
๐
=๐๐๐ฅ๐
๐๐๐ฆ๐
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
๐ฅโ๐ =1
๐ฅ๐ OR 1
๐ฅโ๐ = ๐ฅ๐
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
๐ฅ0 = 1
Any non-zero ๐ฅ-values raised to an exponent of zero is always
equal to 1
Rational Exponents
๐ฅ๐๐ = โ๐ฅ๐๐
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. 3๐ฅโ3
๐ฅโ2
15. (9๐ฅ2๐ฆโ4)โ1
2
16. (16๐ฅ2๐ฆ)3
4
17. (๐ฅ2)
3๐ฅ
๐ฅ7
18. โ4๐ฅโ16
โ(๐ฅโ4)34
19. โ๐ฅ โ โ๐ฅ3 โ ๐ฅ1
6
20. [(๐ฅโ1๐ฆ1
3) (๐ฅโ4
3๐ฆ2)]2
21. (๐ฅ
12๐ฆโ2
๐ฆ๐ฅโ
74
)
4
Topic 4: Factoring
When factoring a trinomial in the form ๐๐ฅ2 + ๐๐ฅ + ๐, remember that you are looking for the two numbers that
multiply to give you ๐ and also add to give you ๐. If there ๐ โ 1, check to see if the coefficient can be factored out first before you continue to factor.
Example: 2๐ฅ2 โ 4๐ฅ โ 30 = 2(๐ฅ2 โ 2๐ฅ โ 15) = 2(๐ฅ โ 5)(๐ฅ + 3)
Special Factorization Methods
General Example(s)
Perfect Square Trinomials
๐2 + 2๐๐ + ๐2 = (๐ + ๐)2 ๐2 โ 2๐๐ + ๐2 = (๐ โ ๐)2
โข ๐ฅ2 + 6๐ฅ + 9 = (๐ฅ + 3)2
โข ๐ฅ2 โ 8๐ฅ + 16 = (๐ฅ โ 4)2
Difference of Squares
๐2 โ ๐2 = (๐ + ๐)(๐ โ ๐)
โข 4๐ฅ2 โ 9 = (2๐ฅ + 3)(2๐ฅ โ 3)
AC Method
๐๐ฅ2 + ๐๐ฅ + ๐ where ๐ โ 1 and ๐ cannot be factored.
โข Find ac.
โข Find factors of ac that add up to b.
โข Write the original expression, but rewrite bx using your new factors.
โข Factor by grouping.
โข 2๐ฅ2 โ 5๐ฅ โ 3
โข ๐๐ = โ6, ๐ = โ6,1
โข 2๐ฅ2 โ 6๐ฅ + ๐ฅ โ 3
โข 2๐ฅ(๐ฅ โ 3) + 1(๐ฅ โ 3)
= (๐ฅ โ 3)(2๐ฅ + 1)
Factoring by Grouping
Only use this method if the polynomial you are trying to
factor has four terms:
๐๐ + ๐๐ โ ๐๐ โ ๐๐ = ๐(๐ + ๐) โ ๐(๐ + ๐)
= (๐ + ๐)(๐ โ ๐)
โข Factor the GCF of the first two factors and factor the GCF of the second two factors.
โข What is in parenthesis will be one of your final factors.
โข What is left outside the parenthesis is the other factor.
โข ๐ฅ3 + 3๐ฅ2 โ 2๐ฅ โ 6
โข (๐ฅ3 + 3๐ฅ2) + (โ2๐ฅ โ 6)
โข ๐ฅ2(๐ฅ + 3) โ 2(๐ฅ + 3)
= (๐ฅ + 3)(๐ฅ2 โ 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. ๐ฅ2 โ 20๐ฅ + 36
23. 2๐ฅ3 โ 2๐ฅ2 โ 4๐ฅ
24. ๐ฅ3 โ 5๐ฅ2 โ ๐ฅ + 5
25. 2๐ฅ2 + ๐ฅ โ 6
26. 9๐ฅ2 โ 4๐ฆ4
27. 3๐ฅ2 + 3๐ฅ โ 60
28. 5๐ฅ2 + 29๐ฅ + 20
29. ๐ฅ6 + 2๐ฅ4 โ 16๐ฅ2 โ 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 ๐ฅ.
Solve by factoring.
30. 4๐ฅ2 + ๐ฅ โ 36 = ๐ฅ2 + 4๐ฅ 31. 9๐ฅ2 + 2๐ฅ โ 4 = 2๐ฅ โ 3
Topic 6: The Quadratic Formula
The solutions to a quadratic equation written in the form ๐๐ฅ2 + ๐๐ฅ + ๐ = 0, where ๐ โ 0, can be found by using the quadratic formula:
๐ฅ =โ๐ ยฑ โ๐2 โ 4๐๐
2๐
** Where the values of ๐ฅ represent the zeros or roots or ๐ฅ-intercepts of the quadratic equation. **
Example: Use the quadratic formula to solve: 4๐ฅ2 + 4๐ฅ โ 8 = 1
Solution:
โข First, you have to set the quadratic equal to zero in order to use the quadratic formula: โ 4๐ฅ2 + 4๐ฅ โ 9 = 0
โข In this case, ๐ = 4, ๐ = 4, and ๐ = โ9. Plug in these values into the quadratic formula.
โข ๐ฅ =โ4ยฑโ42โ4(4)(โ9)
2(4)=
โ4ยฑโ16+144
8=
โ4ยฑโ160
8
โข Simplify the square root, if possible.
โข ๐ฅ =โ4ยฑ4โ10
8โ Simplify your final answer
โข Final answer: โ1ยฑโ10
2
Use the quadratic formula to solve each equation.
32. 4๐ฅ2 + 3 = 8๐ฅ
33. 3๐ฅ2 + ๐ฅ = 2๐ฅ2 + 6๐ฅ โ 4
Topic 7: Functions
In function notation, the symbol ๐(๐ฅ) is read ๐ of ๐ฅ and interpreted as the function value of ๐ at ๐ฅ. In other words, ๐ฆ = ๐(๐ฅ).
** To find a function value, simply plug in the value into every ๐ฅ and simplify. **
Example: If ๐(๐ฅ) = ๐ฅ2 + 3๐ฅ, find ๐(โ5).
Solution:
โข ๐(โ5) means to replace ๐ฅ with the value โ5 and evaluate for ๐(๐ฅ).
โข ๐(โ5) = (โ5)2 + 3(โ5) = 25 โ 15 = 10
โข Final answer: ๐(โ5) = 10
Evaluate each function for the given function value.
Let ๐(๐ฅ) = 5 โ2๐ฅ
3 and ๐(๐ฅ) =
1
2๐ฅ2 + 3๐ฅ
34. ๐ (1
2)
35. โ3๐(โ2)
36. ๐(6) + 2๐(1)
37. ๐(4) โ ๐(5)
Topic 8: Operations with Functions
To write a sum, difference, product, or quotient of two functions, ๐ and ๐, write the sum difference, product, or quotient of the expressions that define ๐ and ๐, then simplify. The domain of the combined
function is restricted to the domains of each original function.
Example: Let ๐(๐ฅ) = ๐ฅ2 + 3๐ฅ + 2 and ๐(๐ฅ) = 5๐ฅ โ 1. Write an expression for each function.
(a) (๐ + ๐)(๐ฅ) (b) (๐ โ ๐)(๐ฅ) (c) (๐๐)(๐ฅ)
(d) (๐
๐) (๐ฅ)
** Keep in mind that the domain of both ๐ and ๐ is (โโ, โ) since both functions are polynomial functions and are continuous everywhere and extend in both directions. **
Solution:
(a) (๐ + ๐)(๐ฅ) = ๐(๐ฅ) + ๐(๐ฅ) = (๐ฅ2 + 3๐ฅ + 2) + (5๐ฅ โ 1) = ๐ฅ2 โ 2๐ฅ + 3
โข Domain: (โโ, โ)
(b) (๐ โ ๐)(๐ฅ) = ๐(๐ฅ) โ ๐(๐ฅ) = (๐ฅ2 + 3๐ฅ + 2) โ (5๐ฅ โ 1) = ๐ฅ2 + 3๐ฅ + 2 โ 5๐ฅ + 1 = ๐ฅ2 โ 2๐ฅ + 3
โข Domain: (โโ, โ)
(c) (๐๐)(๐ฅ) = ๐(๐ฅ) โ ๐(๐ฅ) = (๐ฅ2 + 3๐ฅ + 2)(5๐ฅ โ 1) = 5๐ฅ3 โ ๐ฅ2 + 15๐ฅ2 โ 3๐ฅ + 10๐ฅ โ 2 = 5๐ฅ3 + 14๐ฅ2 + 7๐ฅ โ 2
โข Domain: (โโ, โ)
(d) (๐
๐) (๐ฅ) =
๐(๐ฅ)
๐(๐ฅ)=
๐ฅ2+3๐ฅ+2
5๐ฅโ1, where ๐ฅ โ
1
5 (because that is the value that would make the denominator
equal to zero.
โข Domain: (โโ,1
5) โช (
1
5, โ)
Find each new function and state the combined functionโs domain.
Let ๐(๐ฅ) = 3๐ฅ2 + 2, ๐(๐ฅ) = 2๐ฅ โ 1, and โ(๐ฅ) = ๐ฅ2 + 5๐ฅ.
38. (๐ + ๐)(๐ฅ)
39. (๐โ)(๐ฅ)
40. (๐ โ โ)(๐ฅ)
41. (๐
โ) (๐ฅ)
Topic 9: Composition of Functions
A composition of functions denoted as ๐(๐(๐ฅ)) or ๐[๐(๐ฅ)], is read as โ๐ of ๐ of ๐ฅ,โ and means to plug in
the inside function (in this case ๐(๐ฅ)) in for ๐ฅ in the outside function (in this case ๐(๐ฅ)).
Example:
Given ๐(๐ฅ) = 2๐ฅ2 + 1 and ๐(๐ฅ) = ๐ฅ โ 4, find ๐[๐(1)]. ** To find a composition at a function value, you first need to find ๐[๐(๐ฅ)] and then you plug in the given ๐ฅ-value, in this case ๐ฅ = 1. **
Solution:
โข ๐[๐(๐ฅ)] = ๐(๐ฅ โ 4) = 2(๐ฅ โ 4)2 + 1 โข 2(๐ฅ2 โ 8๐ฅ + 16) + 1 โข 2๐ฅ2 โ 16๐ฅ + 32 + 1 โข ๐[๐(๐ฅ)] = 2๐ฅ2 โ 16๐ฅ + 33
โNow you can plug in ๐ฅ = 1.
โข ๐[๐(1)] = 2(1)2 โ 16(1) + 33 โข Final answer: ๐[๐(1)] = 19
Find the following composition of functions at the indicated function value, if given.
Let ๐(๐ฅ) = ๐ฅ2 โ 1, ๐(๐ฅ) = 3๐ฅ โ ๐ฅ2, and โ(๐ฅ) = 5 โ ๐ฅ
42. โ[๐(๐ฅ)]
43. ๐[โ(2)]
44. ๐[๐(๐ฅ)]
45. ๐[โ(๐(1))]
Topic 10: Evaluating for Piecewise Functions
A piecewise function is exactly what it sounds like. It is a function that is defined in โpieces.โ 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
๐(๐ฅ) = {๐ฅ + 3, ๐ฅ < โ1
๐ฅ2 , โ 1 โค ๐ฅ โค 23, ๐ฅ > 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 โpiece.โ If the < or > symbols are used, then that ๐ฅ-value is NOT included and would be graphed with an open circle. If the โค or โฅ symbols are used, the that ๐ฅ-value IS included and would be graphed with a solid circle.
Example:
Graph the piecewise function given below and find ๐(3).
๐(๐ฅ) = {1
3๐ฅ + 2, ๐ฅ โค 3
โ๐ฅ + 3, 3 < ๐ฅ โค 4
Solution:
Although you could find ๐(3) graphically, if you were to find the function value algebraically, you would have to see where ๐(3) is defined. The value of ๐ฅ = 3 is only included in the first equation and not in the second. Therefore, to find ๐(3), you would plug in ๐ฅ = 3 into the first equation.
๐(3) =1
3(3) + 2 = 3
Graph each function and find the indicated function value algebraically.
46. ๐(๐ฅ) = {๐ฅ + 3, ๐ฅ < 0โ2๐ฅ + 5, ๐ฅ โฅ 0
โข ๐(3) = _____________
47. ๐(๐ฅ) = {1
2๐ฅ, โ 4 โค ๐ฅ โค 2
2๐ฅ โ 3, ๐ฅ > 2
โข ๐(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 ๐ฅ-axis or the ๐ฆ-axis.
** Anything โoutsideโ affects the ๐ฆ-values and does the same as the operation indicates. Anything โinsideโ affects the ๐ฅ-values and does the opposite operation indicated. **
Transformation Notation
Verbal Description
๐(๐ฅ + โ)
Horizontal shift left โ units
๐(๐ฅ โ โ)
Horizontal shift right โ units
๐(๐ฅ) + ๐
Vertical shift up ๐ units
๐(๐ฅ) โ ๐
Vertical shift down ๐ units
๐๐(๐ฅ)
Vertical stretch with a scale factor of ๐
1
๐๐(๐ฅ)
Vertical shrink with a scale factor of 1
๐
๐(๐๐ฅ)
Horizontal shrink with a scale factor of 1
๐
๐ (1
๐๐ฅ)
Horizontal stretch with a scale factor of ๐
โ๐(๐ฅ)
Reflection over the ๐ฅ-axis
๐(โ๐ฅ)
Reflection over the ๐ฆ-axis
Identify the parent function for each of the following. Then, describe the transformations of the parent
functions indicated from the functionโs equation.
48. ๐(๐ฅ) = โ2|๐ฅ + 1| โ 4
โข Parent function: __________________
49. ๐(๐ฅ) =1
3โโ2๐ฅ + 3
โข Parent function: __________________
50. ๐(๐ฅ) = (1
4(๐ฅ โ 1))
2
โข Parent function: __________________
51. ๐(๐ฅ) =3
โ(๐ฅ+2)โ 7
โข Parent function: __________________
Topic 12: Domain and Range
Domain: The set of all the ๐ฅ-values for which a function is defined (input values). Range: The set of all the ๐ฆ-values (output values).
โข All polynomial functions have a domain of (โโ, โ) since their graphs go forever left and right encompassing all possible ๐ฅ-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 functionโs graph and how to find the domain algebraically.
Function Type
Finding Domain Algebraically
Radical Functions (๐(๐ฅ) = โ๐ฅ)
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 ๐ฅ.
Rational Functions (๐(๐ฅ) =1
๐ฅ)
Rational functions are basically fractions (division). Since you can never divide by zero, to find the domain, you set the denominator to โnot equalโ zero and find the restriction.
Trigonometric Functions (๐(๐ฅ) = sin ๐ฅ and cos ๐ฅ only )
The graphs of sine and cosine are waves that oscillate forever in both directions. Therefore, the domains of both functions are
always (โโ, โ).
Natural Log Functions (๐(๐ฅ) = ln ๐ฅ)
The graph of natural log has a vertical asymptote at ๐ฅ = 0. Therefore, to find the domain, like the radical function, you set
what is โinsideโ of ๐๐ 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 functionโs behavior based on the vertical shifts. If the parent function
normally has a range of [0, โ) but the function was shifted up 5 units, then the range would be [5, โ). **
52. ๐(๐ฅ) = ๐ฅ2 โ 3
โข Domain: __________________________
โข Range: ___________________________
53. ๐(๐ฅ) = 3 sin ๐ฅ
โข Domain: __________________________
โข Range: ___________________________
54. ๐(๐ฅ) = โโ๐ฅ + 3
โข Domain: __________________________
โข Range: ___________________________
55. ๐(๐ฅ) =2
๐ฅโ1
โข Domain: __________________________
โข Range: ___________________________
Topic 13: Inverses of Functions
Functions are considered inverses if and only if their ๐ฅ and ๐ฆ values have switched places. Graphically, this means that inverse functions are reflections along the line ๐ฆ = ๐ฅ.
** Inverse function notation is denoted as ๐โ1(๐ฅ). **
To find the inverse of a function, simply interchange the ๐ฅ and ๐ฆ and solve for ๐ฆ. You can also verify your answer by checking that ๐[๐โ1(๐ฅ)] = ๐ฅ and ๐โ1[๐(๐ฅ)] = ๐ฅ.
Example:
Find the inverse of the function ๐(๐ฅ) = โ๐ฅ + 13
. Then, verify that your answer is the inverse by finding at least one of the compositions ๐[๐โ1(๐ฅ)] = ๐ฅ or ๐โ1[๐(๐ฅ)] = ๐ฅ.
Solution:
โข Rewrite ๐(๐ฅ) as ๐ฆ โ ๐ฆ = โ๐ฅ + 13
โข Interchange the ๐ฅ and ๐ฆ โ ๐ฅ = โ๐ฆ + 13
โข Solve for ๐ฆ โ ๐ฅ3 = ๐ฆ + 1 โ ๐ฅ3 โ 1 = ๐ฆ
โข Rewrite using appropriate inverse function notation โ ๐โ1(๐ฅ) = ๐ฅ3 โ 1
Verify your answer by finding one of the compositions indicated.
โข ๐[๐โ1(๐ฅ)] = โ(๐ฅ3 โ 1) + 13
= โ๐ฅ3 + 1 โ 13
โ๐ฅ33= ๐ฅ
Since the composition equals to ๐ฅ, the inverse is
correct!
Find the inverse for each function. Verify that your answer is correct through composition.
56. ๐(๐ฅ) = 2๐ฅ + 1
57. ๐(๐ฅ) =๐ฅ2
3
58. ๐(๐ฅ) =5
๐ฅโ2
59. ๐(๐ฅ) = โ4 โ ๐ฅ + 1
Topic 14: Even and Odd Functions
Even functions are functions that are symmetric over the ๐ฆ-axis.
โข To determine algebraically if a function is even, you must find out if ๐(โ๐ฅ) = ๐(๐ฅ).
โข This means that if you plug in โ๐ฅ and all of the terms remain the same sign, then the function is even.
Odd functions are functions that are symmetric over the origin.
โข To determine algebraically if a function is odd, you must find out if ๐(โ๐ฅ) = โ๐(๐ฅ).
โข This means that if you plug in โ๐ฅ and all of the terms change signs, then the function is odd.
** If you plug in โ๐ฅ into the function and only some terms change signs and others donโt, then the function is neither even nor odd. **
Example of an even function:
๐(๐ฅ) = 2๐ฅ2 + 3|๐ฅ|
โ ๐(โ๐ฅ) = 2(โ๐ฅ)2 + 3|โ๐ฅ| = 2๐ฅ2 + 3|๐ฅ|
Since ๐(โ๐ฅ) looks identical to ๐(๐ฅ), then ๐(๐ฅ) is an even function.
Example of an odd function:
๐(๐ฅ) = โ5๐ฅ3 + 3๐ฅ
โ ๐(โ๐ฅ) = โ5(โ๐ฅ)3 + 3(โ๐ฅ) = 5๐ฅ3 โ 3๐ฅ
Since all of the terms in ๐(โ๐ฅ) are the opposite signs of ๐(๐ฅ), then ๐(๐ฅ) is an odd function.
Example of a function that is neither even nor odd:
๐(๐ฅ) = 2๐ฅ5 + 1
โ ๐(โ๐ฅ) = 2(โ๐ฅ)5 + 1 = โ2๐ฅ5 + 1
Since only one term changed signs and the 1 did not, this means that ๐(๐ฅ) is neither even nor odd.
State whether each of the following functions are even, odd, or neither.
60.
61.
62. ๐(๐ฅ) = 2๐ฅ4 โ 5๐ฅ2
63. ๐(๐ฅ) = ๐ฅ5 โ 3๐ฅ3 + ๐ฅ
64. ๐(๐ฅ) = 2๐ฅ2 โ 5๐ฅ + 3
65. ๐(๐ฅ) = 2 cos ๐ฅ
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 ๐ฅ-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 โ this is a
removable discontinuity aka a hole).
Find the vertical asymptote of ๐(๐ฅ) =๐ฅ2โ5๐ฅโ14
๐ฅ2โ3๐ฅโ28.
โข Factor both the numerator and denominator
โข ๐(๐ฅ) =(๐ฅโ7)(๐ฅ+2)
(๐ฅโ7)(๐ฅ+4)โ Simplify
โข ๐(๐ฅ) =๐ฅ+2
๐ฅ+4
โข At ๐ฅ = โ4 there is a vertical asymptote where at ๐ฅ = 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 ๐ฆ = 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) ๐(๐ฅ) =3๐ฅ
๐ฅ2+2โ HA @ ๐ฆ = 0
(b) ๐(๐ฅ) =2๐ฅ2+3๐ฅโ1
4๐ฅ2+5โ HA @ ๐ฆ =
2
4=
1
2
(c) ๐(๐ฅ) =5๐ฅ3โ๐ฅ
๐ฅ2+2๐ฅโ 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. ๐(๐ฅ) =๐ฅ+4
๐ฅ2+1
67. ๐(๐ฅ) =๐ฅ2+5๐ฅโ6
๐ฅ2โ1
68. ๐(๐ฅ) =2๐ฅ3
๐ฅ2โ4
69. ๐(๐ฅ) =2๐ฅ2โ4๐ฅ+2
๐ฅ2โ3๐ฅโ4
Topic 16: Solving Rational Equations
There are two ways of solving rational equations: 1. Cross multiplication โ 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 โ 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 ๐ฅ that make the denominator equal to zero. **
Example of cross multiplication:
โข ๐ฅโ5
๐ฅ+1=
๐ฅโ9
๐ฅ+5โ Excluded values: ๐ฅ = โ1, โ5
โข Cross multiply
โข (๐ฅ โ 5)(๐ฅ + 5) = (๐ฅ โ 9)(๐ฅ + 1) โ Distribute
โข ๐ฅ2 โ 25 = ๐ฅ2 โ 8๐ฅ โ 9 โ Solve for ๐ฅ.
โข Although the resulting equation is a quadratic, if you subtract ๐ฅ2 from both sides, the term will cancel out.
โข โ16 = 8๐ฅ โ Divide by 8 to both sides.
โข Final answer: ๐ฅ = โ2
** Since ๐ฅ = โ2 is not an excluded value, that is the real solution. **
Example of multiplying the LCD:
โข 12
๐ฅโ1โ
8
๐ฅ= 2 โ Excluded values: ๐ฅ = 1, 0
โข LCD: ๐ฅ(๐ฅ โ 1)
โข Multiply the LCD to the entire equation.
โข ๐ฅ(๐ฅ โ 1) [12
๐ฅโ1โ
8
๐ฅ= 2]
โข 12๐ฅ โ 8(๐ฅ โ 1) = 2๐ฅ(๐ฅ โ 1) โ Distribute
โข 12๐ฅ โ 8๐ฅ + 8 = 2๐ฅ2 โ 2๐ฅ โ Since the resulting equation is a quadratic, set it to equal to zero and solve by factoring.
โข 2๐ฅ2 โ 6๐ฅ โ 8 = 2(๐ฅ2 โ 3๐ฅ โ 4) = 2(๐ฅ โ 4)(๐ฅ + 1) = 0
โข Final answer: ๐ฅ = 4, โ1
** Since ๐ฅ = 4, โ1 are none of the excluded values, both are the real solutions of the equation. **
Solve for ๐ฅ for each rational equation:
70. ๐ฅโ5
๐ฅ+1=
3
5
71. ๐ฅ+3
๐ฅโ4=
๐ฅ+4
๐ฅโ10
72. 1
๐ฅโ2+
1
๐ฅ2โ7๐ฅ+10=
6
๐ฅโ2
73. 2๐ฅ+3
๐ฅโ1=
10
๐ฅ2โ1+
2๐ฅโ3
๐ฅ+1
Topic 17: Logarithms and Exponential Functions
For ๐ฅ > 0 and ๐ > 0, ๐ โ 1 โฆ ๐ฆ = log๐ ๐ฅ is equivalent to ๐๐ฆ = ๐ฅ,
โฆwhere ๐ is the base and ๐ฆ is the exponent.
** Basically, a logarithmic function is the inverse of an exponential function. **
*** The common logarithm, log10 ๐ฅ, is usually written as log ๐ฅ. ***
The logarithmic function can have any base as long as the base is not zero or negative. However, the natural logarithmic function (ln ๐ฅ) has a base of ๐, where ๐ is the exponential base and is
equal to the irrational number 2.718281 โฆ
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 ๐ฅ) functions.
Property
Definition
Example
Product
log๐ ๐๐ = log๐ ๐ + log๐ ๐
โข log3 9๐ฅ = log3 9 + log3 ๐ฅ โข ln ๐ฅ(๐ฅ + 1) = ln ๐ฅ + ln ๐ฅ + 1
Quotient
log๐
๐
๐= log๐ ๐ โ log๐ ๐
โข log2 (4๐ฅ+1
๐ฅ+2)
= log2(4๐ฅ + 1) โ log2(๐ฅ + 2)
โข ln (8
๐ฅ) = ln 8 โ ln ๐ฅ
Power
log๐ ๐๐ = ๐ โ log๐ ๐
โข log2 8๐ฅ = ๐ฅ โ log2 8 โข ln ๐ฅ2 = 2 ln ๐ฅ
Equality
log๐ ๐ = log๐ ๐ , then ๐ = ๐
โข log3(3๐ฅ โ 4) = log3(5๐ฅ + 2), then 3๐ฅ โ 4 = 5๐ฅ + 2
โข This property does not apply to the natural log function.
Logarithmic Exponent with Equal Bases
๐log๐ ๐ฅ
โข 4log4(๐ฅโ1) = ๐ฅ โ 1 โข ๐ln 2๐ฅ = 2๐ฅ
โข This is because the base of ln ๐ฅ is ๐ so they would cancel each other out.
Expand the following logarithmic functions using the properties of logarithms.
74. ln(๐ฅ๐ฆ)6 75. log[(๐ฅ + 1)(๐ฅ2)] 76. ln (5๐ฅ+3
๐ฆ2 )
Topic 18: Natural Log and Exponential Functions with Base ๐
Since the natural logarithmic function has a base of ๐, this means that ln ๐ฅ and ๐ are reciprocal functions and would cancel each other out if they are applied to one another.
When ๐ has an exponent of ln ๐ฅ:
๐ln ๐ฅ = ๐ฅ
Example:
๐ln(2๐ฅ+4) = 2๐ฅ + 4
When ๐ is โinsideโ of ln ๐ฅ:
ln ๐๐ฅ = ๐ฅ
Example:
ln ๐๐ฅ2+1 = ๐ฅ2 + 1
To solve for ๐ฅ when given an exponential equation with base ๐, apply the function ๐๐ to both sides of the equation to cancel out the ๐.
Example: Solve for ๐ฅ given that ๐2๐ฅ = 5
Solution:
โข Apply ๐๐ to both sides โ ln ๐2๐ฅ = ln 5
โข Simplify โ 2๐ฅ = ln 5
โข Divide by 2 โ ๐ฅ =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 ๐ฅ =ln 5
2. **
To solve for ๐ฅ when given a natural logarithmic equation, apply the function ๐ to both sides of the equation to cancel out the ๐๐.
Example: Solve for ๐ฅ given that ln(4๐ฅ + 1) = 3
Solution:
โข Apply ๐ to both sides โ ๐ln(4๐ฅ+1) = ๐3
โข Simplify โ 4๐ฅ + 1 = ๐3
โข Isolate the ๐ฅ โ ๐ฅ =๐3โ1
4
** Just like in the previous example, you would leave your answer as is. **
Solve for ๐ฅ for each of the given equations.
77. ln ๐2๐ฅ = 5๐ฅ โ 3
78. 5 ln ๐3๐ฅ = 30
79. 8๐๐ฅ = 51
80. ln(3๐ฅ โ 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. **
โข All trigonometric function values are positive in
quadrant 1.
โข Only sine and cosecant function values are
positive in quadrant 2.
โข Only tangent and cotangent function values are
positive in quadrant 3.
โข Only cosine and secant function values are
positive in quadrant 4.
On the Unit Circle, the ๐ฅ = cos ๐ and ๐ฆ = sin ๐.
โข Tangent is found by dividing sine by cosine
โ tan ๐ =sin ๐
cos ๐.
โข Secant is the reciprocal of cosine.
โข Cosecant is the reciprocal of sine.
โข Cotangent is the reciprocal of tangent.
Examples of finding a trigonometric function value:
(a) sec๐
3โ Since secant is the reciprocal of cosine,
we would first need to find cos๐
3 and then its
reciprocal.
โข cos๐
3=
1
2โ Therefore sec
๐
3= 2
(b) tan๐
3=
sin๐
3
cos๐
3
=โ3
21
2
โข To divide the fractions, โcopy, change, flip.โ
โข โ3
2(
2
1) = โ3
โข Therefore, tan๐
3= โ3
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, ๐, that would give the value inside the parenthesis.
(a) cosโ1 (โโ2
2) =
3๐
4,
5๐
4
โข Both angles are the solutions because the
cosine value for both angles is equal to โโ2
2.
(b) Evaluate sin (cosโ1 (1
2)) on the interval
0 < ๐ <๐
2.
โข 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.
โข First, we would have to evaluate for the inverse cosine and then find the sine value at the angle that you found.
โข sin (cosโ1 (1
2)) = sin (
๐
3) =
โ3
2
Evaluate the following trigonometric and inverse trigonometric function values.
81. sec11๐
6
82. cot ๐ 83. sin5๐
4
84. sinโ1(โ1)
85. csc (sinโ1 (โ2
2) ) on the
interval ๐
2< ๐ฅ < ๐
86. tan (sinโ1 (โ1
2)) on the
interval ๐ < ๐ฅ <3๐
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 ๐ฅ. 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 [0,2๐).
** Side note: Remember that intervals could be written in either set notation or interval notation. The interval given indicates that ๐ฅ = 0 is included and could be a solution but ๐ฅ = 2๐ 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 ๐ฅ = 2 sin ๐ฅ + โ2
(b) 4 cos2 ๐ฅ + 1 = 4
(c) 2 sin2 ๐ฅ + 3 sin ๐ฅ = โ1
Solutions:
(a) Isolate sin ๐ฅ by first subtracting 2 sin ๐ฅ from both sides of the equation
โ 2 sin ๐ฅ = โ2
โข Divide by 2 โ sin ๐ฅ =โ2
2
โข Apply the inverse of sine to isolate the ๐ฅ โ ๐ฅ = sinโ1 (โ2
2)
โข Final answer: ๐ฅ =๐
4,
3๐
4
(b) Isolate cos2 ๐ฅ by subtracting the 1 and then dividing the 4
โ cos2 ๐ฅ =3
4
โข Square root both sides to eliminate the exponent.
โ โcos2 ๐ฅ = โ3
4=
โ3
โ4=
โ3
2โ cos ๐ฅ =
โ3
2
โข Apply the inverse of cosine to isolate the ๐ฅ โ ๐ฅ = cosโ1 (โ3
2)
โข Final answer: ๐ฅ =๐
6,
11๐
6
(c) Since this equation is quadratic, first set it equal to zero so we can factor โ 2 sin2 ๐ฅ + 3 sin ๐ฅ + 1 = 0
โข Just so that it is easier to see how this can be factored, I will be replacing sin ๐ฅ with ๐ข during the factoring process.
โข If ๐ข = sin ๐ฅ, then 2 sin2 ๐ฅ + 3 sin ๐ฅ + 1 = 2๐ข2 + 3๐ข + 1
โข We can factor the quadratic using the ๐๐ method. In this case, ๐๐ = 2 and ๐ = 2,1. Thereforeโฆ
โข 2๐ข2 + 2๐ข + ๐ข + 1 = 2๐ข(๐ข + 1) + 1(๐ข + 1) = (2๐ข + 1)(๐ข + 1) โข Now replace ๐ข back with sin ๐ฅ.
โข (2 sin ๐ฅ + 1)(sin ๐ฅ + 1) = 0 โข Set each individual factor equal to zero and isolate sin ๐ฅ.
โข 2 sin ๐ฅ + 1 = 0 โ sin ๐ฅ = โ1
2 and sin ๐ฅ + 1 = 0 โ sin ๐ฅ = โ1
โข Apply the inverse of sine to solve for ๐ฅ.
โข ๐ฅ = sinโ1 (โ1
2) =
7๐
6,
11๐
6 and ๐ฅ = sinโ1(โ1) =
3๐
2
โข Final answer: ๐ฅ =7๐
6,
11๐
6,
3๐
2
Solve the following trigonometric equations on the interval [0,2๐).
87. 4 sin2 ๐ฅ โ 1 = 0
88. cos2 ๐ฅ = cos ๐ฅ
89. 2 cos ๐ฅ + โ3 = 0
90. 1 โ sin2 ๐ฅ = 2 + 2 sin ๐ฅ
** Thatโs 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! **