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PENN STATE / Math / MATH 140 / penn state math 41

penn state math 41

penn state math 41

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School: Pennsylvania State University
Department: Math
Course: Calculus With Analytic Geometry
Professor: Albena zikatanova
Term: Fall 2016
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Cost: 25
Description: Trigonometry Review The Sine/Cosine Chart The following chart provides the values of sin and cos for every major angle in the first quadrant ofthe unit circle
Uploaded: 07/20/2017
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For what x-values does y = ---- have honzontal tangent lines?




What is~ -'~derivative"?




For what value of A is the function continuous for all x > O?



Trigonometry Review  The Sine/Cosine Chart  The following chart provides the values of sin and cos for every major angle in the first quadrant of the unit circle.  0 n n n n - - - - 6 4 3 2 sin0 {o :o I I - .; - J~ ,fi J.., I a.... 0- -'.:)_ -- -· d-- ~ cos0 ~, F/ Jd-- /,' ro - - 'd,- d" ~ c} d- (ASTC) All StudentsWe also discuss several other topics like acctg 211 penn state
We also discuss several other topics like acctg 211 penn state
If you want to learn more check out acctg 101
Don't forget about the age old question of entp sales
Don't forget about the age old question of penn state math
We also discuss several other topics like albena zikatanova penn state
Take Calculus  This acronym will help you remember where each trig function is positive and  where it is negative.  s ·e AL,L  C~s  Identities  Every trig function can be expressed in terms of sin and cos by using the identities below:  sine  tane=--cose  1 cose cote=--=-- tane sine  1  sece=--cose  csce=-.-1  Trigonometry Review  The Sine/Cosine Chart  The following chart provides the values of sin and cos for every major angle in the  first quadrant of the unit circle.  0 1C 1C 1C 1C  6 4 3 2  () - ;: ., .., I - - -~  sin8 lo r, £ R J;?  d- ~  cos0 Vii7 JJ7 ~ J,1 f-;1  ~ d" 3-, a- d- (ASTC) All Students Take Calculus  This acronym will help you remember where each trig function is positive and  where it is negative.  S,n-e A~L  Identities  Every trig function can be expressed in terms of sin and cos by using the identities  below:  tanB= sin8  cos8  • l cos8 cot8=--=-- tan8 sin8  l sec8=--cos8  l csc8=--sin8  2 (-1, 0)  The Unit Circle  y  (0, 1)  goo  0 €1 0  ~ '?)~o  (1, 0)  X  3 The Limit of a Function  Limits:  • The limit is they-value that a function approaches as the x's get closer  and closer to a given number.  • The y-value does not necessarily have to exist in order for the limit to  exist.  • Limits may be one-sided. One-sided limits are denoted as follows:  o From the right: lim/(x) ...w...- x___,1+ -, •  o From the left:  lim/(x) x----,J - ~  • If the limit is not one-sided, then you must evaluate the limit coming from  both sides. In order for the overall limit to exist at a point, it must exist  coming from the left and the right, and those two limits must be equal.  Problem 1  Given the graph off (x), compute the following ...  y  6 (6,5) ~  . \ (6, 2)  -4 -2 6 8  -2  -4  -6  e. limf(x)=  x----,6- f. limf(x)=  X---¼6+  c. limf(x)= - J-_ , x, 1+ g. limf(x) = x----,6  d. limf(x) = b N F x---¼1 ,.... h. f(6)= l  s- ;- j  4  ---- --Calculating Limits Algebraically  Evaluating Limits Algebraically  Always start by plugging in the given x value. If you get an answer, stop, you're done.  Most likely, it will not be that easy on the test. Most algebraic limits can be  broken down into three cases as seen below.  We will looks at 3 different cases involving limits:  • Case 1: You plug in and get 0/0  • Case 2: You plug in and get #/0  • Case 3: Limits with Absolute Value  easel: Youpluginandget0/0 .:-> Ottlfa FvBJrrr'-'T'lPIV This tell you two things:  1. You need to simplif.y_  • Factor  • Multiply by the conjugate  • Common denominator  2. Visually, the graph has a removable discontinuitJ!_ (hole) at that x-value.  Problem 2  E l h l. . 1. x 2 - 3x + 2 va uate t e 1m1t: 1m 2 •  x~ 2 X -x-2  a) 0  b) 3  c) 1/3  d) 1/2  e) 2/3  ~ fl v f1 I /1/  (yt-)(><1 ~  liJ  s Problem3 ~-3  Evaluate the limit: !~ x + 4 . •  a) 0  b) -4/3  c) -9/5  d) -8/9  e) 16/5  ,V Problem 4 (x) I _ _le_ (1)  / , . (F} 3 x 1, )  Evaluate the limit: l;~ x2-8x+~l5  1/5 -'- :~ ~/9 2_,,.,, 3 X x->)fyr) c) -1/18 ><--> 1 d) -1  e) ONE  6 Case 2: You plug in and get #/0 ~v,Jf  This tell you that:  • The answer can be oo, -oo, or DNE. ,(  • Visually, the graph has an infinite discontinuity (vertical asymptote) at  that x-value.  • These are typically one-sided limits. To solve, plug in a number, on the  correct side of the limit, close to the x-value that the limit is approaching.  Determine whether the answer will be positive or negative.  • If it is not a one-side limit, you must compute the limit coming from both  sides  o If you get the same thing, then that is the answer  o If you get two different answers, then the limit DNE  Problem 5  Evaluate the limit: lim - 3-. + _:y x-+5+ 5- X  a) 0  b) 5  c) oo  d) -00  e) Does not exist  7 r.?  Problem6  Evaluate the limit: lim x2 -9  .r--+r (x-3)2 .  a) 0  b) 1  c) 00  d) -00  e) Does not exist  ~(x+3)  ¥3}{x-3) ..,,  Z.1 f  -- +  Evaluate the limit: lim £ x-+6 6-x.  a) 0  b) 6  c) 00  d) -00  e) Does not exist  ,R,·r1 + -- ~ - ~ o<J  ?.  ;c  x-+r, - r,-x~ +  r.~  f. t'1'1 x,. +  X ~ (;"" - ~ - •-x - ~-~ ,..  6./ (Ti7!J  8 Case 3: Limits with Absolute Value  First, determine if what's inside the absolute value signs will be positive or  negative:  • If what's inside is positive, remove the absolute value signs and continue.  • If what's inside is negative, remove the absolute value bars and insert a  negative. /) / = )  Problem8 ~  Evaluate the limit liin 2 x- ~ . x--t3+ X + 2x-15  a) 0 'J.I  b) 1/2 J. ,'/VJ x,z!  e) -1/8 [f]  c) -1/2 X-+J ~ (x+t>(x"'2) d) 1/8 / 7 .:7  Problem9  Evaluate the limit: fun 2 Ix-~ . x--.3- X +2x-15  a) 0 'J.'I  I- 5" I= r  -(-5)= s b) 1/2  c) -1/2  d) 1/8  e) -1/8  Problem 10  - {'>71)  (X'-ft)(x/J)  ff!  E I th I. . 1. 3x(2-x) va uate e 1m1t: 1m I I . x--tr x-2  a) 0 l.'f  b) 3 , J.,'r, 3 x{l- X )  c) 6 k - d) -3 -,,. -(x-a)  aj -6 ~~- ,  ---- "J--x  9 Trig Limits  lim SlilX = 1  x-tO X  1. cosx-1 0 lffi =  x-tO X  lim tanx = 1  x-tO X  Squeeze Theorem  or  Trig Limits  lirn sin(f(x))  f(x)-tO f(x)  ~t'/l __ (sot"'1lf~,'ll.j)  )6~/-~1·11 J  The Squeeze Theorem is a type of proof. You are always staring with a fact and  then working towards what you want to prove. Make sure you know the  following graphs and their ranges for Squeeze Theorem problems.  SlilX  cosx  sin2 x  cos2 x  Problem 11  -1 ~ sinx~ 1  -1 ~ cosx ~ 1  0 ~ sin2 x ~ 1  O~cos 2 x ~l  . sin2(5x) Evaluate hm 2 •  x-tO 2x cos(3x)  a) 0 J. i'/'1  b) 25/2 X-iO  c) 5/6  d) 2/3 )t'r1  e) DNE  10 Problem 12  sin(x-2) Evaluate ~~ x2 _ 4  a) 0  b) 1/4 j__, I rl1  c) 1/2 X~l  d) 4  e) 2  ii'l'1  I  -L(  I  X~d Problem 13  csc(2x)  Evaluate :~ cot( 3x) ·  a) 0  X-t-~ ~  I  b) 3/2 j,'/Vl  r;,-;iix  c) 2/3  d) 1/2  e) 1/3  Problem 14  )< ->,()  - I  x2 -sin 2(5x) Evaluate lim 4x2 ·  J x~o ) , y J O ,ri - a K7P Y/ b) 1/4  1  c) -6  d) -21/16  e) -15/4 - - '{ <.  -2'1 C.7  --:;-:CfJ  11 Problem 15 I  Evaluate lim ¾ . x~0~-3 i  a) 0 x /x  b) 1/2 JI '/11  c) 1/3 >< ~" 3 __ ~  d) -1  e) -1/2  /(~ti 5-/  Problem 16  -Ill  Find a nonzero value of a such that limf(x) exists.  ~ f_-vll~~ HO  ~ r"rJC- f(x) = X I X < 01 '1,-,_(c>i ~,~t Tl'rJ {tan(ax)  3x 2 + 2a2 , x ~ 0  a) 1 1 P l v {, / w 2 l ,<, ti  b) 2  c) 1/2  d) ./i/2  e) --/3/2  V"',, ct  X .>tO ") l Z.  .: )X +11--ot  J..  12 Problem 17  Use the Squeeze Theorem to show that Iim ,J-; sin(_!_]= 0. X--?0+ X  -/ ~ S",' 11 {eviyttii"J) ~ /  - I ~ f, ·~ (;) ~ I  _ rx f rx ),'~ r-t: J ~ J?  ± ' J ;  0  -w Lf Fr 11 lrtc-,111 Problem 18 -rvs, pt..v(1 /tAI zt~o 1  F. d 1. f() . h . ( re) f() 3x 5-l2x 3 m +18 1m x , given t at sm x +- :s;- x :s;-----. X--?0 2 18  a) 0 , {'r) ! I(><) :: :_!_ b) 1 >, 11 ---i: 1r  it j ~ c) 2  d) 1/2  e) 18  13 Discontinuities  There are 3 types of discontinuities:  • Hole (Removable)  • Infinite Discontinuity (Non-removable)  • Jump Discontinuity (Non-removable)  Removable Discontinuities  • These are also called "holes"  o Thishappenswhen limf(x)=limf(x)t=f(a).  x~a - x~a+  o To find holes algebraically:  • Factor the top and bottom off (x)  • Identify any common factors  • Set those factors equal to O and solve.  x 2 -x • Example: f(x) = 3 2  x -4x +3x  Non-removable Discontinuities  • Infinite Discontinuity (Vertical Asymptote)  o This happens when lim f(x) = ±00  x~a - or+  o To find vertical asymptotes algebraically:  • Factor the top and bottom off (x)  • Cross out any common factors  • Then set the denominator equal to O and solve . sinx • Except10n: f (x) = - does not have a VA X  x 2 -x • Example: f(x) = 3 2 . x -4x +3x  r f>< J= '  k-J  )<-J::t1  /  • Jump Discontinuity  >r.: :-3  o This happens when lim f(x) t= lim f(x) x~a - x~a +  o Commonly seen with piecewise or absolute value  14 Problem 19  x2 -2x-15 ·  Given f(x) = 3 2 , which of the following statements is true. f(x) has .... x -3x -lOx  A removable discontinuity at x = -2, and infinite discontinuities at x = 0, 5  A removable discontinuity atx = 5, and infinite discontinuities atx = 0, -2  Removable discontinuities at x = 5, 0 and an infinite discontinuity at x= 2  d) A removable discontinuity at x = 5, and an infinite discontinuity at x = 2 ~ f (x) is continuous everywhere /. ~)o  f16L{ X :-5' C tloL c f(x):: r'x;,-r )(><-+ 3)  '1_ (!f/r) ( K+ 2.)  k::- ~ {v.11.)N'~O)  A'= -J....  Problem 20  Which of the following functions has a vertical asymptote at x = 0?  \ sinx Ix/ +  f(x) = - II. f(x) = - + III. f(x) = cotx X X ~- /only  \!§!III only  <SJ. I and II  ~ I and III  "'\ I, II, and I II  :rn= .,  I  I  I  I ,,  .,,,.. I  I  I  I  71""  I  15 Continuity  Continuous  A function is continuous at the point a if the following is true:  • f(a) exists ~  • llinf(x)exists  X-M  • The limit equals f (a)  Problem21  Find the values of A and B that make the function continuous for all real numbers.  ~ o"°' Ir\"~~ { 2x2 + 5 x /1  fO~ .J.f't~' E.f,J ~'\ f(x)= Ax+B -Is;x<2  ~f\., i ~1'~' t. '"" ~,. r"tt ? 8x x~2  t\~lt ~ "1~  "'\" 1~ ~ t1"-' ,o NS·  ~G..." 1r-<  Problem22  ol.>c-'·t s-:A"+ e  ':/ = -A+B  B=A+t  Ax+ B: </ x  olA-t B ~ /IP  'J./1-+f,-+ 1-) -=,,  3A :-q  For what value of A is the function continuous for all x > O?  f(x) = {:X--93'  0<x<9  Ax , x~9  a) 2/3 (6' + 3) x- " b) 7 - =Ax • c) 16/9 (./x'43) Jx'-3  d) 0  e) ) f"i1 +3 {x.,.,;  ~-- - 4;< - s-+ 1 :; 9 A  A-::. ~ - ¼ '1 - . 16 Problem 23  List all values of x where f (x) has a discontinuity. Classify each discontinuity as removable jump, or infinite.  L; 5  ~ : - {  f (Kf J  Let f(x) =  I  !)N £  x+2 x$l I  2x-3 l<x<2.  -2 /1  x-4 x>2  X-:: I (N~O.''J°IJ Mf)  K=J.. (RD, HtJL.E)  ,c 7.. l{ ( v 4 , I/& {...,L  17 Problem24  List all x-values where f(x) has a discontinuity. Classify each discontinuity as  removable, jump, or infinite.  cosx, x<O  x 2 -x /  x 2 -1' O~x<1  f(x) = / 5 -x-2 1<xS4 2 , / x+12  6-x , x>4  If you are still having difficult with this topic, this is a great problem to try by  yourself at home. The answer is given below so you can check your work  x ::t 0: Jump Discontinuity  x = 1: Removable Discontinuity  x = 6: Infinite Discontinuity  18 Intermediate Value Theorem (IVT)  Intermediate Value Theorem  Suppose that f is continuous on the closed interval [ a,b] and let N be any  number between f(a) and f(b), where f(a) * f(b), then there exists at least one  vc:1-lue c in (a, b) such that f (c) = N.  (-3 I 3) (-3 '3) I I  Problem 25  The Intermediate Value Theorem guarantees that there exists a real zero for the  function f(x) = -2x 3 + 4x 2 - x + 1 on which interval(s): -- 'aj [-1,0] hl [0,1]  ~[1,2]  f (- I) :: J. -f '1 "i f (f/) :: I (+)  f (x} ::--o  l-+t::f(1-J  d) [2, 3] f (,) ~ J. -f 'I _ e) The IVT cannot be applied to f ' - 1 1- 1 - ~ /t)  f/J) ~ - I~ + I~ -2 --,. I ~~ - I (-J  19 problem 26  Assume f is a continuous function on [1,4] such that f (1) = 8, f (2) = -5, and  f( 4) = 3. Which one of the following statements must be true?  'Q] f has no zeros in [1, 4]  --b)J has exactly one zero in [1, 4]  ~ f has at least three zeros in [1, 4]  ~ f has at least two zeros in [1, 4]  e) f has exactly two zeros in [1, 4]  ; z  •  I  I  I  'i  (z ,-, )  Problem 27 (True or,alseY  True or False. The Inte~iate Value Theorem guarantees that f (x) = ~ X has a  solution on the interval (-1, 1). --- 20 Derivatives and Rates of Chan~  What is~ -'~derivative"?  • An instantaneous rate of change  • The "slope of the tangent line"  Derivative Notation  • f'(x)  • y'  • dy  dx  f(x) = x 2 f'(x) = 2x  .J '( v:: t  I '(;)J ~ 'I  You will be asked to compute derivatives the "hard way" (limit definition) and the  "easy way" (formulas). In general, you will always want to use the "easy way" unless  you specifically tolq otherwise. You can expect to be asked to compute at least one  derivative the hard way on the free response portion of your exam.  21 The Limit Definition of the Derivative  (aka: "Finding the Derivative the Hard Way")  There are two different limit definitions of the derivative ...  The Definition When to use it:  1. f(x+h)- f(x) lID------ h~O h  1. f(x)- f(a) 1m----- x~a x-a  Problem 28  Use this formula when you are looking for  the general derivative, ((x)  Use this formula when you are asked to  find the derivative at a specific point, f' (a)  (a) Assuming f is a differentiable function, write the limit definition of the  derivative off,· -- _2111 f (x+~)- f(x)  h~o ---- ~  (b) Using only the definition of the derivative from part (a) above, find the  expression for f'(x) if f(x) = x 2 - 3x + 5  1,'/V) (1<-+Jl-3 (><-11,)-+ s 1,--'t 0  v2 2. Y  C +~xh+h -Yx-31, ~ -y +)>< /'  -------~----,----------- ~  22 Problem 29  Use the limit definition of the derivative to find f'(x) if f(x) == vx + 5.  Solution:  I. -Jx+h+5--Jx+5  Jill------ h-.o h  I. -Jx+h+5 --Jx+5 -Jx+h+5 +-Jx+5 1m------x,=====-r===  h-.o h -Jx+h+5 +-Jx+5  1. x+h+5-(x+5) 1m-r=-==-r====- h-.o h( -J X + h + 5 + -J X + 5)  . x+h+S-x-5 hm -r====-r===-- h-.o h(-Jx+h+5 +-Jx+5)  1. h  1m-r====-r===-- h-.o h(-Jx+h+5 +-Jx+5)  I. 1  1m---,.===-----;:.==- h-.o ( -J X + h + 5 + -J X + 5)  1 1  (-Jx+0+5 +-Jx+5) - 2-Jx+S  23 Problem 30  Use the limit definition of the derivative to find f'(x) if f(x) = -1-. x+4  Solution:  1 1 --- - -- lim X + h + 4 X + 4  h-4 0 h  1. x+:+t::) x!4(:::::) Jill  h-4 0 h  x+4 x+h+4  1. (x+4)(x+h+4) (x+4)(x+h+4) 1m  h-40 h  x+4-x-h-4  1. (x+4)(x+h+4) 1m  h-4 0 h  J+J-J-h-J  lim ( X + 4 )( X + h + 4)  h-4 0 h  . -/z 1 hm-------'----x h-4 0 ( X + 4 )( X + h + 4) fi  -1 -1 lim =--- h-4 0 (x+4)(x+h+4) (x+4) 2  24 Problem 3J,  (a) Suppose that f is a differentiable function. State the limit definition of the  derivative of a function f (x) at a point a.  ),·11 [ Cx)--F(Qf)  X~l\ X-0\  (b) Now use your answer from part (a) to find f'( 4) for f (x) = -J2x + 1.  1 °'- }1'rv' ~ - 3  X~'-( X-(/  --!II ~ J  25 Basic Rules  d -(c)=O  dx  d d d -[f(x)±g(x)]= -(f(x))±-(g(x)) dx dx dx  d d  dx (cf(x)) = c dx (f(x))  The Product Rule  Differentiation Formulas  Examples  !!:...(1) = 0  dx  d d [f(x)g(x)] = f(x)g'(x) + g(x)f'(x)  x Lt v {,(vi+ ulv  !!_ [f(x)l = g(x)f'(x) - f(x)g'(x)  The Quotient Rule  dx g(x) (g(x) )2  Problem 32  d  dx ((x 2 + 1)(6x + 2)) =  {'1<2-1-U(, + Jx(6Kt2)  d ( x2 ) J._,>< (x-+,) -{t}{J..x:)  dx X + l = X-f tj l.  For each of the following function, find the derivative f'(x).  f(x) = -,Jx ~ [x) Yi  ' (; ) I • I/ , f )< ~ - '){ / Z, - i}..  1  f(x) = rr3 + xrr + - T[  r'() 1r-, , X: 1'r)<. ,  26 The Chain Rul~  dxf(g(x)) = f'(g(x))xg'(x) ) .,  d  X =-S>< ·I  Trig Derivatives (Memorize these!)  d . -smx = cosx dx d . -cosx = -smx dx  d  -(2x + 1) 5 = dx  !J(1-K·t-1)'f b)  ~S//JX '\_,  - CuJx ~c?fK  .!!:..tanx = sec 2 x dx  -secx = sec x tan x dx  d  Problem 33  If f (x) = 1/,-3-x2=----9-x, find f' (x).  -cotx = - csc 2 x dx  d  -cscx = - cscxcotx dx  d  a) ,f (x):: ( z ~ 3(3x2 -9x) 213 3x - 'Ix)  2x-3  b) 6x-9 '() , I -2/j  (3x 2 - 9x) 213 { X ~ J )x z -1)< )  c) 1  - 7 r~x-'J) & 2x-3 7 f J-x-< - er x) 2./3 (3x 2 -9x) 213  3(3x2 - 9x) 213 ...  e) 2x-3  (3x 2 - 9x)1'3  { &x - er)  27 Problem 34  Let f(x)= ~. Thenf'(x) =  X  6x-I ,f.(x).:: X  4x-1 (&,x-~1/)  a) (6x- I)2 13 .f 1 (><) -- ( Y3 !) ( (,><-t) _ _!_ ( - Z/)  6x+ 1 -- ) (px-,) ((;) (x)  b) (6x-1)213 (c~)(-,)'-j)~  2 !'f_,,J, u X -~ (h-1)' _ (fr: 1  c) 6 c- -2./3{( -J-(-1) ;:-I  (6x -1)413 / )  d) ( 6x -1 )213 (G X- I) 1/J •  e) ( 6x-1)413 J"trJ~  4x-1  Find the derivative of the functions below:  f(x) = sin(5x)  J '(x) :S-C of (,x)  f(x) = sin(sin(5x))  f l ( )<) :: l O S [~I·~ { 5' )()) • (_ (l) ( f" )< ) • )  3  f(x) = sin3 ( &) f-(~J = (s ,'/I { Ji7))  f" [;c) :, '3 ( S t'n f 'J-x ') ~ . { c, > hx ..  28 Common Test Questions  Tangent Lines  To find the equation of a tangent line:  • Point: (xi, y 1). The point will be given to you.  • Slope: f' (x). Take the derivative off (x), then plug in the x - value of  the point.  • Equation: y - y1 = m(x - x1)  Horizontal Tangent Line  • Where f'(x) = 0  • In other words, set the numerator off' (x) = 0  • They can only occur at x-values that are in the domain off (x)  Vertical Tangent Line  • Where f'(x) = undefined  • In other words, set the denominator off' (x) = 0  • They can only occur at x-values that are in the domain off (x)  Parallel & Perpendicular Lines  Parallel Lines  • Parallel lines have the same slope  Perpendicular Lines  • Perpendicular lines have opposite reciprocal slopes (flip the sign, flip  the fraction)  • Also called "normal" and "orthogonal"  Secant Line (Average Rate of Change)  To find the slope of the secant line, also known as the average rate of change  between two points, use your basic slope formula below. Do not take a  derivative!  29 Problem 36  Find the slope of the tangent line to the curve f (x) = x sin x 2 atthe point ( ../ii, O ).  a) -1  b) 2rr  c) 1/2  d) -2rr  e) -5  Problem 37  Find the equation of the tangent line f(x) = x3 + 5 at the point (1, 2). 3x  a) y=-x+3  b)y=3x-l  c) y=3x  d) y=x-3  e) y=2x+8  ~ q XJ - JxJ -/ S- ~  ) x-3  - J  ---)x ..  5 X <- d--. ')(3 -- r-- s-- f '(,) :: 2 - )  -J C - :. _ 1 i:::-- ;> l. df r  5 3  )"'-2::-/(y-t}  Y-) ::--x-f-t  30 Problem 38 {Answer: EJ  Forwhatvalue(s) ofx does the graph ofy = x 3 + 4x 2 + x -1 have a tangent line  that is perpendicular to the line y = ../5 + ~ x?  L----L.J a) x = 0 and x = 1 4/VJ: ..!_  b) x = 1 and x = 1/2 . ..,  c) X = 3 l'1 :: - 't  e) x = -1 andx = -5/3 1 1== "3 x: 2 + r >< ...,. 1  d) x = -1 and x = -3  - 'i -=1x2 -+i'X -ft  Jx 2-+ 9' 'X + > = (J}  (;x + ~ )(,x + I )  [3_c -~ I-/-]  Problem 39 {Answer: BJ  x2 +4x-4 . For what x-values does y = ---- have honzontal tangent lines? x-l  a) x = 0 and x = 1  b) x = 0 and x = 2  c) x=l  d) x = 1 and x = 4  e) x = 2 andx = 4  31 Problem 40 (Answer: BJ  Find the average rate of change of the function f (x) = sin x on the interval [ ~, ~].  a) 2-Jl f (xz) -f (x,) n  b) 4-2J2  n  c) 4+2J2  n  d) -2J2  n  e) -J2  2n  Problem41  tan( n +h )- I  Find lim ---'-- 4----"--- h-40 h  Kz -Xr  s ,',;, {f) - ), 1/1 ( if-)  -L.- ) 1'!1 f (x+~) - -J (x)  "1-'l t? ~  Hint: This limit represents the derivative of some function f at some point a.  a) 4 f (><) = i-P1,1 [x J  =  b) 0  c) 1  d) 2  e) 3  f 1(K) .. - S'-e t z(x)  f '(!!;-) ~ S,e c2(f)  I -2  ::  32 Problem42  Suppose that f is a differentiable function such that f (2) == 12 and f' (2) == 8. Find  the slope of the tangent line to the graph of h(x) = f(:) at the point (2,3).  h '(x) ~ x z; '(K) - l(x-) 2x  a) 3  b) 5  X  ><'t  c) -1  d) -2  e) 2  Problem43  ~ I, '/2) : ~  '-f f y, )- f (2) ( 'f J  y ( f) - I J (t_;)  ;: ~  - )2_'-f/'  ---:--.._I - - le,  v - --;;: ~ -1  If g(2) == 3, g' (2) == 7, g' (3) == -1, g(3) == 5, and f (x) == g(g(2x) ), then find f' (1). a) -14  b) -10  c) 15  d) 12  e) -5  .f'(.x}' './'(1(,;c)) ' 7'(Jx) · J.  f'(,):: J '(y'(,J) · f(•) ·2  :/'(3) . , 7 . 1  -- -( . t.;). /0_i]  33 Posi_tiol!fVelQclty /Acceleration  Position  • Usually denoted by s(t) or h(t)  • Gives the position of the particle at any point in time t  Velocity  • The first derivative of position  • Denoted by v(t)  • At rest: v(t) = 0  • Moving forward: v(t) > 0  • Moving backward: v(t) < 0  Acceleration  • The first derivative of velocity  • The second derivative of position  • Denoted by a(t)  • Speeding up: v(t) and a(t) have the same sign  • Slowing down: v(t) and a(t) have different signs  Projectile Motion  Maximum Height:  • Set v(t) = 0  • Then plug in h(t)  Velocity at the ground:  • Set h(t) = 0  • Then plug into v(t)  34 Problem44  The position of an object is given by the function s(t) = t +cost meters. Find its  velocity at the moment when its acceleration is 1 m/ s 2 .  a) 1 m/s  b) Om/s  c) rrm/s  d) -1 m/s  e) -2rrm/s  Problem45  ) ' ( t) :-_ / - )1 1)? t  ) '/tJ ~ - C o_J t :: I  cost~-;  t-=-tJY  V ('i';J)"!- s I ,:-.._.  I// I I  A particle moving in a straight line has the position s(t) = t 3 - 6t 2 + 9t + 1 for  time t ~ 0, where tis measured in seconds ands in feet. When is the particle  moving in a positive direction?  i))o < t < 1 and t > 3  15) 0<t<3  c) t > 3  d) 1 < t < 3  e) 0 < t < 1  v(t) :: "3 i - I 'J. i. -f 1 ..: 0  J(t-'-(t:-fJ)::: ()  7 (i: -3)/t:--1)  t:. / ) 3  I -f- 7  0 {,)  35 Problem46  If a ball is thrown vertically upward with a velocity of 32 ft/sec, then its height h  after t seconds is h(t) = -16t 2 + 32t + 48.  a) What is the maximum height the ball reaches?  v~ CJ  V (-t-} -=-°? 2 t:: -f 5 2 -:; &  -Jj:t:- ~ - 3-t:- -c -:: I  ~ (t} :: - / (p -f J 2 +-y 7  -:: 1,~'1j':r;"(Jf  b) With what velocity does the ball hit the ground?  h-:: (/  CJ~ -f(pl::.'2.-+ 3J.i +y 0  () - - -f~{t:i -z& -3)  0-: -le,(1,,-3) {t+,J  t ~ 3 J -½z  36 Differentiability and Graphing the Derivative  Interpreting Graphs  • If f(x) is increasing, then the y-coordinate off' (x) will be positive.  • If f (x) is decreasing, then the y-coordinate off' (x) will be negative.  • If f (x) has a slope of zero at a point (top of a "hill" or bottom of a "dip"), then  the graph off' (x) will either touch or cross the x-axis at that point.  Differentiability: A function f (x) is said to be differentiable at a if f'(a) exists.  The rules ...  • If f (x) is differentiable at x = a, then f (x) is continuous at x = a.  • If f (x) is not continuous at x = a, then f (x) is not differentiable atx = a.  • NoteJhere is no rule for when f (x) is continuous! This can go either way.  • There are 3 cases where f'(x) does not exist:  o Where there is a hole or break in the function (discontinuity)  o Where there is a sharp turn, corner, or cusp  o Where there is a vertical tangent line  Problem47  Which of the following functions is differentiable at x = O?  \  37 Problem48  The graph off is given below. State all values of x at which f is not differentiable.  b C  Problem49  Which of the following functions is differentiable at x = O?  ~ NPrf-f yv~N ~ f(x) = lxl  v. A· ~tcx) =;  c) f(x) = xl/3  @~(x) = x4/3 _-----sr  ~(x) = cscx  g  38 ,  oblem 50  f . Find the values at which f ( x) is not diff erentiab I e.  sin(nx)  f(x)= x2 +1  x<O  ..,_<"l  O~x~l ~  { cy::~  2x  c) x=Oandx=l  ~ x=tr  ~ x is differentiable everywhere  x~o  - l-: CJ  ~:  f (x): /  R ·,  Problem 51  Given the graph off (x) below, draw the graph f' (x).  y =f(x)  --- x> 1  39 2.6 Implicit _Differentiation  Implicit Differentiation  • Use implicit differentiation when you do not have y explicitly defined as a  function of x.  • In other words, you have a whole bunch of x' sandy' s jumbled together.  • It is likely that you will see a product rule, quotient rule, or chain rule with  implicit differentiation.  Let's go over some basics first. Find the derivative of...  X  y ( dj  y I Jy ' J1{x)  (3x 2 + 1) 5 r { 3 X 1. -f I ) '1 • G )(  y5 f 'I '1 . !Y d~  40 --:!z  {/,,X  l.E y-C#J fx9 - f,4 y z  :: """'=~--~-------...;_  41

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