calculus summer tutorial summer tutorial june, 2018 page 1 calculus summer tutorial the purpose of...

CALCULUS SUMMER TUTORIAL JUNE, 2018

Calculus Summer TUTORIAL

The purpose of this tutorial is to have you practice the mathematical skills necessary to be successful in Calculus. All of the skills covered in this tutorial are from Pre-Calculus, Algebra 2, and Algebra 1. The material covered is from our district approved, Pre-calculus, textbook. If you need to, you may use reference materials to refresh your memory (old notes, textbooks, online resources, etc.). While graphing calculators will be used during a few tests and quizzes, the majority of in class assessments are non-calculator. You are encouraged to learn how to be calculator-independent. At the end of this page, there are links to some suggested online calculators.

Calculus is a fast-paced course that is taught at the college level to prepare you for AP Calculus. There is a lot of material in the curriculum that must be covered before the end of the year. Therefore, we cannot spend a lot of class time re-teaching prerequisite skills. This is why you have this tutorial. Spend some time with it and make sure you are clear on everything covered in this tutorial so that you will be successful in Calculus. Of course, you are always welcomed to seek help from your teacher if necessary.

For assistance with the assignment part, you may contact me at [email protected]. Emails may take a few days during summer for a response. Please be specific in your email for what you need assistance with, include the section and the question number as well. For each question in the assignment, there is an example in this tutorial. The summer assignment packet is posted online on our school’s website. Reference to the corresponding section is listed under each question.

Calculators Links

Online Calculator

https://www.desmos.com/calculator

https://mathway.com/graph

Emulator for Download

https://wabbit.codeplex.com/

http://lpg.ticalc.org/prj_tilem/download.html

Good Luck!

mailto:[email protected]

https://www.desmos.com/calculator

https://mathway.com/graph

https://wabbit.codeplex.com/

http://lpg.ticalc.org/prj_tilem/download.html


I. Lines The slope intercept form of a line is y = mx + b where m is the slope and b is the y-intercept. The point slope

form of a line is y - y

1= m(x - x

1) where m is the slope and

(x

1,y

1) is a point on the line. You should be very

comfortable using the point slope form of the line. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes. Example1:

Find the equations of (a) line parallel and (b) perpendicular to y =

-1

3x + 5 that contains the point (−2,1).

Solution: Part a

1=

-1

3(-2) + b Using the slope-intercept form with

m =

-1

3and point (−2,1).

1=

2

3+ b Multiply −1and −2.

1-

2

3= b Subtract

2

3from both sides.

3

3-

2

3= b Get a common denominator of 3.

1

3= b Combine like terms.

y =

-1

3x +

1

3 This is the equation of the line parallel to the given line that contains (−2,1)

Part b

1= 3(-2) + b Using the slope-intercept form with m = 3 and point (−2,1).

1= -6+ b Multiply 3 and −2.

1- -6 = b Subtract −6 from both sides.

7 = b Subtract.

y = 3x + 7 This is the equation of the line perpendicular to the given line that contains

(−2,1).

Example 2: Find the slope and y-intercept of 6x -5y =15.

Solution: First you must get the line in slope-intercept form.

-5y =15- 6x Subtract 6x form both sides.

y =

15- 6x

-5 Divide by −5.

y =

6

5x - 3 Simplify.

The slope is m=

6

5 and the y-intercept is −3


Example 3:

Find the equation of the line that passes through the point (1, −2) and has slope m= −3. Solution: Since we are given a point and slope it is easier to use the point slope form of a line.

y - -2 = -3(x -1) Substitute into point slope form for

(x

1,y

1) and m.

y + 2 = -3x +1 Minus a negative makes + and distribute −3.

y = -3x -1 Subtract 2 from both sides.

Example 4:

Find the equation of the line that passes through (−1,3) and (4,5).

Solution:

m =

5- 3

4 - -1=

2

5 You will need to find slope using

m =y

2- y

1

x2- x

1

.

5 =

2

5(4) + b Choose one point to substitute back into either the point slope or slope-intercept

form of a line. Using the slope-intercept form with m =

2

5 and point (4,5).

5 =

8

5+ b Multiply 2 and 4.

5-

8

5= b Subtract

8

5from both sides.

25

5-

8

5= b Get a common denominator of 5.

17

5= b Combine like terms.

y =

2

5x -

17

5 Equation of the line in slope intercepts form.

Example 5: Graph the following equation: 3y + 3= x

Solution:

3y = x - 3 First you must get the equation in slope-intercept form by isolating y.

y =

1

3x -1 The slope =

1

3 and the y-intercept = −1

Plot the point (0,−1). From the first point go up 1 and over 3 to the right to get a second point.

Now connect the two points to get the line.


II. Intercepts The x-intercept is where the graph crosses the x-axis. You can find the x-intercept by setting y = 0.

The y-intercept is where the graph crosses the y-axis. You can find the y-intercept by setting x = 0. Example1:

Find the intercepts for y = (x + 3)2 - 4

Solution: X-intercept

0 = (x + 3)2 - 4 Set y = 0

4 = (x + 3)2 Add 4 to both sides

±2 = (x + 3) Take square root of both sides

-2 = (x + 3) or 2 = (x + 3) Write as 2 equations

-5= x or -1= x Subtract 3 from both sides Y-intercept

y = (0 + 3)2 - 4 Set x = 0

y = 32 - 4 Add 0+3

y = 9- 4 Square 3

y = 5 Add 4 to both sides


III. System of Equations Example1: Use substitution or elimination method to solve the system of equations.

x2 + y2 -16x + 39 = 0

x2 - y2 - 9 = 0

ìíï

îï

Solution: Elimination Method:

2x2 -16x + 30 = 0 Add the two equations and combine like terms.

x2 - 8x +15 = 0 Divide by 2. (x - 3)(x - 5) = 0 Factor the trinomial.

x = 3 and x = 5 Use zero product property to find the values of x.

32 - y2 - 9 = 0 52 - y2 - 9 = 0 Plug x=3 and x = 5 into oneof theoriginal equations..

-y2 = 0 16 = y2 Solve y = 0 y = ±4

Points of Intersection (5,4), (5,-4) and (3,0)

Substitution Method:

y2 = -x2 +16x - 39 Solve one equation for one variable. (1st equation solved for y)

x2 - (-x2 +16x - 39) - 9 = 0 Plug what y2 is equal to into second equation.

2x2 -16x + 30 = 0 Combine like terms.

x2 -8x +15 = 0 Divide by 2.

(x - 3)(x -5) = 0 Factor the trinomial

x = 3 and x = 5 Use zero product property to find the values of x.

32 - y2 - 9 = 0 52 - y2 - 9 = 0 Plug x=3 and x = 5 into oneof theoriginal equations..

-y2 = 0 16 = y2 Solve y = 0 y = ±4

Points of Intersection (5,4), (5,-4) and (3,0)


IV. Functions

To evaluate a function for a given value, simply plug the value into the function for x.

Reads “f of g of x”. Means to plug the inside function (in this case g(x) )

in for x in the outside function (in this case, f(x)). Example1:

Given f (x) =2x2 +1 and g(x) = x - 4 find f(g(x)).

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

Example 2: Given: f(x) = 3x + 5 and g(x) = 2x − 1. Find: f(g(2)), g(f(2)) and f(g(x)) Solution: To find f(g(2)) we must first find g(2):

g(2) = 2(2) −1 = 4 – 1 = 3

Since g(2) = 3 we can find f(g(2)) = f(3) = 3(3) + 5 = 9 + 5 = 14 To find g(f(2)) we must first find f(2): f(2) = 3(2) + 5 = 6 + 5 = 11

Since f(2) = 11 we can find g(f(2)) = (11) = 2(11) – 1 = 22 − 1 = 21

To find f(g(x)) we must put the function g(x) into f(x) equation in place of each x.

f(g(x)) = f(2x − 1) = 3(2x − 1) + 5 = 6x – 3 + 5 = 6x + 2

The domain of a function is the set of x values for which the function is defined. The range of a function is the set of y values that a function can return. In Calculus we usually write domains and ranges in interval notation. If the domain were -1 < x ≤ 7 then in interval notation the domain would be (-1,7]. Example 3:

Find the domain and range for f (x) = x - 3

Solution: Since we can only take the square root of positive numbers x – 3 ≥ 0 which means that x ≥ 3. So we would say the domain is [3,∞). Note that we have used a [ to indicate that 3 is included. If 3 was not to be included we would have used (3,∞). The smallest y value that the function can return is 0 so the range is (0,∞).


V. Symmetry:

x-axis substitute in –y for y into the equation. If this yields an equivalent equation then the graph has x-axis symmetry. If this is the case, this is not a function as it would fail the vertical line test. y-axis substitute in –x for x into the equation. If this yields an equivalent equation then the graph has y-axis symmetry. A function that has y-axis symmetry is called an even function. Origin substitute in –x for x into the equation and substitute in –y for y into the equation. If this yields an equivalent equation then the graph has origin symmetry. If a function has origin symmetry, it is called an odd function. In order for a graph to represent a function it must be true that for every x value in the domain there is exactly one y value. To test to see if an equation is a function we can graph it and then do the vertical lines test.

Example 1: Is x - y2 = 2 a function?

The graph is below: Solution: When a vertical line is drawn it will cross the graph more than one time so it is NOT a function. Example 2: Test for symmetry with respect to each axis and the origin.

Given equation: xy - 4 - x2 = 0

Solution: x-axis (change all y to –y):

x(-y) - 4 - x2 = 0

-xy - 4 - x2 = 0 Since there is no way to make this look like the original it is NOT

symmetric to the x-axis. y-axis: (change all x to –x)

-xy - 4 - (-x)2 = 0

-xy - 4 - x2 = 0 Since there is no way to make this look like the original it is NOT

symmetric to the y-axis. Origin: (change all x to –x and change all y to –y )

-x(-y) - 4 - (-x)2 = 0

xy - 4 - x2 = 0 Since this does look like the original it is symmetric to the origin.


Example 3:

The figure to the right shows the graph of xy - 4 - x2 = 0 .

It is symmetric only to the origin. Example 4: Determine algebraically whether f(x) = –3x2 + 4 is even, odd, or neither. Solution: If I graph this, I will see that this is "symmetric about the y-axis"; in other words, whatever the graph is doing on one side of the y-axis is mirrored on the other side: This mirroring about the axis is true for even functions. But the question asks me to make the determination algebraically, which means that I need to do it with algebra, not with graphs as follows: f(–x) = –3(–x)2 + 4 Plug –x in for x. = –3(x2) + 4 Square –x. = –3x2 + 4 Multiply by –3. The final expression is the same thing I'd started with, which means that f(x) is even.


VI. Asymptotes and Holes Given a rational function if a number causes the denominator and the numerator to be 0 then both the numerator and denominator can be factored and the common zero can be cancelled out. This means there is a hole in the function at this point. Example1:

Find the holes in the following function f (x) =

x - 2

x2 - x - 2.

Solution:

f (x) =x - 2

(x +1)(x - 2) Factoring and canceling. When x=2 is substituted into the function the

denominator and numerator both are 0.

f (x) =1

(x +1) But (x≠2) this restriction is from the original function before canceling.

The graph of the function f(x) will look identical to

f (x) =1

(x +1)except for the hole at x=2.

f (x) =

x - 2

x2 - x - 2

f (x) =1

(x +1)

Note the hole at x=2 Given a rational function if a number causes the denominator to be 0 but not the numerator to be 0 then there is a vertical asymptote at that x value. Example 2:

Find the vertical asymptotes for the function f (x) =

x - 2

x2 - x - 2.

Solution: When x = –1 is substituted into f(x) then the numerator is –1 and the denominator is 0 therefore there is an asymptote at x = –1. See the graphs above. Given a rational function if a number causes the numerator to be 0 but not the denominator to be 0 then the value is an x-intercept for the rational function.


Example 3: Discuss the zeroes in the numerator and denominator f (x) =

x + 3

2x

Solution: See the graph is to the right. When x = –3 is substituted into the function the numerator is 0 and the denominator is –6 so the value of the function is f(–3)=0 and the graph crosses the x-axis at x = –3. Also note that for x = 0 the numerator is 3 and the denominator is 0 so there is a vertical asymptote at x = 0. Example 4:

Find the holes, vertical asymptotes and x-intercepts for the function f (x) =

x2 - 3x

3x2 + 6x.

Solution:

f (x) =x(x - 3)

3x(x + 2) Factor to find all the zeroes for both the numerator and denominator.

x = 0 and x = 3 Numerator has zeroes. x = 0 and x = –2 Denominator has zeroes. x = 0 Is a hole. x = –2 Is a vertical asymptote. x = 3 Is a x-intercept.


VII. Inverses • To find the inverse of a function, simply switch the x and the y and solve for the new “y” value.

• To PROVE one function is an inverse of another function, you need to show that:

f (g(x)) = g( f (x)) = x

Example 1:

Find the inverse function for f (x) = x +13

Solution:

f (x) = x +13 Rewrite f(x) as y.

y = x +13 Switch x and y.

x= y +13 Solve for the new y.

x( )

3

= y +13( )3

Cube both sides.

x3 = y +1 Simplify.

y = x3 -1 Solve for y.

f-1(x) = x3 -1 Rewrite in inverse notation.

Example 2:

If: f (x) =

x - 9

4and g(x) = 4x + 9 show f(x) and g(x) are inverses of each other.

Solution:

f (g(x)) = 4x - 9

4

æ

èçö

ø÷+ 9 g( f (x)) =

4x + 9( ) - 9

4

= x - 9 + 9 =4x + 9 - 9

4

= x =4x

4

f (g(x)) = g( f (x)) = x Therefore they are inverses of each other.


VIII. Finding Solutions

• Factoring or using the quadratic formula to solve equations of the form ax2 + bx + c = 0 .

x =

-b ± b2 - 4ac

2a

• Solving Inequalities by factoring, creating a number line, and checking regions.

• Simplify Rational expressions by finding the common denominator.

Example 1:

Solve x2 - 2x =10.

Solution:

x2 - 2x =10 Add 7 to both sides to get all terms to one side of equation = 0.

x2 - 2x -10 = 0 Determine a, b and c. a = 1, b = –2, and c = –10.

x =- -2( ) ± -2( )

2

- 4 1( ) -10( )2 1( )

Use quadratic formula.

=

4 ± 4 - -40( )2

=

4 ± 4 + 40

2

=

4 ± 44

2 Simplify your answer.

x =

4 ± 44

2=

4 ± 2 ×2 ×11

2=

4 ± 2 11

2=

4

2±

2 11

2= 2 ± 11

x = 2 ± 11 These are the Solutions.

Example 2:

Solving Inequalities y £ 2x2 + x - 4 .

Solution: The two x-intercepts are x = –1.69 and x = 1.19. To find out if your Solutions are in between or outside, we need a test point. Choose a point in between. • If the inequality is true for your test point, the solution is all the

numbers in between. • If the inequality is false for your test point, the solution is all the

numbers outside. 2(0) ²+(0) – 4 ³0 Let’s pick x = 0. – 4 ³0 False! So the solutions are outside. x £ -1.69 or x ³1.19 Solutions.


Example 3:

Simplify:

2x

3+

1

5x

Solution:

2x

3+

1

5x LCD: 15x since 3 * 5x = 15x.

2x

3•

5x

5x+

1

5x•

3

3 Multiply each fraction by missing part of denominator over itself to get a

common denominator.

10x2

15x+

3

15x Multiply numerators and denominators of each fraction.

10x2 + 3

15x Add numerators and keep the same denominator.

Example 4:

Simplify:

2x

x2 - 9+

1

x - 3

Solution:

2x

x2 - 9+

1

x - 3 LCD: (x−3)(x+3) – Factored form of x2 – 9.

2x

x2 - 9+

(x + 3)

(x + 3)

1

x - 3

æ

èçö

ø÷ Multiply each fraction by missing part of denominator over itself (only need to

do this on right fraction) to get a common denominator.

2x

x2 - 9+

x + 3

x2 - 9 Multiply numerators and denominators of each fraction as needed.

3x + 3

x2 - 9 Add numerators and keep the same denominator.


IX. Absolute Value and Piecewise Functions In order to remove the absolute value sign from a function you must: • Find the zeroes of the expression inside of the absolute value. • Make sign chart of the expression inside the absolute value. • Rewrite the equation without the absolute value as a piecewise function. For each interval where the

expression is positive we can write that interval by just dropping the absolute value. For each interval that is negative we must take the opposite sign.

Example1: Rewrite the following equation without using absolute value.

f (x) = 2x + 4

Solution:

2x + 4 = 0 Find where the expression is 0.

2x = -4 Subtract 4.

x =

-4

2 Divide by 2.

x = -2 Simplify. Plug in any value less than –2 into 2x + 4, you will get a negative value. Plug in any value more than –2 into 2x + 4, you will get a positive value.

Write as a piecewise function. Be sure to change signs of each term for any part of the graph that was negative on the sign chart.

Example 2: Rewrite the following equation without using absolute value.

f (x) = 2x2 + 5x - 3

Solution:

2x2 +5x - 3= 0 Find where the expression is 0.

(2x -1)(x + 3) = 0 Factor.

2x -1= 0 or x + 3= 0 Set each factor equal to 0.

x =

1

2 or x = -3 Solve each equation.

Plug in any value less than –3 into (2x -1)(x + 3) , you will get a

positive value. Plug in any value more than –3 and less than

1

2

into (2x -1)(x + 3) , you will get a negative number. Plug in any

value more than

1

2 into

(2x -1)(x + 3) , you will get a positive

number.


_-

-2

+

-3

_- + +

1/2

f (x) =

2x2 + 5x - 3, (-¥,-3)U(1

2,¥)

-2x2 - 5x + 3, - 3£ x £1

2

ì

íïï

îïï

f (x) =-2x - 4, x < -2

2x + 4, x ³ -2

ìíî


Example 3: Rewrite the following equation without using absolute value.

f (x) = 3x - 9 + 2

Solution:

3x -9 = 0 Find where the expression is 0 (For the part in the absolute value only.)

3x = 9 Add 9

x = 3 Divide by 3 Plug in any value less than 3 into 3x - 9 , you will get a negative value. Plug in any value more than 3 into 3x - 9 , you will get a positive value.


f (x) =-3x +11

3x - 7

ìíî

x < 3

x ³ 3

üýþ

Simplify

Absolute value inequalities require you to write two separate inequalities. You were probably taught to Keep Flip Change. One inequality will be identical to the inequality, just without the absolute value sign. The second inequality will have a flipped inequality sign and the opposite value.

3

_- +

f (x) =-3x + 9 + 2, x < 3

3x - 9 + 2, x ³ 3

ìíî


X. Exponents

• A fractional exponent means you are taking a root. For example x1/2 is the same as x .

• Negative exponents mean that you need to take the reciprocal. For example x-2 means

1/ x2 and 2 / x-3

means 2x3 .

• When factoring, always factor out the lowest exponent for each term.

• When dividing two terms with the same base, we subtract the exponents (numerator exponent-

denominator exponent). If the difference is negative then the term goes in the denominator. If the

difference is positive then the term goes in the numerator.

Example1:

Write without fractional exponent: y = x2/3

Solution:

y = x23 Notice that the root is the bottom number in the fraction and the power is the top number in

the fraction. Example 2:

Write with positive exponents: y =

2

5x-4

Solution:

y =

2x4

5

Example 3:

Write with positive exponents and without fractional exponents:

f (x) =(x +1)-2(x - 3)

1

2

(2x - 3)-

1

2

Solution:

f (x) =x - 3 * 2x - 3

(x +1)2

Example 4:

Factor: y = 3x-2 + 6x - 33x-1

Solution:

The lowest exponent for x is –2 so 3x-2 can be factored from each term. Leaving y = 3x-2(1+ 2x3 -11x)

Notice that for the exponent for the 6x term we take 1– (–2) and get 3.

For the 33x-1term we take –1– (–2) and get 1 as our new exponent.


Example 5:

Simplify f (x) =

(2x)3

x8

Solution:

First you must distribute the exponent. f (x) =

8x3

x8. Then since we have two terms with x as the base

we can subtract the exponents. Since 3-8 results in -5 we know that we will have x5in the

denominator. f (x) =

8

x5.

Example 6:

Simplify $$

f (x) =x2 - 2x +1

(x2 -1)

Solution: First we must factor both the numerator and denominator.

f (x) =(x -1)2

(x +1)(x -1). Then we

can see that we have the term (x-1) in both the numerator and denominator. Subtracting exponents

we get 2-1=1 so the term will go in the numerator with 1 as it’s exponent.

f (x) =(x -1)

(x +1).

Example 7:

Factor and simplify f (x) = 4x(x - 3)1/2 + x2(x - 3)-1/2

Solution: The common terms are x and (x-3). The lowest exponent for x is 1. The lowest exponent for (x-3) is -

1/2. So factor out x(x - 3)-1/2 and obtain f (x) = x(x - 3)-1/2[4(x - 3) + x]. This will simplify to

f (x) = x(x - 3)-1/2[4x -12 + x]. Leaving a final solution of

x(5x -12)

x - 3.


XI. Rational Expressions When simplifying complex fractions, multiply by a fraction equal to 1, which has a numerator and denominator composed of the common denominator of all the denominators in the complex fraction.

Example1:

-7 -6

x + 1

5

x + 1

=

-7 -6

x + 1

5

x + 1

ix + 1

x + 1

= -7x - 7 - 6

5

= -7x - 13

5

Example 2:

-2

x+

3x

x - 4

5 -1

x - 4

=

-2

x+

3x

x - 4

5 -1

x - 4

i x(x - 4)

x(x - 4)

= -2(x - 4) + 3x(x)

5(x)(x - 4) - 1(x)

= -2x + 8 + 3x

2

5x2

- 20x - x

= 3x

2- 2x + 8

5x2

- 21x

Example 3: Simplify, using factoring of binomial expressions. Leave answers in factored form.

(x +1)3(4x -9)- (16x +9)(x +1)2

(x -6)(x +1)=

(x +1)2 (x +1)(4x -9)- (16x +9)éë

ùû

(x -6)(x +1)

=(x +1)2(4x2 -5x -9-16x -9)

(x -6)(x +1)

=(x +1)2(4x2 - 21x -18)

(x -6)(x +1)

=(x +1)2(4x +3)(x -6)

(x -6)(x +1)

= (x +1)(4x +3)

Example 4: Simplify by rationalizing the numerator.

( )

( )

4 2 4 2 4 2

4 2

4 4

4 2

4 2

1

4 2

x x x

x x x

x

x x

x

x x

x

+ − + − + += •

+ +

+ −=

+ +

=+ +

=+ +


XII. Exponents and logarithms

bx = yÛ log

by = x

• bxby = bx+y

log

bx + log

by = log

bxy

•

bx

by= bx- y

logb

x - logb

y = logb

x

y

æ

èçö

ø÷

• bx( )

y

= bxy log

bx y = y log

bx

• bx = by or if

log

bx = log

by , then

x = y

• log

bbx = x and b

logbx

= x

• log x = log

10x and

ln x = log

ex

•

logba =

logca

logcb


XIII. Natural Logarithms

Recall that y = ln x( ) and

y = ex (exponential function) are inverse to each other properties of the Natural Log:

ln AB( ) = ln A( )+ ln B( )

Example1: ln 2( )+ ln 5( ) = ln 10( )

lnA

B

æ

èçö

ø÷= ln A( ) - ln B( )

Example 2:

ln 6( ) - ln 2( ) = ln6

2

æ

èçö

ø÷= ln 3( )

ln Ap( ) = p ln A( )

Example 3: ln x4( ) = 4ln x( ) and

3ln 2( ) = ln 23( ) = ln 8( )

ln ex( ) = x ,

ln e( ) = 1,

ln 1( ) = 0,

y = - f (x)

Example 4: Use the properties of natural logs to solve for x:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

( ) ( )

( ) ( )

2 5 11 7

5 11

7 2

5 11ln ln

7 2

ln 5 ln 7 ln 11 ln 2



ln 11 ln 2

ln 5 ln 7

x x

x

x

x

x

x x

x x

x

x

• = •

=

=

− = −

− = −

− = −

−=

−


XIV. Graphing Trig Functions

( ) sin( )f x x=

Domain: ( , )− Range: [-1,1]

Period: 2 ODD

Maxima @ 2

2x n = +

Minima @ 32

2x n = +

x-intercepts @ x n=

( ) cos( )f x x=

Domain: ( , )− Range: [-1,1]

Period: 2 EVEN

Maxima @ 2x n=

Minima @ 2x n = +

x-intercepts @ 2

x n

= +

( ) tan( )f x x=

Domain: 2

x n

+ Range: ( , )−

Period: ODD No maxima or minima x-intercepts @ x n=

( ) cot( )f x x=

Domain: x n Range: ( , )−

Period: ODD No maxima or minima

x-intercepts @ 2

x n = +


( ) csc( )f x x=

Domain: x n Range: ( , 1] [1, )− −

Period: 2 ODD

Maxima @ 32

2x n = +

Minima @ 2

2x n = +

No x-intercepts

( ) sec( )f x x=

Domain: 2

x n

+ Range: ( , 1] [1, )− −

Period: 2 EVEN

Maxima @ 2x n = +

Minima @ 2x n= No x-intercepts

For , A = amplitude, period =

= Phase Shift (positive C/B shift left, negative C/B shift right) and K = vertical shift.

XV. Transformations of graphs

f (x) = Asin(Bx + C) + K2p

BC

B


• y = f (x - a) is the graph of

y = f (x) shifted horizontally a units (to the right if a > 0 and to the left if

a < 0 )

• y = f (x) + a is the graph of

y = f (x) shifted vertically a units (up if a > 0 and down if a < 0 )

• y = af (x) is the graph of

y = f (x) stretched or shrunk vertically by a factor of a (stretched if a >1 and

shrunk if 0 < a <1)

• y = f (ax) is the graph of

y = f (x) stretched or shrunk horizontally by a factor of a (stretched if 0 < a <1

and shrunk if a >1)

• y = - f (x) is the graph of

y = f (x) reflected over the x-axis

• y = f (-x) is the graph of

y = f (x) reflected over the y-axis

Example1: Identify the basic shape and then list all transformations in their proper order

f x( ) = - 2 x +1( )

3

- 5

Solution: Shape: “cubic” [from exponent, 3] Horizontal Shift: 1 left [from x + 1] Vertical Stretch by a factor of 2 [from coefficient, - 2] x-axis reflection [from negative on coefficient, -2] Vertical Shift: 5 down [from - 5 on outside]

XVI. Basic graphs


y x=

1y

x=

y x=

ny x= , n even

ny x= , n odd

xy a= , 1a

, 0 1xy a a=

log , 1ay x a=

log , 0 1ay x a=

XVII. Equation Of Circles


The general equation for a circle that has center (h, k) and radius r is: (x – h)2 + (y – k)2 = r2.

For example, the circle pictured above has this equation: (x – 5)2 + (y – 4)2 = 9. For many circles, the h and/or k are negative; remember that a double negative is often written as a +. Putting Equations in Standard Form Example1: Find the standard form, center, and radius of the following circles: x2 + y2 + 6x – 8y – 11 = 0 Solution: (x2 + 6x) + (y2 – 8y) = 11 Combine like terms. (x2 + 6x + 9) + (y2 – 8y + 16) = 11 + 9 + 16 Complete the squares. (x + 3)2 + (y – 4)2 = 36 Rewrite in standard. Center: (-3, 4) and Radius: 6

XVIII. Lines I. Special products and factoring

Sum/difference of two cubes: ( ) ( )

( ) ( )

3 3 2 2

3 3 2 2

a b a b a ab b

a b a b a ab b

+ = + − +

− = − + +

XIX. Geometry formulas


The following formulas are needed for a topic in AP calculus called related rate. Area:

Triangle: 1

2bh Trapezoid: 1 2

1( )

2h b b+ Circle: 2r

Surface area of Sphere: 24 r Lateral area of cylinder: 2 rh Volume:

Cone: 21

3r h Sphere: 34

3r Cylinder: 2r h

Prism: Bh where B is the area of the base Pyramid: 1

3Bh where B is the area of the base..

XX. Trig. Equations and Special Values You are expected to know the special values for trigonometric functions.

Use

180

p radians to get rid of radians and convert to degrees.

Use

p radians

180 to get rid of degrees and convert to radians.

You can determine the sine or cosine of a quadrantal angle by using the unit circle. The x-coordinate of the circle is the cosine and the y-coordinate is the sine of the angle.

Example1: sin90 = 1 cos

p

2= 0

You should study the following trig identities and memorize them before school starts: Reciprocal identities

xx

csc

1sin =

xx

sec

1cos =

xx

cot

1tan =

xx

sin

1csc =

xx

cos

1sec =

xx

tan

1cot =

Tangent Identities

x

xx

cos

sintan =

x

xx

sin

coscot =


Pythagorean Identities

1cossin 22 =+ xx xx 22 sec1tan =+ xx 22 csc1cot =+ Double angle Identities

xxx cossin22sin = xxxxx 2222 sin211cos2sincos2cos −=−=−= ..

Reduction Identities

xx sin)sin( −=− xx cos)cos( =− xx tan)tan( −=−

We use these special values and identities to solve equations involving trig functions. Solve each of the equations for 0 £ x < 2p . Isolate the variable, sketch a reference triangle, find all the Solutions within the given domain, 0 £ x < 2p . Remember to double the domain when solving for a double angle. Use trig identities, if needed, to rewrite the trig functions.

Example 2: Find all Solutions to 1sinsin2 2 =+ xx Solution:

1sinsin2 2 =+ xx Original Problem

01sinsin2 2 =−+ xx Get one side equal to 0. 0)1)(sin1sin2( =+− xx Factor

0)1sin2( =−x and 0)1(sin =+x Set each factor equal to 0

2

1sin =x

and 1sin −=x Get the trig function by itself

x =p

6+ 2pk

x =5p

6+ 2pk

and kx

22

3+= Solve for x

.

A - All trig functions positive in Quad I S - sin & csc functions positive in Quad II T – tan & cot functions positive in Quad III C – cos & sec function positive in Quad IV


XXI. Inverse Trigonometric Functions Inverse Trig Functions can be written in either of these ways:

arcsin x( ) sin-1 x( )

Inverse trig functions are defined only in the quadrants as indicated below due to their restricted domains. sin-1x cos-1x cos-1 x tan-1 x

tan-1 x sin-1 x Example 1: Express the value of “y” in radians.

y = arctan-1

3

Solution: Draw a reference triangle. 2 -1

3

This means the reference angle is 30 or p

6. So, y = –

p

6 so that it falls in the interval from

-p

2< y <

p

2

Answer: y = – p

6

Example 2: Find the value without a calculator.

cos arctan5

6

æ

èçö

ø÷

Solution: Draw the reference triangle in the correct quadrant first. Find the missing side using Pythagorean Theorem. Find the ratio of the cosine of the reference triangle.

cosq =6

61

6

5

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