AP Calculus INFORMATION SHEET 2019-2020

Name Per Date

Summer Assignment As advanced placement students, your first assignment for the 2019-2020 school year is to come to class the very first day with the math concepts covered in this assignment mastered. The following assignment will help you practice the concepts needed to ready for the first day of school. Make sure: 1. Copy down the original problem (unless it is a graph) 2. Show all steps to support your answer 3. Circle or box your answers *If these above steps are not done, then significant points will be docked from your overall score. Basically, ALL of your work must be in logical order while indicating your answers clearly. Write neatly in pencil. I will not give credit to anything that is sloppy or has no supporting work. In addition, please do not hand in work on paper torn from spiral notebooks. Hole punched plain paper or graph paper is preferred. This packet will be turned in as credit/no credit based on having the assignment fully completed and all supporting work clearly shown. DO NOT cheat on this assignment. If you do not complete it on your own and think through the concepts covered in this packet then you are only hurting yourself. It will come back and hurt your future test scores. This does not imply that you cannot work with other students on this. Just do it in a smart way making sure that you are mastering the topics in this packet.

The next 11 pages of this packet contain formulas, background information, and example problems with steps to help you understand the topics covered in the assignment portion. Your assignment starts on page 13. There is not space for you to show your work on it. As stated above, you must show all work on a separate piece of paper. You can do google searches for any of these topics. Here is a good site for most algebra topics: https://www.purplemath.com/modules/index.htm Algebraic topics -Exponents -Negative and fractional exponents -Domain -Factoring -Solving quadratic equations -Even and odd functions -Complex fractions -Composition of functions -Solving rational (fractional) equations -Graphing piecewise functions Trigonometry -Evaluating trig functions -Evaluating inverse trig functions -Solving Limits There will be a quiz on trigonometric values on the third day of school.

Exponents

Simplify and write with positive exponents.

1. 28x−− 2. ( ) 235x−

− 3. 2

4

3x

− −

4. ( )1

10 236x 5. ( )2

3 327x−

6. ( )3

2 416x−

In calculus, you will be required to perform algebraic manipulations with negative exponents as well as fractional exponents. By

definition: 1n

nxx

− = and 1 aa

b bx x

=

. When evaluating, 1 aa

b bx x

=

it will almost always be best to apply the denominator of the

exponent to the base FIRST, then apply the numerator. It is better to “shrink” the number, because making it larger. As a reminder, rules of exponents are as follows: -When we multiply, we add exponents: a b a bx x x +⋅ =

-When we divide, we subtract exponents: a

a bb

x xx

−= , 0x ≠

-When we raise powers, we multiply: ( )ba a bx x ⋅=

2

8x

( )( )

2 3 22 66

1 15255

xxx

− ⋅−− = =−

( )( ) ( )

2 8

2 2 84

3 193x

xx

−

− −

−= =

−

( )1 1

10 52 236 6x x= ( ) ( )

2 2 221 13 3 33 3

1 1 1927 27

xx x

= =

33 6 1

4 4 43 32 2

1 816 16xx x

− = =

Factoring

1. 2 81x − 2. 216 4x − 3. 325 9x x− 4. 2 6 40x x− − 5. 26 5 4x x+ − 6. 4 3 23 11 6x x x− +

You must be extremely knowledgeable on factoring expressions of the form: -Factoring out the GCF: ( )3 2 2 1x x x x x x+ + = + +

-Difference of squares: ( )( )2 2x a x a x a− = + −

-Trinomials w/ leading coefficient 1: ( )( )2 __ __x bx c x x+ + = ± ± ← need two numbers that multiply to c and add to b.

-Trinomials w/ leading coefficient not 1: 2ax bx c+ + ← can factor by guess and check, diamond method, box method, etc

( )( )9 9x x+ − ( )( )4 2 4 2x x+ − ( ) ( )( )225 9 5 3 5 3x x x x x− = + −

( )( )4 10x x+ − ( )( )3 4 2 1x x+ − ( ) ( )( )2 2 23 11 6 3 2 3x x x x x x− + = − −

Complex Fractions/Simplifying Rational Expressions

1. ( )

2 23 3x h x

−+

2.

1 14 4 x

x

−+ 3.

3 2

2

2 16 404 48 80x x xx x+ −+ +

4.

3

7214

x

x

Common strategies: -Get a common denominator -To divide, multiply by the reciprocal

( )

( )( )

2 23 3

x hxx h x x x h

+⋅ − ⋅

+ +

( )( )( )

223 3

x hxx x h x x h

+−

+ +

( ) ( )2 2 2

3 3x x h

x x h x x h+

−+ +

( )2 2 2

3x x hx x h− −

+

( )2

3h

x x h−+

( )( )

( )( )

4 41 14 4 4 4

xx x

x

+⋅ − ⋅

+ +

( ) ( )4 4

4 4 4 4xx x

x

+−

+ +

( )4 44 4

xx

x

+ −+

= ( )4 4

xx

x+

( )1

4 4x

x x⋅

+ =

( )1

4 4 x+

( )( )

2

2

2 8 20

4 12 20

x x x

x x

+ −

+ +

( )( )( )( )

2 10 24 10 2x x x

x x+ −+ +

( )( )

22 2x x

x−+

372 14

xx⋅

212 2

x⋅

2

4x

Solving

1. 2 10 20 4x x+ − = 2. 5 315 36 0x x x− + = 3. 3 5

2 x=

4. 3 2 1 4x− − = 5. 2 3 6 2 18 3

4 5 20x x x− + − −

− =

You must know how to solve equations that involve factoring, simplifying rational expressions, and fractions. -The goal is always to isolate the variable -Sometimes, the “law of flip flop” will be useful -Always isolate the variable following this order: undo addition/subtraction, undo multiplication/division, undo exponents, and undo what is in parentheses -You can only solve quadratics by factoring when one side of the equal sign is ZERO!

-Zero product property: ( )3 34 0 42 2

x x x and x + − = ⇒ = − =

-Common errors: don’t forget ± when square rooting in the solving process and don’t forget to distribute a negative!

2 10 24 0x x+ − = ( )( )12 2 0x x− + =

12, 2x x= = −

2 1 1x− − = 112

x − = −

114

x − =

1116

x − =

1716

x =

3 25

x= (law of flip flop!)

310

x=

9100

x=

( )4 215 36 0x x x− + =

( )( )2 212 3 0x x x− − =

( ) ( )2 20, 12 0, 3 0x x x= − = − =

0, 2 3, 3x x x= = ± = ±

2 3 5 6 2 4 18 34 5 5 4 20

x x x− + − −⋅ − ⋅ =

10 15 24 8 18 320 20 20x x x− + − −

− =

( ) ( )10 15 24 8 18 3x x x− − + = − − 14 23 18 3x x− − = − −

4 20x = 5x =

Linear Equations 1. Find the equation of a line with the given info in point-slope form then convert to standard form

(a) ( )1 , 3,105

m = − (b) ( ) ( )4,2 , 6, 2− −

2. Write the equations of the line through the given point parallel to the given line then perpendicular to the given line (a) ( )3,5 ; 2 4x y− − + = (b) ( )3, 4,2x = − (c) ( )3 6 10, 1, 4x y− = −

It is very important that you are comfortable with writing linear equations and to be able to convert one from to another. -Slope: Given two points, ( )1 1,x y and ( )2 2,x y , the line passing through the points can be written as

2 2

1 1

y xymx y x

−∆= =∆ −

-Point-Slope: (by far the most common and efficient form you will use in calculus this year)

( )1 1y y m x x− = − ; where m is the slope and ( )1 1,x y the point your line passes through

-Slope-Intercept: (not commonly used because point-slope is so much quicker)

y mx b= + ; where m is the slope and b is the y-intercept of the line

-Standard form: (typically, you would not choose to write your equation in this form but answers on a multiple choice question might be of this form)

Ax By C+ = ; where , ,A B and C are whole numbers and A is positive

-Parallel lines: two distinct lines are parallel if and only if 1 2m m=

-Normal lines: two lines are normal (perpendicular) if their slopes are opposite reciprocals 1 2 1m m⋅ = −

-Horizontal lines: have slope of zero and form #y =

-Vertical lines: have undefined slope and form #x =

( )110 35

y x− = +

1 3105 5

y x− = + (p.s)

1 535 5

x y− + =

( )1 53 55 5

x y − + = ⋅ −

5 53x y− = − (std)

( )2 2 4 2

6 4 10 5m − −= = − = −

− −

22 45

y x− = − + (p.s.)

2 65

x y+ =

2 6 55

x y + = ⋅

2 5 30x y+ = (std)

2 4y x= +

*use 2m = for

( )5 2 3y x+ = −

*use 12

m = − for ⊥

( )15 32

y x+ = − −

*all vertical lines are

to each other 4x =

*all horizontal lines are ⊥ to vertical lines

2y =

6 3 10y x− = − +

1 52 3

y x= −

*use 12

m = for

( )14 12

y x+ = −

*use 2m = − for ⊥

( )4 2 1y x+ = − −

Logarithms and Exponentials

1. log 2 log50+ 2. ln192 ln 3− 3. 5 3ln e

4. ( )2ln 3 3 0x x− + = 5. ( ) 1ln ln 32

x x+ − = 5. 5 20x =

6. ( )ln ln 1 1x x− − = 7. 2 5xe− = 8. 12 3x x−=

Calculus spends a great deal of time on exponential functions in the form xb . We need to have them practiced now so that when we begin working with them this year, we do not have to stop and review. It is important to keep a rigorous pace throughout the year so we have time review in class before your AP test. Here is what you need to know

- ( )log log loga b a b⋅ = + log log loga a bb

= −

log logba b a=

-logarithmic functions and exponential functions are inverses (they undo each other in the solving process):

( )ln xe x= and ln xe x=

-the natural log is log base e and know the following values: ln1 0= 1e e= 0 1e = -in calculus, you will notice that the natural log is much more commonly used

( )log 2 50⋅ log100

192ln3

ln 64

35ln e

35

( )2ln 3 3 0x xe e− +=

2 3 3 1x x− + = 2 3 2 0x x− + =

( )( )1 2 0x x− − = 1, 2x x= =

( )( )ln 3 ln 4x x⋅ − = ( )( )ln 3 ln 4x xe e⋅ − = ( )3 4x x⋅ − =

2 3 4 0x x− − = ( )( )1 4 0x x+ − =

1, 4x x= − =

ln 11

xx

=−

ln 11x

xe e− =

1x e

x=

−

( )1x e x= − x ex e= −

x ex e− = − ( )1x e e− = −

( )1exe

−=

−

( ) ( )ln 5 ln 20x =

( )ln 5 ln 20x = ln 20ln 5

x =

( ) ( )2ln ln 5xe− = 2 ln 5x− =

ln 52

x = −

1 ln 52

x = − 12ln 5x

−=

1ln5

x =

*the last 3 steps not necessary but sometimes you have to keep

manipulating your expression to match an answer on a multiple choice question

( ) ( )1ln 2 ln 3x x−=

( )ln 2 1 ln 3x x= − ln 2 ln 3 ln 3x x= −

ln 2 ln 3 ln 3x x− = − ( )ln 2 ln 3 ln 3x − = −

ln 3ln 2 ln 3

x −=

−

Function Composition

1. Given ( ) 23 2xxxf += and ( ) 13 2 −= xxg , find

(a) f + g (b) f – g (c) fg (d) f/g 2. Find both gf and fg

(a) ( )x

xf 1= , ( ) xxxg 43 −= (b) ( ) 3+= xxf , ( ) 122 −= xxg

3. Given the table below, find the following: (a) ( )( )1gf (b) ( )( )1ff (c) ( )( )3fg

4. Given ( ) 3 5f x x= − − , find ( ) ( )1 5f f−

x 1 2 3 4 5 6 f(x) 3 1 4 2 2 5 g(x) 6 3 2 1 2 3

Know the notation for function composition: ( )( )f g x is equivalent to ( )( )f g x

-To do function composition, plug the innermost function into the input (x) for the outside function

( ) ( )3 2 22 3 1x x x+ + − 3 25 1x x+ −

( ) ( )3 2 22 3 1x x x+ − − 3 2 1x x− +

( )( )3 2 22 3 1x x x+ − 5 4 3 22 6 2x x x x+ − −

3 2

2

23 1

x xx+−

( )( )f g f g x=

*plug ( )g x into ( )f x

( )( ) ( )3

14

f g xx x

=−

( )( )g f g f x=

*plug ( )f x into ( )g x

( )( )31 14g f x

x x = −

( )( ) 3

1 4g f xx x

= −

( )( )f g f g x=

*plug ( )g x into ( )f x

( )( ) ( )2 12 3f g x x= − +

( )( )g f g f x=

*plug ( )f x into ( )g x

( )( ) ( )23 12g f x x= + −

3 12x + − 9x −

( )1 6g = *now plug 6 into f

( )( )1f g

( )6f 5

( )1 3f = *now plug 3 into f

( )( )1f f

( )3f 4

( )( ) ( )( )3 3g f f g=

( )3 2g = *now plug 2 into g

( )( )3f g

( )2f 1

( ) ( )( ) ( )

1 1 3 5 4 5 4 5 1

5 5 3 5 2 5 2 5 3

f

f

− = − − − = − − = − = −

= − − = − = − = −

So ( ) ( ) ( )1 5 1 3 2f f− = − − − =

Piecewise Functions

1. Write the following as a piecewise function (a) ( ) 2 6f x x= + (b) ( ) 3 5f x x= − +

2. Graph the following piecewise function ( )

2 if 13 2 if 1

if 1

x xf x x x

x x

≤ −

= + <

≥

Piecewise functions are very common to work with in our first unit, especially in the context of limits -One common function to know is the absolute value function, x

Graph of x Piecewise function of x

; 0; 0

x xx

x x− ≤

= >

*Factor out common factors of the input of your absolute value function

( ) ( )2 3f x x= +

*Your piecewise is always split around the zero of the input of the absolute value function

( )2 6 ; 32 6

2 6; 3x x

xx x

− + ≤ −+ = + > −

2 6; 32 6

2 6; 3x x

xx x

− − ≤ −+ = + > −

*Factor out common factors of the input of your absolute value function

no common factors *Your piecewise is always split around the zero of the input of the absolute value function

( )( )

3 5; 33 5

3 5; 3

x xx

x x

− − + ≤− + = − + >

8; 33 5

2; 3x x

xx x− + ≤

− + = + >

Note: the bound 1x <

means 1 1x− < < Remember: and≤ ≥ means you need a closed dot Remember: and< > means you need an open dot

Trigonometry

1. Evaluate the following trigonometric functions (WITHOUT a calculator!)

(a) 4

7sin π (b) 3

cot π (c) 3

5tan π (d) 4

sec π (e) 6

7cos π (f) πcsc

2. Evaluate the following inverse trig functions (WITHOUT a calculator!)

(a) 1 3sin2

− (b) 1 3cos2

− (c) 1 1sin2

− −

(d) 1 1cos2

− (e) 1tan 3−

3. Solve the following equations involving trig functions

(a) 2sin 1 0x − = (b) 2sin cos sin 0x x x+ =

Knowing your trig values is probably the most important thing to have mastered before starting your year in calculus. It is also equally important to know inverse trig values. You also need to know how to solve equations that involve trig functions. The following problems illustrate what is expected to be mastered.

For example, knowing 1sin6 2π =

within a matter of seconds is extremely important and helpful. Also, it is important to know

the reverse direction in something such as 1 2 7cos ,2 4 4

π π− =

. Any trigonometric function and its inverse “undo” each other.

For example, ( )1sin sin x x− = . -There are two main approaches to knowing/figuring out your trig values: knowing your unit circle solid and using reference triangles (which does go hand in hand with the unit circle). The benefit to the unit circle is that it is VERY quick when working with sine and cosine. Outside of these functions, the manipulation needed for trig functions such as cosecant and cotangent can be cumbersome to deal with. The benefit of reference triangles is that it can more efficient when working with any trig value. Go to the following link: https://www.youtube.com/watch?v=xgIVqLLFN4s It is a very helpful video on evaluating trig expressions using reference triangles. It is based on knowing your 30 60 90° − °− ° and 45 45 90° − °− ° . I will say, I never convert the radian measure to degrees as he does in the video. I do think it is helpful to see the conversion to visualize exactly where a radian measure places you in order to apply the trig function as you are watching the video. The video is done by a guy named Steve Chow and he has a great youtube channel called blackpenredpen.

1 222

− = − 1 333

= 3− 2 32

− und

2,3 3π π

4,3 3π π 5,

3 3π π 5 7,

4 4π π

11,6 6π π

2sin 1x = 1sin2

x =

*In order to solve for x, you must “undo” the sine function next to it by applying inverse sine as shown next.

( )1 1 1sin sin sin2

x− − =

1 1sin2

x − =

5,6 6

x π π=

*Factor out the common factor, sin x

( )sin 2cos 1 0x x + = *Set both factors equal to 0 and solve each individually

sin 0, 2cos 1 0x x= + = ( ) ( )1 1sin sin sin 0 , 2cos 1x x− −= = −

( )1 1sin 0 , cos2

x x− −= =

( )1 1 10, cos cos cos2

x x− − − = =

2 40, ,3 3

x x π π= =

Limits

1. Evaluate each limit using the graph of f below (a) ( )

0limx

f x→

(b) ( )4

limx

f x→−

(c) ( )5

limx

f x→

(d) ( )2

limx

f x→

(e) ( )

10lim

xf x

+→ (f) ( )

7limx

f x→

(g) ( )1

limx

f x→−

(h) ( )10

limx

f x→

2. Evaluate the following limits at infinity (covered in Pre-AP Math 3)

(a) 2

2

8 3lim2 5 1x

x xx x→∞

−− + −

(b) 3

3

8 3lim2 5 1x

x xx x→−∞

− −+ −

(c) 2

3

8 3lim2 5 1x

x xx x→−∞

− −+ −

Here is a very helpful site that walks you through what is limit is, when does a limit exist, evaluating limits, evaluating limits at infinity. https://www.calculus-help.com/phobedemo/ -DO NOT watch this video in chrome. It worked for me in explorer. -When you go to this site, it will automatically begin playing the video for 1.1 (as shown to the right). Watch 1.1-1.4. On the right side down a little on the webpage, You will see the list of tutorials as shown to the right. Notation: -If you see ( )

3limx

f x→

, then that means you are approaching 3x = from BOTH sides of the graph of ( )f x

-If you see ( )3

limx

f x+→

, then that means you are approaching 3x = from the RIGHT side of the graph of ( )f x

--If you see ( )3

limx

f x−→

, then that means you are approaching 3x = from the LEFT side of the graph of ( )f x

1 2

*Note: you see here that the limit exists at -4 but there is no value defined at 2.

DNE (does not exist) *Note: the limit does not exist at x=2 because the limit from the left side of 2 is 2 and the limit from the right side of 2 is -1. The limit from both sides of an x-value must be the same for the limit to exist.

3

0

3 *Note: the limit at x=7 is 3 at the open circle, not the value ( )7 4f =

−∞ DNE (does not exist) *Note: the right handed limit at 10 is 0 and the left handed limit at 10 is −∞ .

-4 *recall: if the degree of the numerator and denominator are the same, then the limit at infinity is the ratio of their coefficients

4 *here you must be careful because we are evaluating our limit at −∞ and the leading terms have odd degrees. For

example ( )3−∞ = −∞ and

0 *recall: if the degree of the numerator is less than the degree of the denominator then the limit at infinity is 0. If the degree of the numerator is greater than the denominator, then the limit at infinity is +/- infinity

Assignment 1. Simplify and write with positive exponents.

(a) ( ) 32 23x x−− −− (b) 4 312x y− (c)

45

32x

− −

(d) ( )2

2 364x x−− (e) ( )3

2 4 532x xy−

(f)

32 4

3

81xx−

(g) ( )4

6 38x−

(h)

2 33 2

1 12 2

a b

b a

(i) ( )

1

1

ab a

−

−

2. Factor the following expressions.

(a) 4 81x − (b) 2 7 10x x+ + (c) 38 50x x−

(d) 3 24 32x x x− − (e) 25 13 6x x− − (f) 4 3 27 5 2x x x− −

(g) 3 29 15 6x x x− + (h) 2144 16x − (i) 32 98x x− 3. Simplify the following expressions.

(a) 1 1

x h x−

+ (b)

1 13 3x

x

−+ (c)

2

5

2

10x

x

(d)

25

5

aa

a

−

+ (e)

( )( ) ( )21 3 11

x x xx

− + − ++

(f) 2

2

5 64 4

x xx x− +− +

4. Solve each equation.

(a) 6 7 3 5 5 78

4 7 28x x x− − +

+ = (b) 3 26 27 0x x x− − = (c) 9 3 3 1x x+ − =

(d) 2

2 2

5 10 3 316 4 16

b b bb b b− −

+ =− − −

(e) 4 26 8 0x x− + = (f) 1 13 2 1x=

+

5. Solve the following equation for the indicated unknown quantity.

(a) 2 2 6 0;dy dy dyx x y ydx dx dx

+ + + =

(b) ' 1 '; 'xy y y y+ = + (c) ;A P nrP P= +

6. Use the given information to write the equation in point slope form then convert it to standard form.

(a) ( ) ( )3,6 , 1, 2− − − (b) 2 1, 6,3 3

m = −

(c) ( ) ( )7,1 , 3, 4− −

7. Write the equations of the line through the given point parallel and normal to the given line. (a) ( )5, 3 , 4x y− + = (b) ( )6, 2 , 5 2 7x y− + = (c) ( )3, 4 , 2y− − =

8. Find k if the lines 3 5 9x y− = and 2 11x ky+ = are parallel and perpendicular. 9. Evaluate each expression in (a)-(c) and solve the equation in (d)-(k).

(a) 2 3log log3 32+ (b)

4ln ln123− (c) ( )5

log 3

(d) ( )2ln 3 3 0x x− + = (e) 23 18x− = (f) 3 1 10xe + =

(g) ( )4 52 3 4 11x− + = (h) ( )ln 2 1 5xe x+ = (i) 2 2

4x

x

ee

+

=

(j) ( ) ( ) ( )2ln 2 3 ln 4 ln 2 11x x x− + + = + (k) ( ) ( ) ( )ln 3 ln 7 23 ln 1x x x− = − − +

10. Given ( ) 23 2f x x x= − , ( ) 2 1g x x= − , and ( ) 2 1h x x= − , find the following:

(a) ( )( )f g x (b) ( )( )h g x (c) ( )( )f h x

(d) f h⋅ (e) ( )( )g g x (f) ( )( )g f x

11. Given the table below, find the following: (a) ( )( )1g f (b) ( )( )1g g (c) ( )( )6f g

12. Find ( ) ( )f x h f x

h+ −

for the given function f .

(a) ( ) 9 3f x x= + (b) ( ) 5 2f x x= − (c) ( ) 1f xx

=

X 1 2 3 4 5 6 f(x) 3 1 4 2 2 5 g(x) 6 3 2 1 2 3

13. Evaluate the following trigonometric functions.

(a) cosπ (b) tan2π

(c) 3csc2π

(d) 5sin4π

(e) 1 3sin2

−

(f) ( )1tan 3− − (g) cot2π

(h) 1 3tan2

− − (i)

5cos6π

(j) ( )arcsin 1−

(k) ( )1tan 1− − (l) ( )1cos 1− − (m) 1 3cot sin2

−

(n) 5sin4π

(o) 1 4cos sin3π−

14. Solve each equation on the domain )0,2π . Be sure to check your answers to make sure they work.

(a) 22sin 1 sinθ θ= − (b) 22 tan sec 0θ θ− = (c) tan sec 2 tanx x x=

(d) 29 tan 3 0x − = (e) 2 sin 3 cos sin 32xx x⋅ = (f) 2 23sin cosx x=

15. Model the following scenarios with an equation. You must define all variables used. Then solve the equation if applicable.

(a) A seven foot ladder, leaning against a wall, touches the wall x feet above the ground. Write an expression (in terms of x) for the distance from the foot of the ladder to the base of the wall.

(b) Molly is designing a poster. The length of the poster is 6 inches longer than the width. If the poster requires 160 square inches of poster board, find the width w of the poster. (write an equation in terms of w; then use the equation to solve the problem).

(c) Find the surface area of a box of height h whose base dimensions are p and q, and that satisfies the following conditions:

-the box is closed -the box has an open top -the box has an open top and a square base with side length p.

(d) A piece of wire 5 inches long is to be cut into two pieces. One piece is x inches long and is to be bent into the shape of a square. The other piece is to be bent into the shape of a circle. Find an expression for the total area made up by the square and circle as a function of x. (e) A rectangular garden is 12 feet long and 10 feet wide. Part of the garden is torn up to install a sidewalk of uniform width around the garden. If the area of the new garden is 48 square feet, find the width of the sidewalk. (f) A kite is 100m above the ground. If there are 200m of string out, what is the angle between the string and the horizontal? (assume the string is perfectly straight). (g) The second angle in a triangle is 3 less than twice the first angle. The third angle measure is 8 more than twice the first angle. Write an equation using one variable to represent this problem. Then find the measure of each angle.

16. Sketch the graph of each function. On the graph, indicate on the x-intercept(s), y-intercept, and period (if applicable). Then state the domain, range, x-intercept, y-intercept, and period (for the trig functions).

(a) ( ) xxf = (b) ( ) 2xxf = (c) ( ) 3xxf = (d) ( )x

xf 1=

(e) ( ) xxf = (f) ( ) xxf 3= (g) ( ) xxf ln= (h) ( ) xxf sin= (i) ( ) xxf cos= (j) ( ) xxf tan= 17. Rewrite each function as a piecewise function then graph. Use a table of values centered at the zero of the input of the absolute value expression. Include 7 values on your table.

(a) ( ) 2f x x= + (b) ( ) 11

xf x

x−

=−

(c) ( ) 33x

f xx

−=

−

18. Graph the following piecewise functions.

(a) ( ) 2 13

xf x x

−= +

forforfor

2

200

>≤≤

<

xx

x (b) ( )

−≥+−<+

=2 if 12 if 32

xxxx

xf

(c) ( )

≥−

<+

−≤

=

1 if 271 if 231 if 1

xxxxx

xf

19. Find a piecewise function formula for the given graph. (a) (b) 20. Graph the function, state its domain and range, and rewrite the function as a piecewise function. (a) ( ) 2 6f x x= + (b) ( ) 3 5f x x= + −

−4 −3 −2 −1 1 2 3 4 5 6

−4

−3

−2

−1

1

2

3

4

−5 −4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

21. Evaluate each limit using the graph of ( )f x below.

(a) ( )

2lim

xf x

+→− (b) ( )

0limx

f x−→

(c) ( )0

limx

f x+→

(d) ( )0

limx

f x→

(e) ( )

2limx

f x+→

(f) ( )2

limx

f x−→

(g) ( )2

limx

f x→

(h) ( )4

limx

f x→

(i) ( )

5limx

f x+→

(j) ( )5

limx

f x−→

(k) ( )5

limx

f x→

(l) ( )7

limx

f x−→

22. Evaluate the limit.

(a) 2

2 5lim3 1x

xx→∞

++

(b) 7 8lim2 9x

xx→∞

+−

(c) 2

lim 3x

x→

+

(d) 2

2

5 6lim2x

x xx→

+ ++

(e) 2

2

8 4 1lim16 7 2x

x xx x→∞

− ++ −

(f) 5 4

2

1lim2 3x

x x xx x→−∞

− + −+ −

(g) 2

4

5 6lim2x

x xx→∞

+ ++

(f) 3

2 3

4 8 1lim16 7 2x

x xx x→∞

− − ++ −

(g) 3 22 7 4lim

4x

x x xx→∞

− −−