Oakton Community College
MAT 250    Calculus I
Section 002:   MW  10:30 - 12:50    Room 2129   DP Campus
Paul Boisvert     Professor of Mathematics 
Spring  Semester  2012
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Homework Assignments
Note:  Below, exponents are written with the wedge ^ notation.  So "x raised to the third power" would be "x^3". 
Fractions are written with slashes and parentheses:  A fraction with a + b in the top and c - d in the bottom is written as  ( a + b )  /  ( c - d ) .
Square Roots are written as "sqt(x)", and higher roots as fractional exponents, i.e.,  cube root of x would be  "x^(1/3)"
Date Page.  Problems  Read
Section:
Wed
1/18		 1067 Also, read and review Trig section in Chapter 1, i.e., section 1.4
1, 2, 15, 16
  
Even Answers:    2)  {x | x < 2 }          
16) -5 + 1/5 =  -24/25        		 Appendix A
Mon
1/23		 1067 17-23 ALL,
37-44 ALL,
Board Problem:  
A)  Simplify using fractional exponents:  sqrt(x) [ cube root of (x^5) ] / [ sixth root of (x^11) ]

Answers:  18)  2a + h        20)  -12 / (x + 3)(x - 3)         22)  x = 0 ,  x = 3 ,  x = -3  
38)  y-axis,   i.e. ,    x = 0        40) Horizontal ,  so:    y = 3
42)  2x - 5 = 0 is missing y, so it is vertical.  Parallel line to it is also vertical.   Vertical line through (-6 , 0)  is:   x = -6.
44)  Line through  (-9, 2) and (3, -5) has slope (-5 - 2) / (3 - -9)  =  -7/12.   Perp. slope would be  +12/7.  
    Note:  the rest of this problem is more involved than would be on a test--it asks you to find a bisector, rather than just telling you the point it wants you to use.  Other than this extra issue, the problem is simple and straightforward.  Let's proceed:
"Bisector" means it goes through the point halfway between our two points, which is the midpoint, which is  ( average x-value, average y-value).    To average numbers, add them and      divide by 2.    
So,   Average x  =  (-9 + 3) / 2  =  -3         Average y =  (2 + -5) / 2   =  -3/2
So our line goes through  (-3, -3/2) , the bisector,  with slope 12/7.  
Point-Slope Equation:    y + 3/2  =  (12/7) ( x + 3)   Stop, do not simplify.   

A)  x^ ( 3/6 + 10/6 - 11/6)  =  x^ (2/6)  =  REDUCE!  =   x^(1/3)   OR   cube root of x.  
Wed
1/25		 p. 8 13, 14, 16, 17, 18
21-29 odd,
35-39 odd (in these, skip finding the Domains)
Board Problems:  find the Domain for each of these functions
A)  y = (x + 2) / (x^2 + 5x - 14)           B)  y = 5 / (x^7 - 24x^5)        
C)  y = sqrt( 5 - 8x)                             D)  y = cube root of (2x + 1)

Answers:     14)  Note, since this is a function of y, for this function, "y" plays the role of x.  So I'm going to state Domain and Range without using "x" or "y":              Domain = All Real Number EXCEPT  3 and -2.    
From the graph (on calculator, and re-stating it as  y = g(x) )  , Range looks like (-oo, oo), as it appears that all "y"-values (on the calculator) are hit on various parts of the graph.
16)  Domain:  w <  2 , or (-oo, 2]    Range looks like [0, oo)            
18)  Domain: x >  -5  ,  or  [-5, oo)     Range looks like lowest value is hit at x = 0, subst. into g(x) and get y = -4sqrt(5),  all values above that to oo look like they are also hit, so
       Range  =  [-4sqrt(5) , oo )

Board Problems:
A)  All x except x = -7, x = 2     B)  All x except  x = 0, x = 4sqrt(6), x = -4sqrt(6)          C)  (-oo , 5/8 ]    
D)  All Real Numbers (since odd roots have no problems...)		 1.1
Mon
1/27		 p. 19 11, 17, 19, 20

Answers:  20)  All 3 parts are lines   First line, 2x + 2, has a ghost point at (0, 2) and a real point at (-1, 0), then continues to the left...
                  2nd line, x + 2, has a real point at (0,2) , which is inside the ghost point in the first line above, and another real point at (2, 4).  
It goes only in between those points, does NOT extend to right or left beyond them.     3rd line, 3 - x/2 , has a ghost point at (2, 2) and a real point at (3, 1.5) or (4, 1) , and continues to the right. 		 1.2

p. 32 17, 18, 33-36 ALL
Board Problems:     A)  Find f-inverse (x)  if  f(x) = 2(x - 5)^7 + 3
B)  Find f-inverse (x)  if  f(x) = 5 - [ cube root of (2x + 1)]          
C)  Find f-inverse (x)  if  f(x) = (4x - 11 ) / (3x - 10)

Answers:  18)  f-inverse of x  =  (x/3) ^ (1/3)       34)  x = 5^(-1)  =  1/5        36)  b^3  =  125 ,  so  b = 3        

Answers to Board Problems:  A)  f-inverse of x  =  [ cube root of ( [x - 3] / 2 ) ]  + 5      
B)  f-inverse of x  = [ (5 - y)^3  -  1 ]  /  2
C)  Solve for y in this equation:     x = (4y - 11) / (3y - 10)  -->   multiply both sides by LCD of  (3y -10), cancel and simplify,
getting:  3xy - 10 x  =  4y - 11.   Then group all y-terms on one side, all non-y terms on the other side, and factor out y from its side,  
getting y(    )  =   [   ] .  Then divide by the group factor (   ) on both sides to get final answer:   f-inverse of x  =  (10x - 11) / (3x - 4)
Note, if all signs are changed in a fractional answer, it is always still ok--that is, the answer of (-10x +11) / -3x + 4) would be ok. 		 1.3
Wed
2/1		 p. 44 15-22 ALL
ALSO:  on Handout Sheet given in class, do:  Problems I and II (on ln and e^x equations) , AND, for TRIG:  0, 1, 2, 3, 5, 10, 11, 13
ANSWERS on Back of Handout Sheet for its problems.

BOOK EVEN ANSWERS:
 16)  alpha for 2pi/3 in II  is pi/3, so sin(2pi/3)  =  +sin(pi/3)  =  sqrt(3) / 2    
18)  15pi/4 has coterminal angle of 7pi/4 or -pi/4  in IV, where tan is negative.   So alpha is +pi/4     Answer is  - tan(pi/4),   or -1.  
20)  7pi/6 in III where sec is negative,  has alpha = pi/6, so answer is  -sec(pi/6) =  -1/cos(pi/6)   =  -1/ [ sqrt(3) / 2 ]   =  -2/sqrt(3)
22)  16pi/3 has coterminal 4pi/3 in III, where sin is negative.  alpha is pi/3,  so answer is  -sin(pi/3)  =  -sqrt(3) / 2 .		 1.4
Mon
2/6		   REVIEW:  Before coming to class (i.e., over the weekend) REVIEW all Homework problems.  
Any HW questions you have had trouble with you should be ready to ask me for help with during the review session.  
Bring your Book to class.  
ALSO, there will be an IN-CLASS Review Assignment, which will count as a regular HW assignment, and be handed in with your other HW after the Test.
  
Wed
2/8		   TEST 1    Covers all material on which HW was asssigned above.  Covers up through the part of Section 1.4 that we did in class on Wednesday 2/1, and the HW assigned for it, including the handout sheet.
Closed Book, No Notes.  Make sure you have your special trig function values memorized (as well as everything else you need to know.)
BRING SOME PAPER of your own.   This can be paper you will rip out of a notebook (as long as it has nothing written on it.)  Or any other kind of blank paper you wish.  Bring several sheets.
BRING your OWN Calculator.  NO SHARING of calculators during the test 		  
Mon
2/13		   All HW and Review Assigned before Test 1 is now Due  
  p. 44  49-53 odd,  66-69 ALL,
74, 75, 77 --For these graphs, show New Period, 4 spaces marked off, Two Full periods (one could be negative if you like), and Tail Ends
                     Also relabel y-axis to show vertical shift, if there is one.
  AND:  On old Handout sheet, do #8 and 9.  --See Answers on Handout sheet for those.

Answers:  
66)   Theta is in III, draw picture of 3, 4, 5, right triangle,  get cos = -3/5 , sec = -5/3 , csc = -5/4, tan = 4/3 , cot = 3/4 .     
68)  Same picture as for 66), but now theta is in IV.    cos = 3/5 , sin = -4/5 , csc = -5/4, tan = -4/3 , cot = -3/4
74)  Amplitude from 3 to -3.  New Period 2pi/2 = pi.   Marks are at pi/4, pi/2, 3pi/4, and pi.  Function starts at (0,0) and goes upwards from there.  Points alternate top, 0, bottom, 0, top, 0, etc...   Second period goes out to +2pi (or backwards to -pi).  Tail ends go up on right side, down on left.  No vertical shift or relabeling.		  1.4
Wed
2/15		 p. 44  15-22 ALL, 29-31 ALL, 
Also, on old Handout sheet, do 6, 7, 15  --NOTE, answer on Handout sheet for 15 is partially wrong, the two points with x = -5.73 do NOT exist after all, since the square root has a negative number inside for the y-values in those points.  (My y-values for those points are in error, there are no real y-values at all.)

Answers:
16)  alpha for 2pi/3 in II  is pi/3, so sin(2pi/3)  =  +sin(pi/3)  =  sqrt(3) / 2    
18)  15pi/4 has coterminal angle of 7pi/4 or -pi/4  in IV, where tan is negative.   So alpha is +pi/4     Answer is  - tan(pi/4),   or -1.  
20)  7pi/6 in III where sec is negative,  has alpha = pi/6, so answer is  -sec(pi/6) =  -1/cos(pi/6)   =  -1/ [ sqrt(3) / 2 ]   =  -2/sqrt(3)
22)  16pi/3 has coterminal 4pi/3 in III, where sin is negative.  alpha is pi/3,  so answer is  -sin(pi/3)  =  -sqrt(3) / 2 .

30)  Factor out theta, set it = 0 , and then set other factor = 0, get:     cos (theta) = -1/2.  
Solution to the first equation is simply theta = 0.  
Solution to the second equation is theta = 120 degrees in II, and the angle with the same ref. angle in III, namely 240 deg.
I'll use radians, though, so solutions are { 2pi/3 ,  4pi/3  ,   +  2kpi  }  
Total set of solutions includes the 0 as well:  { 2pi/3 ,  4pi/3  ,   +  2kpi  }  U  {0}

  		  
Mon
2/20		   College Closed   
Wed
2/22 		 p. 73 27, 29, 37-47 odd 2.1, 2.3 
  p. 82  19, 20, 22    Infinite Limits   Notice that x here approches a finite "c" value, x does NOT approach infinity.  It is only the Limit L that may be +oo or - oo.  Therefore, this does NOT use the Ratio of the Highest Powers (RHP) approach.  That is used in the next section below.

For these infinite limits, make a table showing an x-value slightly to the left and right of the limit , and decide if the function goes to plus or minus oo.   If both sides agree, use that answer as the limit.  If they disagree on the sign, say "No Limit".

Answers:        20)   40         22)   negative infinity (on both sides)		  2.4
p. 92 15-26 ALL   Here, use RHP, since the limits are as x approaches infinity.  Do NOT substitute oo in until you have formed RHP and simplified!

Answers:
16)  negative infinity       18)  +0, approaches 0 from above         20)  0 + (-oo)  =  negative infinity  
22)    Limit as x approaches either oo or -oo  is 1/2.  So Horiz. Asymp. in each direction is  y = 1/2
24)     Limit as x approaches either oo or -oo  is 4.  So Horiz. Asymp. in each direction is  y = 4
26)    Limit as x approaches either oo or -oo  is  negative infinity.   So NO  Horiz. Asymp.'s		  
Mon
2/27 		 p. 73    33, 31, 32,  44, 48    (44 was not originally assigned, but has now been assigned.  Use LCD...)
  Note, 48 is OPTIONAL, no need to do it if you don't want to...answer below.

Answers:  32)  a)  0      b) 0       c)  0      d)  0       e)  15       f)  No Limit
44)  -1/25       48)  Multiply by conjugate of top, top will simplify to x^2, cancel with bottom x^2, substitute x = 0 and you will get:   Answer:  1/(2a)		  
  p. 82    9, 11, 17, 18, 21, 27, 29    For these infinite limits, make a table showing an x-value slightly to the left and right of the limit , and decide if the function goes to plus or minus oo.   If both sides agree, use that answer as the limit.  If they disagree on the sign, say "No Limit".

Answers:  18) a)   positive infinity    b)  negative infinity      c)  LIMIT does not exist (since one-sided limits do NOT agree!)          		  
  p. 92  9, 10, 12,   [not originally assigned in class, but has now been assigned.  11, 13, 14 are NOT assigned after all.]
31-36 ALL  -- here, when it talks about "End behavior", it means find the Limit as x --> oo, and the Limit as x --> -oo.  If the functions are not polynomials, that is, if the involve exponentials, ln's, or trig, you may use your graphing calculator to get an idea of what it happening. 

39-47 odd, part a) ONLY for each one   NOTE:  [47 is OPTIONAL, only look at if going to Calc. 2--hint, multiply by conjugate top and bottom, and only do it for Lim as x --> +oo, skip the -oo limit.  Therefore, you can drop the absolute value signs.  (The limit at -oo would be handled by using, for negative numbers, -x for |x| , and -x + 1 for |x - 1|, but you can skip that part.)  ]  

Answers: 10)   5      12)  4        14)  5         
32)   As x --> oo ,   Limit is oo.    As  x --> -oo ,  Limit is +0 (from above.)  
         So there is a Horiz. Asymp. of  y = 0 (or the x-axis)  as  x -->  -oo
34)   Limit as x approaches either oo or -oo  is  positive infinity (because of the absolute value signs.)  So NO Horiz. Asymp.
36)   Limit as x approaches oo is  +0 .      Limit as x --> -oo  is  positive infinity.    
        So there is a Horiz. Asymp. of  y = 0 (from above) as x --> oo
  
Wed
2/29		 p. 132  11, 12, 13, 18   Use the Definition (mine only, which is one of two in the bood--do not use the other one in the book .)
That is, use     Lim      f(x + h) - f(x)     You may check by using the shortcut rules, but must show all work for the long definition process.
                    h --> 0           h

Answers:  12) deriv = ... =  Lim as h--> 0  (-6xh - 3h^2 - 5h) / h   =  Lim as h--> 0  (-6x - 3h - 5)  =  -6x - 3(0) - 5  =  -6x - 5
                    At x = 1, slope =  f '(1)  =  -6(1) - 5  =  -11  ,  and   Point slope form:    y + 7  =  -11(x - 1)             

18)  deriv = ... =  Lim as h--> 0  (6xh + 3h^2 - 4h) / h   =  Lim as h--> 0  (6x + 3h - 4)  =  6x + 3(0) - 4  =  6x - 4
       At x = 1, slope =  f '(1)  =  6(1) - 4  =  2  ,  and   Point slope form:   y + 1  =  2(x - 1)  		 3.1 
  p. 143  7-24 ALL, 35, 36, 37  ,  25-33 odd,  42-46 ALL

Some answers are below, others coming soon...
Answers:    42)  dy/dx = 9x^2 + 10x  + 6               d2y/dx2  =  18x + 10               d3y/dx3  =  18  
44)  dy/dx =  6x + 5e^x              d2y/dx2  =  6 + 5e^x              d3y/dx3  =  5e^x  
46)  dy/dx =  10e^x                 d2y/dx2  =  10e^x                  d3y/dx3  =  10e^x soon...		  3.2
Mon
3/5		   REVIEW:  Before coming to class (i.e., over the weekend) REVIEW all Homework problems.  
Any HW questions you have had trouble with you should be ready to ask me for help with during the review session.  
Bring your Book to class.  ALSO, there will be an IN-CLASS Review Assignment, which will count as a regular HW assignment, and be handed in with your other HW after the Test.		  
Wed
3/7		    TEST 2    Covers all material on which HW was asssigned above, since the first test.  Covers up through Section 3.2
Closed Book, No Notes.  
BRING SOME PAPER of your own.   This can be paper you will rip out of a notebook (as long as it has nothing written on it.)  Or any other kind of blank paper you wish.  Bring several sheets.
BRING your OWN Calculator.  NO SHARING of calculators during the test 		  
     SPRING BREAK   MAR  12 -18   College Closed  
Mon
3/19		   All HW (and Review) since Test 1 is Due   
  p. 152 9, 11, 12, 17, 19, 20, 21, 22,
15 a) only    23) a) , 25 a) ,  26a) ,  27a)
31, 32, 33, 35, 37-42 ALL ,
43 a) b) only ,  45 a) b)
47  

Answers:  12)  after factoring answer, you get  4e^t [ 1 / ( 2sqrt(t) )   +  sqrt(t) ]      
20)  4w/ (w^2 + 1)^2   
22) Move (  )^-1  down to denom, use Quotient Rule, get...  [ (4x + 1)( 1/sqrt(x) )  -  4( 2sqrt(x) - 1) ] /  (4x + 1)^2
26a)  After Q.R., top simplifies to  x^2 - 2ax + a^2 , and bottom is (x - a)^2 .  But these are the same thing (factored or expanded), so quotient = 1.
32)  -12 / p^4               38)   Q. R., factor e^x out of top and cancel with one power of e^x in the bottom, get:  (1 - x) / e^x
40)  Simplify, factor, get:  (2x - 3) e^[-x]                42)     187.5 e^0.075t		  3.2
  p. 161 15-21 odd, 27-31 odd, 33-36 ALL, 51, 53, 55

Answers:
34) dy/dx  =  sec^2 (x),   d2y/dx2  =  2 [ sec(x)] [ sec x  tan x ]  , note we are using the chain rule here on the inside function of sec x.       
       Then simplify answer to  2 sec^2 (x) tan(x)                        
36)  I am going to use the variable "t" here instead of "theta".
      dy/dt  =  Prod. Rule  =  cos t (cos t) + sin t ( -sin t)  =  cos^2 (t)  -  sin^2 (t) .  But this should be simplified via the double-angle formula to:
      dy/dt = cos (2t)    Now take d2y/dt2  easily, though still using the chain rule on the little (2t) inside:
      d2ydt2  =  -sin (2t) (2)  =  -2 sin (2t)		  3.4
Wed
3/21 		 p. 180 7-27 odd, 29, 33, 35, 37-43 odd, 49, 51
55 --SKIP graph  
Not all the above were assigned on Wed, some were assigned in the following class Mon 3/26.  But all must be done eventually 		  3.6
Mon
3/26		 p  180 See previous class above (Wed 3/21)  for total assignment for both days.  All problems assigned either day are listed there.   
Wed
3/28 		 p. 171 11, 13, 15  In these, do c) and d) only
17 , 37
19 b) only
 43  b) c) only   This is Simple Harmonic Motion (SHM)		  3.5
Mon
4/2		 p. 189 5-19 odd   21, 23, 25 3.7 
Wed
4/4 		 p. 189  27-31 odd,  Find implicit 2nd derivative.  
41  1st derivative only (then evaluate to find slope)		  
  p. 199  9-16 ALL    These are  ln derivatives...  HINT:  whenever possible, it is best to SIMPLIFY the ln function BEFORE taking Derivatives.  
                    Use precalculus  laws of logs to simplify when possible:    ln (AB/C)  =  ln A + ln B - ln C    ,   and  ln (B^P)  =  P ln (B)
                    Then take derivative, which should now be much easier.

Answers:  10)  ln (2x^8)  =  simplified  =  ln 2  +  8 ln x    THEN, Derivative =  8/x        
                 10)  Alternate method:   Deriv.  =  1 / (2x^8) [ 16x^7] , using chaing rule.  Simplify to  same answer,  8/x.

12)  Prod. rule:    (e^x) / x  +  ln(x)  e^x   =   Factor out e^x, and get:     e^x [1/x  +  ln x ]

14)  I will ignore the absolute value signs here, and only do it for x > 1, where x^2 - 1 is already positive.
       Deriv. of  ln (x^2 - 1)  =    [ 1 / (x^2 - 1) ]  (2x)  ,  using the chain rule.   Simplify to   2x / (x^2 - 1)  

16)  ln [cos^2 (x) ]  =  Simplify using  ln (B^P)  =  P ln (B)   So this simplifies to, pulling out the "2" power,   2 ln [ cos x].
       Now, the Deriv. of   2 ln [ cos x]  is:   2[ 1 / cos x ]  [-sin x]  using the chain rule.   This simplifies to   -2 tan x      		 3.8 
Mon
4/9		   REVIEW:  Before coming to class (i.e., over the weekend) REVIEW all Homework problems.  
Any HW questions you have had trouble with you should be ready to ask me for help with during the review session.  
Bring your Book to class.  ALSO, there will be an IN-CLASS Review Assignment		  
Wed
4/11		    TEST 3    Covers all material on which HW was asssigned above, since the first test. 
Covers up through part of Section 3.8 -- derivatives of  ln (  )  functions.  
Closed Book, No Notes.  
BRING SOME PAPER of your own.   This can be paper you will rip out of a notebook (as long as it has nothing written on it.)  Or any other kind of blank paper you wish.  Bring several sheets.
BRING your OWN Calculator.  NO SHARING of calculators during the test 		   
Mon
4/16 		 p. 199  17-21 ALL  ,  27-31 ALL  , 37-41 ALL [SKIP 39]  ,   53, 55
67-73 odd, 72		 3.8 
  p. 209 7-25 odd  3.9
Wed
4/18		   Handout Sheet on Related Rates  Click on link  
  p. 230  31, 33, 38, 42

Answers:  38)  Critical point at x = 2.  Also check endpoints of 0 and 5.   Get  (0,0) ,   (2, 0.736)  , and  (5, 0.41)    Abs. Min at (0,0) , Abs Max at (2, 0.736)

42)  dy/dx = 3x^2 - 4x -5 = 0  for Crit. pts.   Does not factor, use Q.F.,  get ...  x = 2.1,  and x  = -.8   for critical points, possible local max or min's.
BUT neither of those is in the interval [4 , 8], so throw them out.  Evaluate ONLY the endpoints now, and get (4,18) (Abs. Min)  and (8, 350) (Abs. Max)		 4.1 
Mon
4/23		   Handout HW sheet on Graphs ( Solutions also given out on Wed 4/25 in class )  
  p. 301 Integration   Don't forget the + C  !
11-18 ALL
19-25 ALL
27 - 31 ALL
33, 34
Answers:     12)  x^11 + C       14)   - sin( 4x) + C          16)    -cot(s) + C         18)   1/z^6  + C
20)  -3/u - (4/3)u^3  + u  + C          22)   -5/t + (4/3) t^3  + C         
24) Distribute and simplify function BEFORE integrating:  Integrate [ 60m^4  - 50 m^2 ] dm  and get:    12m^5  - (50/3) m^3  + C
28)   Notice that  for sin(t/4) = sin ( [1/4] t ) ,  the constant "k" inside is "1/4".  So when you divide by that 1/4, you will invert and multiply it to "4/1" at the end when you simplify.    Answer:   -(1/4) cos (4t) + 4 cos (t/4)  + C
30)  tan (2v)  + C    
34)   Write 2 sqrt (t) as  2 t^(1/2), use power rule...   Answer:  (1/2)e^(2t)  + (4/3) t^(3/2)   + C		 4.8
Wed
4/25		  p. 301 39-47 odd       Initial Condition Problems,  use initial condition to evaluate the " + C "  4.8 
  p. 347  31-39 odd    Definite Integrals:  Integrate, then evaluate (top value result) - (bottom value result)    
                    Note, thes are not presented as "Areas", even though they could be thought of that way, and are done by the same method.		 5.3
  p. 347 45, 41, 42, 47  Areas  Integrate f(x) from lower limit to upper limit of the area.  Same process as for 31-39 odd
Board Problem:  Find the area under the curve  y = sin(7x) + e^(3x)  from x = 0 to x = 1

Answers:  Board Problem:  After evaluating, you get:  -(1/7) cos(7) + (e^3) / 3  - [ -1/7 + 1/3 ]  =  make sure you use radians  =
                 =  -0.1077  +  6.6952 + 0.1429 - 0.3333     =  6.3971 or  6.4  square units of Area

42)  Need to find limits of integration by finding out where curve y = 4 - x^2  hits the x-axis, i.e., where it = 0.  
        So solve 0 = 4 - x^2, and get  ...  x = +2 and x = -2.  
Now, integrate  4 - x^2  from x = -2 to x = +2, and get ...  4(2) - (2^3)/3 - [ 4(-2) - (-2^3)/3 ]  =  8 - 8/3 + 8 - 8/3  =  32/3		 5.3
Mon
4/30		    REVIEW for Test 4:  Before coming to class, make sure you have reviewed all HW problems SINCE Test 3.  
Bring all HW questions you have had trouble with to class for help.  Bring your Book to class.  

ALSO, there will be an IN-CLASS Review Assignment, which will count as a regular HW assignment, and be handed in with your other HW after the Test.		  
Wed
5/2 		    TEST 4    Covers all material on which HW was asssigned above.  Covers up through Sections 4.8 and 5.3
   Closed Book, No Notes.  Make sure you have your special trig function values memorized (as well as everything else you need to know.)
    BRING SOME PAPER of your own.  BRING your OWN Calculator.  NO SHARING of calculators during the test

IMPORTANT NOTE:   If you miss Test 4, you CANNOT take a makeup in the normal way.  You have only two options:
      1.  Take a zero for your grade, and make that the one that you raise to the (Final Exam minus 10%).
           OR:
      2.  Take an Incomplete in the class, and make up Test 4 very soon after Finals week.   YOU MUST STILL TAKE THE FINAL ON TIME, though!
            If you choose this second option, there will still be the usual late penalty on Test 4.  Once you take it, I will grade it and then change your Incomplete to the new grade.
     
So, if you miss Test 4, talk to me about these options very soon afterwards!  But, I urge you NOT TO MISS TEST 4 !		  
Mon
5/7 		   Return Test 4    

REVIEW for Final Exam  [you should already have reviewed all old HW and old Tests and old Reviews before you come into class this day.]
Bring your BOOK to class for the review.  Bring a GRAPHING CALCULATOR, too		  
Wed
5/9 		    Last HW Assignments Due.

FINAL EXAM:  Bring Paper, 2 Pencils or Pens, and a GRAPHING Calculator of your Own!

YOU MAY USE ONE 8 1/2 BY 11 NOTESHEET FOR THIS TEST.  (OR SMALLER.)  
BOTH SIDES OF THE PAPER IS FINE.    No other notes or books allowed

Starts on Time.  If you are late, you get no extra time. 
Plan to get here 1 hour early, just in case of unforeseen events!

If you miss the Final Exam for any reason other than a serious emergency, you will get a ZERO for it, which will destroy your grade.  

If you have an emergency, call me IMMEDIATELY at 847-635-1935 and leave a detailed message with the nature of the emergency, YOUR PHONE NUMBER, and when you can make up the test.   Also email me the same information.   Emergency situations will still need to be made up soon, with a late penalty.
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Final Grades will be available online through MyOakton shortly after they are due, which is TUE, May 15.  You can check starting WED, May 16.   I am not allowed to give out grades by email or phone, which are not secure.  If you wish to see your final exam, email me after Wed May 16 and set up a time to come in and look at it sometime over the summer.  		  
     
   
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