AP CALC SUMMER PACKET 13-14 - Abraham Clark High Schoolroselleachs.sharpschool.net/UserFiles/Servers/Server... · ACHS%AP%CALCULUS%SUMMERPACKET%%%%%MR.%NAEM%%201342014% Page1%!! Name:_____!

ACHS AP CALCULUS SUMMER PACKET MR. NAEM 2013-‐2014 Page 1

Name:______________________________________________________________

AP Calculus AB Summer Assignment

Due Date: The beginning of class on the last class day of the first week of school.

The purpose of this assignment is to have you practice the mathematical skills necessary to be successful in Calculus AB. All of the skills covered in this packet are skills from Algebra 2 and Pre-‐Calculus. If you need to, you may use reference materials to assist you and refresh your memory (old notes, textbooks, online resources, etc.). While the graphing calculators will be used in class, there are no calculators allowed on this packet. You should be able to do everything without a calculator.

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. This is why you have this packet. Spend some time with it and make sure you are clear on everything covered in the packet so that you will be successful in Calculus. Of course, you are always encouraged to seek help from your teacher if necessary.

This assignment will be collected and graded as your first test, the last class day of the first week of school. Be sure to show all appropriate work to support your answers. In addition, there may be a quiz on this material during the first quarter. All questions must be complete with the correct work. You must return in September knowing how to do all the material in this packet.

For assistant with the packet you may contact me at [email protected] . Emails may take a few days for a response. Please be specific in your email what you need assistance with, include the section and the question number as well. Sample online websites are giving at the end of the packet.

Good Luck!


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 )( 11 xxmyy −=− where m is the slope and )( 1,1 yx 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.

Example 1: Find the equations of (a) line parallel and (b) perpendicular to 5

31

+−

= xy that contains the point (-‐2,1)

Solution:

Part a (using slope from example above)

b+−−

= )2(311 Using the slope-‐intercept form with

31−

=m and point (-‐2,1)

b+=321 Multiply -‐1and -‐2

b=−321 Subtract

32 from both sides

b=−32

33 Get a common denominator of 3

b=31 Combine like terms

31

31

+−

= xy This is the equation of the line parallel to the

given line that contains (-‐2,1)

Part b (using slope from example above)

b+−= )2(31 Using the slope-‐intercept form with 3=m and point (-‐2,1)

b+−= 61 Multiply 3 and -‐2

b=−− 61 Subtract -‐6 from both sides

b=7 Subtract

73 += xy This is the equation of the line perpendicular to the given line that contains (-‐2,1)

Write an equation of a line through the point (a) parallel to the given line and (b) perpendicular to the given line:

1. Point: (1,5) line: 6x-‐2y=8

2. Point: (-‐2,2) line: 3x+5y=8

Find the slope and y-‐intercept of the line:

3. 2x-‐3y = 12

4. 8. 4x+y=1

5. y = 2

6. x = -‐51


Example 2:

Find the slope and y-‐intercept of 1556 =− yx

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

xy 6155 −=− Subtract 6x form both sides

5615

−−

=xy Divide by -‐5

356

−= xy Simplify

The slope is m=56 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.

)1(32 −−=−− xy Substitute into point slope form for )( 1,1 yx and

m

132 +−=+ xy Minus a negative makes + and distribute -‐3

13 −−= xy Subtract 2 from both sides

Find the equation of a line in slope intercept form:

7. Contains (1,2) and (-‐2,4)

8. Contains (2,1) and m=4

9. Contains (1,7) and m=0

10. Contains (6,5) and m is undefined


Example 4: Find the equation of the line that passes through (-‐1,3) and (4,5).

Solution: You will need to find slope using 12

12

xxyym

−

−=

52

1435=

−−−

=m

Choose one point to substitute back into either the point slope or slope-‐intercept form of a line.

b+= )4(525 Using the slope-‐intercept form with

52

=m

and point (4,5)

b+=585 Multiply 2 and 4

b=−585 Subtract

58 from both sides

b=−58

525 Get a common denominator of 5

b=517 Combine like terms

517

52

−= xy Equation of the line in slope intercepts form.

Write the following equations in point slope form

11. contains (-‐1,4) and (3,8)

12. Passes through (6,2) and had a y-‐intercept of 5


Example 5: Graph the following equation: xy =+ 33

Solution:

First you must get the equation in slope-‐intercept form:

33 −= xy

131

−= xy

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.

Sketch a graph of the equation:

13. y=-‐3x+ 2

14. 12. x=4 – y

15. y-‐1=3x+12

16. x = 7


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.

Example:

Find the intercepts for 4)3( 2 −+= xy

Solution:

X-‐intercept

4)3(0 2 −+= x Set y=0

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

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

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

x=− 5 or x=−1 Subtract 3 from both sides

Y-‐intercept

4)30( 2 −+=y Set x=0

432 −=y Add 0+3

49 −=y Square 3

5=y Add 4 to both sides

Find the intercepts for each of the following.

1. 3 2y x= − +

2. 3 2y x= +

3. 2

2

3(3 1)x xyx+

=+

4. 2 3 4y x x= −


III. System of Equations

Use substitution or elimination method to solve the system of equations.

Example:

x2 + y!16x +39 = 0x2 ! y2 ! 9 = 0

Elimination Method2x2 !16x +30 = 0x2 !8x +15= 0(x !3)(x ! 5) = 0x = 3 and x = 5Plug x=3 and x = 5 into one original32 ! y2 ! 9 = 0 52 ! y2 ! 9 = 0!y2 = 0 16 = y2

y = 0 y = ±4Points of Intersection (5, 4), (5,!4) and (3, 0)

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

1. x + y = 84x ! y = 7

2. x2 + y = 6x + y = 4

3. x2 ! 4y2 ! 20x ! 64y !172 = 016x2 + 4y2 ! 320x + 64y +1600 = 0

Substitution MethodSolve one equation for one variable.

y2 = !x2 +16x !39 (1st equation solved for y)x2 ! (!x2 +16x !39)! 9 = 0 Plug what y2 is equal

to into second equation.2x2 !16x +30 = 0 (The rest is the same asx2 !8x +15= 0 previous example)(x !3)(x ! 5) = 0x = 3 or x ! 5


IV. Functions

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

f ! g( )(x) = f (g(x)) OR f [g(x)] read “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)).

Example 1: 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))=g(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

Let f(x)=3x+2 and g(x)=1+x2 find each of the following:

1. f(g(0))=

2. g(g(2))=

3. g(f(x))=

4. f(g(x))=

If ! ! = !!,! ! = 2! − 1. !"# ℎ ! = 2! , find the following:

5. f(h(-‐1))=

For f(x) = 8x – 3, find

6. ! !!! !!(!)!

=

For f(x) = x2 , find

7. ! !!! !!(!)!

=


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 3)( −= xxf

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,∞).

Find the domain and range for each function give your answer using interval notation:

8. 2( ) 9h x x= −

9. h(x) = sin x

10. 2( )2 3

f xx

=+

11. 2 1, 0( )

2 2, 0x x

f xx x+ <⎧

= ⎨+ ≥⎩

12. ! = !!!!!!!!

13. ! = !!!!!!!!

14. ! = !!!!!!!!!!!!!!"


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 22 =− yx 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.

Test for symmetry with respect to each axis and the origin.

1. ! = ! ! + 2

2. ! = 6 − !)


Example 2:


Given equation: 04 2 =−− xxy

Solution:

x-‐axis (change all y to –y):

04)( 2 =−−− xyx

04 2 =−−− xxy 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)

0)(4 2 =−−−− xxy

04 2 =−−− xxy 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 )

0)(4)( 2 =−−−−− xyx

04 2 =−− xxy since this does look like the original it is symmetric to the origin.

Example 3:

The figure to the right shows the graph of 04 2 =−− xxy .

It is symmetric only to the origin.


3. ! = !!!!!

4. ! = !! − !


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 a hallmark of 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.

So I'll plug –x in for x, and simplify:

f(–x) = –3(–x)2 + 4 = –3(x2) + 4 = –3x2 + 4

My final expression is the same thing I'd started with, which means that f(x) is even.

Show work to determine if the relation is even, odd, or neither:

5. ! ! = 2!! − 7

6. ! ! = −4!! − 2!

7. ! ! = 4!! − 4! + 4


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.

Example 1: Find the holes in the following function

22)( 2 −−

−=

xxxxf

Solution: When x=2 is substituted into the function the denominator and numerator both are 0.

Factoring and canceling )2)(1(

2)(−+

−=

xxxxf

)1(1)(+

=x

xf but (x≠2) this restriction is from the original

function before canceling. The graph of the function f(x) will

look identical to 1( )( 1)

f xx

=+

except for the hole at x=2.

22)( 2 −−

−=

xxxxf 1( )

( 1)f x

x=

+

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

22)( 2 −−

−=

xxxxf

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.

For each function below list all holes, vertical asymptotes and x-‐intercepts

1. 1. )12)(3()2)(3()(

+−

+−=

xxxxxf

2. 2. 12

12

2

−+

−=

xxxy


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 x

xxf23)( +

=

Solution: 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. The graph is to the right.

Example 4: Find the holes, vertical asymptotes and x-‐intercepts for the given function:

xxxxxf633)( 2

2

+

−=

Solution: First we must factor to find all the zeroes for both the numerator and denominator:

)2(3)3()(

+

−=

xxxxxf

Numerator has zeroes x=0 and x=3

Denominator has zeroes x=0 and x=-‐2.

x=0 is a hole

x=-‐2 is a vertical asymptote

x=3 is a x-‐intercept

3. 3. 823212)( 2

23

−−

+−=

xxxxxxf

4. 4. 23149)( 2

2

++

+−=

xxxxxg


VII. Inverses

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

Example 1:

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

y = x +13 Switch x and y

x = y +13 Solve for your new y

x( )3 = y +13( )3 Cube both sides

x3 = y +1 Simplifyy = x3 !1 Solve for yf !1(x) = x3 !1 Rewrite in inverse notation

Also, recall that to PROVE one function is an inverse of another function, you need to show that:

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

Example 2:

If: f (x) = x ! 94

and g(x) = 4x + 9 show f(x) and g(x) are

inverses of each other.

f (g(x)) = 4 x ! 94

"

#$

%

&'+ 9 g( f (x)) =

4x + 9( )! 94

= x ! 9+ 9 =4x + 9! 9

4

= x =4x4

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

Find the inverse for each function.

1. f (x) = 2x +1

2. f (x) = x2

3

Prove f and g are inverses of each other.

3. f (x) = x3

2g(x) = 2x3

4. f (x) = 9 ! x2 , x " 0 g(x) = 9 ! x


VIII. Finding Solutions

A. Factoring or using the quadratic formula to solve equations.

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

C. Solve by finding the common denominator.

A. Solve each equation:

1. 7!! − 3! = 0

2. 4! ! − 2 − 5! ! − 1 = 2

3. !! + 6! + 4 = 0

4. 2!! − 3! + 3 = 0

5. 2!! − ! + 2 ! − 3 = 12

6. ! + !!= !"

!

B. Solve each inequality:

7. !! − 16 > 0

8. !! + 6! − 16 > 0


9. !! − 3! ≥ 10

10. 2!! + 4! ≤ 3

11. !! + 4!! − ! ≥ 4

12. 2 sin! ! ≥ sin !, 0 ≤ ! < 2!

C. Solve for x:

13. !!− !

!= !

!

14. ! + !!= 5

15. !!!!− !!!

!= 1


IX. Absolute Value and Piecewise Functions

A. In order to remove the absolute value sign from a function you must: 1. Find the zeroes of the expression inside of the

absolute value.

2. Make sign chart of the expression inside the absolute value.

3. 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.

Example 1: Rewrite the following equation without using absolute value.

42)( += xxf

Solution:

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

2x=-‐4 Subtract 4

x=-‐4/2 Divide by 2

x=-‐2 Simplify

⎩⎨⎧

+

−−=

4242

)(xx

xf ⎭⎬⎫

−≥

−<

22

xx

A. Write the following absolute value expressions as piecewise expressions (by remove the absolute value): 1. ! = 2! − 4

2. ! = 6 + 2! + 1

3. ! = 4! + 1 + 2! − 3

_-‐

Put in any value less than -‐2 into 2x+4 and you get a negative.

Put in any value more than -‐2 into 2x+4 and you get a positive. -‐2

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

+



352)( 2 −+= xxxf

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

2

2

2 5 3( )

2 5 3

x xf x

x x

⎧ + −⎪= ⎨

− − +⎪⎩

3 1/ 2

3 1/ 2

x and x

x

< − > ⎫⎪⎬

− ≤ ≤ ⎪⎭

B. Solve the following absolute value inequalities: 4. ! − 3 > 12

5. ! − 3 ≤ 4

6. 10! + 8 > 2

-‐3

_-‐ + +

1/2

Put in any value less than -‐3 into (2x-‐1)(x+3) and you get a positive number.

Put in any value more than -‐3 and less than ½ into (2x-‐1)(x+3) and you get a negative number.

Put in any value more than 1/2 into (2x-‐1)(x+3) and you get a positive number.

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



293)( +−= xxf

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

⎩⎨⎧

+−

++−=

293293

)(xx

xf ⎭⎬⎫

≥

<

33

xx

⎩⎨⎧

−

+−=

73113

)(xx

xf ⎭⎬⎫

≥

<

33

xx Simplify

B. 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.

7. 3! − 4 > −2

8. ! − 6 > −8

3

_-‐ +

Put in any value less than 3 into 3x-‐9 and you get a negative.

Put in any value more than 3 into 3x-‐9 and you get a positive.

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


X. Exponents

A fractional exponent means you are taking a root. For

example 2/1x is the same as x .

Example 1: Write without fractional exponent: 3/2xy =

Solution: 3 2xy = Notice that the root is the bottom number in the fraction and the power is the top number in the fraction.

Negative exponents mean that you need to take the

reciprocal. For example 2−x means 21 x and 32 −x means 32x .

Example 2: Write with positive exponents: 452

−= xy

Solution: 52 4xy =

Example 3: Write with positive exponents and without

fractional exponents: 2/1

2/12

)32()3()1()( −

−

−−+= xxxxf

Solution: 2)1(323)(

+−−= xxxxf

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

Write without fractional exponents:

1. 3/12xy =

2. 4/12 )16()( xxf =

3. 4/33/127 xy =

4. 9!! =

5. 64!! =

6. 8!! =

7. 27!! =

Write with positive exponents:

8. 32)( −= xxf

9. 22 )4( −= xy

10. ! = !!!!!

!!

11. ! ! = !!! !!

!!!! !!


Example 4: 12 3363 −− −+= xxxy

Solution: The lowest exponent for x is -‐2 so 23 −x can be

factored from each term. Leaving )1121(3 32 xxxy −+= − . Notice that for the exponent for the 6x term we take 1-‐ (-‐2) and get 3. For the 133 −x term we take -‐1-‐(-‐2) and get 1 as our new exponent.

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.

Example 5: Simplify 8

3)2()(xxxf =

Solution: First you must distribute the exponent.

8

38)(xxxf = . 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 5x

in the denominator. 5

8)(x

xf = .

Example 6: Simplify )1(12)( 2

2

−

+−=

xxxxf

Solution: First we must factor both the numerator and

denominator. )1)(1(

)1()(2

−+

−=

xxxxf . 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.

)1()1()(

+

−=xxxf .

Factor then simplify:

12. 23 1824)( −− −+= xxxxf

13. xxxxxf 3)2()2(5)( 2/12/12 −+−= −

14. )12(4)12(6)( 1 −−−= − xxxxf


Example 7: Factor and simplify 2/122/1 )3()3(4)( −−+−= xxxxxf

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 2/1)3( −−xx and obtain

])3(4[)3()( 2/.1 xxxxxf +−−= − . This will simplify to

]124[)3()( 2/1 xxxxxf +−−= − . Leaving a final solution of

3)125(

−

−

xxx

.

.

Simplify rational expressions:

15. xxxf2)4()(32

=

16. )12()3()3)(12(

4

2

−−

−+=

xxxxy

17. 1341816)( 2

2

−+

+−=

xxxxxf

18. 2510

252

2

+−

−=

xxxy


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.

Example 1:

!7! 6x +1

5x +1

= !7! 6

x +15x +1

i x +1x +1

= !7x !7!65

= !7x !135

Example 2:

!2x+

3xx ! 4

5! 1x ! 4

=

!2x+

3xx ! 4

5! 1x ! 4

i x(x ! 4)x(x ! 4)

= !2(x ! 4)+3x(x)5(x)(x ! 4)!1(x)

= !2x +8+3x2

5x2 ! 20x ! x

= 3x2 ! 2x +85x2 ! 21x

1. !!+ !

!=

2. !!− !

!=

3. !!∙ !!=

4. !!!!=

5. !!!!!!!

=

6. !!!!"!!!"!!!!!!!"

=

7. !!!

!!

!!!=


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)!" #$

(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 24 2

4 44 2

4 2

14 2

x x xx x x

xx x

xx x

x

+ − + − + += •

+ ++ −

=+ +

=+ +

=+ +

8. !!!!

− !!=

9. !!!!!!!!

=

10. !!!!!!!!

=

11. !

!!!!=

12. 9 3xx+ −

=

13. x h xh

+ −=


XII. Natural Logarithms Recall that ( )lny x= and xy e= (exponential function) are

inverse to each other

Properties of the Natural Log:

( ) ( ) ( )ln ln lnAB A B= +

Example 1: ( ) ( ) ( )ln 2 ln 5 ln 10+ =

( ) ( )ln ln lnA A BB

⎛ ⎞ = −⎜ ⎟⎝ ⎠

Example 2: ( ) ( ) ( )6ln 6 ln 2 ln ln 32⎛ ⎞− = =⎜ ⎟⎝ ⎠

( ) ( )ln lnpA p A=

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

( ) ( ) ( )33ln 2 ln 2 ln 8= =

( )ln xe x= , ( )ln 1e = , ( )ln 1 0= , 0 1e =

Example 4:

Use the properties of natural logs to solve for x:

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

2 5 11 75 117 25 11ln ln7 2

ln 5 ln 7 ln 11 ln 2



ln 11 ln 2ln 5 ln 7

x x

x

x

x

x

x x

x x

x

x

• = •

=

⎛ ⎞=⎜ ⎟

⎝ ⎠

− = −

− = −

− = −

−=

−

Express as a single logarithm:

1. 3 ln ! + 2 ln ! − 4 ln ! =

2. 3 ln ! = 1

3. !!!! = 7


2

- 2

- 5 5

f x( ) = sinx( )2

- 2

- 5 5

f x( ) = cosx( )

XIII. Graphing Trig Functions

y = sin x and y = cos x have a period of 2 and an amplitude of 1. Use the parent graphs above to help you sketch a graph of the functions below. For , A =

amplitude, = period,

= Phase Shift (positive C/B shift left, negative C/B shift

right) and K = vertical shift.

Graph two complete periods of the function.

1.

2.

3.

4.

!

f (x) = Asin(Bx + C) + K2!B

CB

f (x) = 5sin x

f (x) = sin2x

f (x) = ! cos x !"4

#$%

&'(

f (x) = cos x ! 3


2

-2

(-1,0)

(0,-1)

(0,1)

(1,0)

XIV. Trig. Equations and Special Values

You are expected to know the special values for trigonometric functions. Fill in the table to the right and study it.

Use

180!

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

Use ! 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.

Example 1: sin90! = 1 cos

!2= 0

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

Reciprocal identities

xx

csc1sin =

xxsec1cos =

xxcot1tan =

xxsin1csc =

xxcos1sec =

xx

tan1cot =

Tangent Identities

xxx

cossintan =

xxx

sincoscot =

Pythagorean Identities

1cossin 22 =+ xx xx 22 sec1tan =+ xx 22 csc1cot =+

Double angle Identities

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

1.

Degrees Radians)

Cos Sin Quadrant

0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360°

Find all solutions to the equations. You should not need a calculator. (Hint: one of these has NO solution.)

2. 1cos4cos4 2 −=− xx

3. 01sin3sin2 2 =++ xx


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 < 2" . Isolate the variable, sketch a reference triangle, find all the solutions within the given domain, 0 ! x < 2" . 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

21sin =x and 1sin −=x Get the trig function by itself

kx

kx

ππ

ππ

265

26

+=

+=and kx π

π 223

+= Solve for x (these

are special values)

4. 0sin2sin 2 =− xx

5. xx 2cos2sin3 =

6. 2sin3sin2 2 =+ xx

7. 2cos52cos =+ xx (hint: use double angle identity)

8. 1)sin(cos =x

9. 03sin2sin 2 =−− xx


XV. Inverse Trigonometric Functions

Inverse Trig Functions can be written in one of 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 >0

cos-‐1x < 0 cos-‐1 x >0

tan-‐1 x >0

sin-‐1 x <0

tan-‐1 x <0

Example 1:

Express the value of “y” in radians.

y = arctan !13

Solution:

Draw a reference triangle. 3

2 -‐1

This means the reference angle is 30° or !6. So, y = –

!6 so

that it falls in the interval from !"2< y < "

2

Answer: y = – !6

For each of the following, express the value for “y” in radians.

1. y = arcsin ! 3

2

2. y = arccos !1( )

3. y = arctan(!1)


Example 2:

Find the value without a calculator.

cos arctan 56

!"#

$%&

Solution:

Draw the reference triangle in the correct quadrant first. Find the missing side using Pythagorean Thm.

Find the ratio of the cosine of the reference triangle.

cos! = 661

For each of the following give the value without a calculator.

4. tan arccos

23

!

"#$

%&

5. sec sin!1 12

13"

#$%

&'

6. sin arctan12

5!

"#$

%&

7. sin sin!1 7

8"

#$%

&'

6

5


WEBSITES:

Using the TI-‐83/84 Graphing Calculator [highly recommended site] http://www.prenhall.com/divisions/esm/app/graphing/ti83/ Just Math Tutoring (click on Algebra/SAT videos on the left and scroll down to “Solving equations and inequalities”) http://www.justmathtutoring.com/ Purple Math http://www.purplemath.com/modules/compfrac.htm

Solving by the Substitution method: http://www.purplemath.com/modules/systlin4.htm Solving by the Elimination method: http://www.purplemath.com/modules/systlin5.htm LAWS OF EXPONENTS http://www.mathsisfun.com/algebra/exponent-‐laws.html http://www.algebralab.org/practice/practice.aspx SLOPE OF A LINE http://www.ck12.org/algebra/Slope-‐of-‐a-‐Line-‐Using-‐Two-‐Points/ WRITE THE EQUATION OF A LINE IN SLOPE-‐INTERCEPT FORM GIVEN TWO POINTS http://www.ck12.org/algebra/Standard-‐Form-‐of-‐Linear-‐Equations/ GRAPH THE EQUATION OF A LINE http://www.ck12.org/algebra/Graphs-‐Using-‐Slope-‐Intercept-‐Form/ OPERATIONS WITH FRACTIONS http://www.aaamath.com/fra.html http://www.ck12.org/arithmetic/Equivalent-‐Fractions/ http://www.ck12.org/arithmetic/Fractions-‐in-‐Simplest-‐Form/ http://www.ck12.org/arithmetic/Sums-‐of-‐Fractions-‐with-‐Like-‐Denominators/ http://www.ck12.org/arithmetic/Differences-‐of-‐Fractions-‐with-‐Different-‐Denominators/ http://www.ck12.org/arithmetic/Products-‐of-‐Two-‐Fractions/ http://www.ck12.org/arithmetic/Quotients-‐of-‐Fractions/ Unit Circle and Trig Equations http://tutorial.math.lamar.edu/Classes/CalcI/TrigEquations.aspx http://www.analyzemath.com/TrigEqExplore/TrigEqExplore.html

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