Functions A B C D E F G H I J K L. Contents:


 Giles Barber
 2 years ago
 Views:
Transcription
1 Funtions Contents: A reltion is n set of points whih onnet two vriles. A funtion, sometimes lled mpping, is reltion in whih no two different ordered pirs hve the sme oordinte or first omponent. Algeri Test: If reltion is given s n eqution, nd the sustitution of n vlue for results in one nd onl one vlue of, then the reltion is funtion. Geometri Test or Vertil Line Test: A B C D E F G H I J K L Reltions nd funtions Funtion nottion Domin nd rnge Composite funtions Even nd odd funtions Sign digrms Inequlities (inequtions) The modulus funtion Rtionl funtions Inverse funtions Grphing funtions Finding where grphs meet If we drw ll possile vertil lines on the grph of reltion, the reltion: ² is funtion if eh line uts the grph no more thn one ² is not funtion if t lest one line uts the grph more thn one. f : 7! {z + } funtion f suh tht is mpped to + equivlent forms f() = + nd = +. The domin of reltion is the set of vlues of in the reltion. The rnge of reltion is the set of vlues of in the reltion. Set nottion Intervl nottion Numer line grph Mening f j > g [, [ the set of ll suh tht is greter thn or equl to f j <g ], [ the set of ll suh tht is less thn f j <6 g ], ]  the set of ll suh tht is etween nd, inluding f j 6 0 [ >4g ], 0] [ ]4, [ 0 4 the set of ll suh tht is less thn or equl to 0, or greter thn 4
2 Given funtions f : 7! f() nd g : 7! g(), the omposite funtion of f nd g will onvert into f(g()). f ± g is used to represent the omposite funtion of f nd g. It mens f following g. (f ± g)() =f(g()) or f ± g : 7! f(g()). In generl, (f ± g)() 6= (g ± f)(). f(g()) 6= g(f()). A funtion f() is even if f( ) = f() for ll in the domin of f. A funtion f() is odd if f( ) = f() for ll in the domin of f. The grph of n even funtion is smmetril out the is. The grph of n odd funtion hs rottionl smmetr out the origin. SIGN DIAGRAMS Drw sign digrm for +. + is zero when = nd undefined when = \Qw INEQUALITIES (INEQUATIONS)
3 solve + > 4 ws: If + > 4, then +> 4( ) ) +> 4 4 ) 7 > WRONG ) > 7 RIGHT > _ ) 7 <<. Solve : +5 > ) +5 > 0 ) ( )( +) > 0 Sign digrm of LHS is ) ], ] or [, [.  Qe This leds us to the lgeri definition of modulus: The modulus of, jj = ½ if 0 if <0 The reltion = jj is in ft funtion. We ll it the modulus funtion, nd it hs the grph shown. This rnh is = , < 0. = This rnh is =, > 0. An equivlent definition of jj is: jj = p.
4 ² jj > 0 for ll ² j j = jj for ll ² jj = for ll ² jj = jjjj for ll nd ² = jj for ll nd, 6= 0 ² j j = j j for ll nd. jj Modulus is distne. j j = j j ss tht the distne from to on the numer line equls the distne from to. If jj = jj then =. Solve for : j +j = 7 + = 7 + = 7 or += 7 ) = ) = 5 So, = or 5 Solve for : j +j = j j + = ( ) + = or + = ( ) 4= ) = So, = or 4.
5 RATIONAL FUNCTIONS OF THE FORM The grph of f() = + = +  is shown elow. = + + d, 6= 0 =, f() is undefined. = is vertil smptote. = \Qw  = s! from the left, f()! s! from the right, f()! or s!, f()! s! +, f()!. The sign digrm of = + is \Qw We n write: s!,! from ove or s!,! + s!,! from elow s!,!. = is horizontl smptote. ONETOONE AND MANYTOONE FUNCTIONS A onetoone funtion is n funtion where: ² for eh there is onl one vlue of nd ² for eh there is onl one vlue of. Onetoone funtions stisf oth the vertil line test nd the horizontl line test. This mens tht: ² no vertil line n meet the grph more thn one ² no horizontl line n meet the grph more thn one. If the funtion f() is onetoone, it will hve n inverse funtion whih we denote f (). Funtions tht re not onetoone re lled mntoone. While these funtions must stisf the vertil line test the do not stisf the horizontl line test. At lest one vlue hs more thn one orresponding vlue. If funtion f() is mntoone, it does not hve n inverse funtion. However, for mntoone funtion we n often define new funtion using the sme formul ut with restrited domin to mke it onetoone funtion. This new funtion will hve n inverse funtion.
6 PROPERTIES OF THE INVERSE FUNCTION If = f() hs n inverse funtion, this new funtion: ² is denoted f () ² must indeed e funtion, nd so must stisf the vertil line test ² hs grph whih is the refletion of = f() in the line = ² stisfies (f ± f )() = nd (f ± f)() =. The domin of f is equl to the rnge of f. The rnge of f is equl to the domin of f. SELFINVERSE FUNCTIONS An funtion whih hs n inverse, nd whose grph is smmetril out the line =, i s selfinverse funtion. For emple: ² The funtion f() = is the identif funtion, nd is lso selfinverse funtion. ² The reiprol funtion f() =, 6= 0, is lso selfinverse funtion, s f = f. GRAPHING FUNCTIONS Drwing the grph of funtion llows us to identif fetures suh s: ² the es interepts where the grph uts the nd es ² turning points whih ould e lol minimum or lol mimum ² vlues of for whih the funtion does not eist ² the presene of smptotes, whih re lines or urves tht the grph pprohes.
7 REVIEW SET A NONCALCULATOR For eh grph, stte: i the domin ii the rnge iii whether the grph shows funtion. 4 d If f() =, find: f() f( ) f( ) Suppose f() = + where nd re onstnts. If f() = 7 nd f() = 5, find nd. 4 Stte funtions f nd g for whih: f(g()) = p g(f()) = 5 Solve for : j4 j = j +7j +6>6 ³ + 6 If g() =, find in simplest form: g( +) g( ) 7 For eh of the following grphs determine: i iii the domin nd rnge whether it is funtion  5 \Wl_T_ (, 5) ii iv the nd interepts if it hs n inverse funtion Determine whether the following funtions re even, odd, or neither: f() = 4 f() = f() = p Find f () given tht f() is: 0 Drw sign digrm for: ( + )(4 ) If f() = +, f() =, nd f () = 4, find nd.
8 Drw sign digrm for ( + )( ). Hene, solve for : 6 < 0. Consider f() = nd g() = 6. Show tht f( ) = g( 4 ). Find (f ± g)( ). Find suh tht g() =f(5). 4 Given f : 7! +6 nd h : 7!, show tht (f ± h )() =(h ± f) (). 5 Suppose h() =( 4) +, [4, [. Find the defining eqution of h. Show tht (h ± h )() =(h ± h)() =. REVIEW SET B For eh of the following grphs, find the domin nd rnge: = (  )(  5) If f() = nd g() = +, find in simplest form: (f ± g)() (g ± f)() (, ) CALCULATOR =
9 Drw sign digrm for: Consider f() = For wht vlue of is f() undefined, or not rel numer? Sketh the grph of this funtion using tehnolog. Stte the domin nd rnge of the funtion. 5 Consider the funtion f() = +. Find nd given tht = f() hs smptotes with equtions = nd =. Write down the domin nd rnge of f (). 6 For eh of the following grphs, find the domin nd rnge: (, 5) (, 5) = 7 Solve for : + (,) = = j j > j +j =+ 8 Solve for : > Cop the following grphs nd drw the grph of eh inverse funtion on the sme set of es: 5 0 Consider the funtion f : 7! d e 4 +. Determine the equtions of the smptotes. Stte the domin nd rnge of the funtion. Disuss the ehviour of the funtion s it pprohes its smptotes. Determine the es interepts. Sketh the funtion. Consider the funtion f() =( ) + where is rel onstnt. Find given tht f() is n even funtion. Solve grphill: j 6j > +. Grph the funtion f() = jj +. Hene find ll vlues of for whih jj + >.
10 Consider the funtions f() = + nd g() =. Find (g ± f)(). Given (g ± f)() = 4, solve for. Let h() =(g ± f)(), 6=. i Write down the equtions of the smptotes of h(). ii Sketh the grph of h() for 6 6. iii Stte the rnge of h() for the domin Consider f : 7! 7. On the sme set of es grph =, = f(), nd = f (). Find f () using vrile interhnge. Show tht (f ± f )() =(f ± f)() =, the identit funtion. 5 The grph of the funtion f() =, is shown longside. 5 Sketh the grph of = f (). Stte the rnge of f. Solve: i f() = 0 ii f () = (, ) (, ) 5 6 Consider the funtion f() =0: Use tehnolog to help sketh the funtion. Determine the position nd nture of n turning points. Hene, find the mimum nd minimum vlues of f() on the intervl REVIEW SET C For eh of the following grphs, find the domin nd rnge: (, )  (, ) (, ) Given f() = +, find: f( ) suh tht f() =4. Stte the vlue(s) of for whih f() is undefined: f() = 0 + f() = p +7
11 4 Drw sign digrm for: f() =( + 4)( +) f() = 5 Given h() =7, find: ( + )( +8) h( ) in simplest form suh tht h( ) =. 6 Suppose f() = nd g() = p. Find in simplest form: i (f ± g)() ii (g ± f)() Stte the domin nd rnge of: i (f ± g)() ii (g ± f)() 7 Suppose f() = ++. Find,, nd if f(0) = 5, f( ) =, nd f() = 4. 8 Cop the following grphs nd drw the grph of eh inverse funtion on the sme set of es:  9 Suppose f() is n odd funtion. Prove tht g() =jf()j is n even funtion. 0 Find the inverse funtion f () for: f() =7 4 f() = + 5 Given f : 7! 5 nd h : 7! 4, show tht (f ± h )() =(h ± f) (). Solve for : > 0 Given f() = + nd g() =, find (g ± f )(). 4 Sketh funtion with domin f j 6= 4g, rnge f j 6= g, nd sign digrm. 4 5 Sketh the grph of g : 7! for ], ]. Eplin wh g hs n inverse funtion g. Find lgerill, formul for g. d Sketh the grph of = g (). e Find the rnge of g. f Find the domin nd rnge of g. 6 Consider the funtion = p Find the es interepts. Find n turning points of the funtion. d e Find n smptotes of the funtion. Stte the domin nd rnge of the funtion. Sketh the funtion, showing its ke fetures.
12 REVIEW SET A i Domin = f j R g ii Rnge = f j > 4g iii Yes i Domin = f j R g ii Rnge = fg iii Yes i Domin = f j R g ii Rnge = f j 6 or > g iii No d i Domin = f j R g ii Rnge = f j g iii Yes = 6, = 4 f() = p, g() = g() =, f() = + 5 = or ], 8] or [, [ i Domin = f j R g, Rnge = f j > 5g 5 ii int, 5, int 9 iii is funtion iv No i Domin = f j R g, Rnge = f j = or g ii no interepts, interept iii is funtion iv No 8 odd neither even 9 f () = f 4 () = We 4 =, = ], [ or ], [ f( ) = ( ) =9 g( 4 )= 6( 4 )=9 4 (f h )() =(h ± f) () = ± 5 h () =4+ p, > 69 = 4 REVIEW SET B Domin = f j R g, Rnge = f j > 4g Domin = f j 6= 0, 6= g, Rnge = f j 6 or >0g = f() = + Domin = f j 6= 0g, Rnge = f j >0g 5 =, = Domin = f j 6= g, Rnge = f j 6= g 6 Domin = f j > g, Rnge = f j <<5g Domin = f j 6= g, Rnge = f j 6 or > 5g 7 =or 7 ], ] or [5, [ 5 8 [, 5] ] 8, [ or ], [ 9 = 5 f f vertil smptote =, horizontl smptote = 4 Domin = f j 6= g, Rnge = f j 6= 4g s!,! s!,! 4 s! +,! s!,! 4 + d interept 4, interept e = 4 f Qw  Qr =6 ], [ or ]9, [ = ) ( Qw ' Qe ) = > for [ jj +, [ f() = (g ± f)() = = + i vertil smptote =, horizontl smptote =0 = =Qe = f  = + =0
13 ii =0 (, Q r ) = Qe h() = + (, W u ) REVIEW SET C Domin = f j > g, Rnge = f j 6 <g Domin = f R g, Rnge = f,, g = = < Qe iii Rnge = f j 6 4 or > 7 g 4 f () = +7 Qw 5 f  7 f = f () (, ) Qw (, ) = f() (, ) = 0 (, ) = 6 i p ii p i Domin = f j > 0g, Rnge = f j 6 g ii Domin = f j 6 0:5g, Rnge = f j > 0g 7 =, = 6, =5 8 f f  f  = = 0 f () = 7 f () = 5 4 (f ± h )() =(h ± f) 4 +6 () = 5 [ 5, ] ], [ or ]4, [ 6 4 =4  f Rnge = f j g i ¼ :8 ii = 6 (0.558,.) = = .5 (4.77, 4.) lol mimum t (0:558, :) lol minimum t (4:77, 4:) mimum is :, minimum is 4: 5, d An horizontl line = = uts the grph t g most one. g()= p +, > (,) g  (,) e Rnge of g f j > g f Domin of g f j > g, Rnge of g f j 6 g 6 interept :6, interept 4:9 lol mimum t ( 0:97, 4:47) vertil smptote =4, horizontl smptotes =, =7 d Domin = f j 6= 4g, Rnge = f j 6 4:47, >7g e (0. 97,4.47) =4 =7 =
14 Trnsforming funtions Contents: A B C D E F G H Trnsformtion of grphs Trnsltions Strethes Refletions Misellneous trnsformtions Simple rtionl funtions The reiprol of funtion Modulus funtions In prtiulr, we n perform trnsformtions of grphs to give the grph of relted funtion. These trnsformtions inlude trnsltions, strethes, nd refletions. In this hpter we will onsider trnsformtions of the funtion = f() into: ² = f()+, is onstnt ² = pf(), p is positive onstnt ² = f() ² = f( ) ² = f( ), is onstnt ² = f(q), q is positive onstnt When we perform trnsformtion on funtion, point whih does not move is lled n invrint point. TRANSLATIONS ² For = f()+, the effet of is to trnslte the grph vertill through units. I If >0 it moves upwrds. I If <0 it moves downwrds. ² For = f( ), the effet of is to trnslte the grph horizontll through units. I If >0 it moves to the right. I If <0 it moves to the left. ² For = f( )+, the grph is tr ns l ted horizontll units nd vertill units. We s it is trnslted the vetor STRETCHES. ² For = pf(), p>0, the effet of p is to vertill streth the grph the sle ftor p. I I If p> it moves points of = f() further w from the is. If 0 <p< it moves points of = f() loser to the is. ² For = f(q), q>0, the effet of q is to horizontll streth the grph the sle ftor q. I I If q> it moves points of = f() loser to the is. If 0 <q< it moves points of = f() further w from the is.
15 REFLECTIONS ² For = f(), we reflet = f() in the is. ² For = f( ), w e reflet = f() in the is. ² For = f (), we reflet = f() in the line =. SIMPLE RATIONAL FUNCTIONS A funtion of the form = + + d, is lled simple rtionl funtion. 6= d where,,, nd d re onstnts, These funtions re hrterised the presene of oth horizontl smptote nd vertil smptote. An grph of simple rtionl funtion n e otined from the reiprol funtion = omintion of trnsformtions inluding: ² trnsltion (vertil nd/or horizontl) ² strethes (vertil nd/or horizontl). = For emple: ² = k is vertil streth of = with sle ftor k. ² = k is horizontl trnsltion of = through k units.
16 Consider the funtion f() = 6 +. d Find the smptotes of = f(). Disuss the ehviour of the grph ner these smptotes. Find the es interepts of = f(). Sketh the grph of the funtion. e Desrie the trnsformtions whih trnsform = into = f(). f Desrie the trnsformtions whih trnsform = f() into =. f() = 6 + = ( +) 8 + = As!,!. As! +,!. As!,! +. As!,!. When =0, = 8 += 6. ) the interept is 6. When =0, 6=0 ) = ) the interept is. e eomes eomes = f() is trnsltion of = 8 Now = 8 through. hs smptotes =0 nd =0. ) = f() hs vertil smptote = nd horizontl smptote =. under vertil streth with sle ftor 8. under refletion in the is. 8 eomes under trnsltion through d So, = is trnsformed to = f() under vertil streth with sle ftor 8, followed refletion in the is, followed trnsltion through. f To trnsform = f() into =, we need to reverse the proess in e. We need trnsltion through, followed refletion in the is, followed vertil streth with sle ftor 8. f() = 6 + =. 6 =
17 THE RECIPROCAL OF A FUNCTION For funtion f(), the reiprol of the funtion is When = f() is grphed from = f(): f(). ² the zeros of = f() eome vertil smptotes of = f() ² the vertil smptotes of = f() eome zeros of = f() ² the lol mim of = f() eome lol minim of = f() ² the lol minim of = f() eome lol mim of = f() ² when f() > 0, ² when f()! 0, f() > 0 nd when f() < 0, f() < 0 f()! nd when f()!, f()! 0. Grph on the sme set of es: = nd = = nd = =  =  Qw  =  = =
18 MODULUS FUNCTIONS ½ if > 0 We hve seen tht the modulus funtion is defined f : 7! jj = if <0. To otin the grph of = f(jj) from the grph of = f(): ² disrd the grph for <0 ² reflet the grph for > 0 in the is, keeping wht ws there ² points on the is re invrint. The modulus of the funtion f() is jf()j = ½ f() if f() > 0 f() if f() < 0. To otin the grph of = jf()j from the grph of = f(): ² keep the grph for f() > 0 ² reflet the grph in the is for f() < 0, disrding wht ws there ² points on the is re invrint. Drw the grph of f() = ( ) nd on the sme set of es drw the grph of = jf()j nd = f(jj). = jf()j = ½ f() if f() > 0 f() if f() < 0 The grph is unhnged for f() > 0 nd refleted in the is for f() < 0. = jf()j = f(jj) = ½ f() if > 0 f( ) if <0 The grph is unhnged for > 0 nd refleted in the is for <0. = f(jj) = f() = f()
19 REVIEW SET 5A NONCALCULATOR If f() =, find in simplest form: f() f() f( ) d f() If f() =5, find in simplest form: f( ) f( ) f ³ d f() f( ) The grph of f() = + + is trnslted to its imge g() the vetor. Write the eqution of g() in the form g() = d. 4 The grph of = f() is shown longside. The is is tngent to f() t = nd f() uts the is t =. On the sme digrm, sketh the grph of = f( ) where 0 <<. Indite the interepts of = f( ). =f ( )
20 5 For the grph of = f() given, sketh grphs of: = f( ) = f() = f( +) d = f()+ ( 58, ) (, ) =f ( ) 6 Consider the funtion f : 7!. On the sme set of es grph: = f() = f( ) =f( ) d =f( ) + 7 The grph of = f() is shown longside. Sketh the grph of = g() where g() =f( +). Stte the eqution of the vertil smptote of = g(). Identif the point A 0 on the grph of = g() whih orresponds to point A. (, ) = =f ( ) A( 0, ) 8 The grph of = f() is drwn longside. Drw the grphs of = f() nd = jf()j on the sme set of es. Find the interept of f(). Show on the digrm the points tht re invrint for the funtion f(). d Drw the grphs of = f() nd = on the sme set of es. f()  = f() (4, ) 4 (, ) (, ) 9 Let f() =, 6=, >0. + On set of es like those shown, sketh the grph of = f(). Lel lerl n points of intersetion with the es nd n smptotes. On the sme set of es, sketh the grph of = f(). Lel lerl n points of intersetion with the es.  0 Consider f() = where is positive rel numer. Find epressions for jf()j nd f(jj). Sketh = jf()j nd = f(jj) on the sme set of es. Solve for given is positive rel numer: j j = jj.
21 REVIEW SET 5B CALCULATOR Use our lultor to help grph f() =( +) 4. Inlude ll es interepts, nd the oordintes of the turning point of the funtion. Consider the funtion f : 7!. On the sme set of es grph: = f() = f( +) =f( +) d =f( +) Consider f : 7!. Does the funtion hve n es interepts? Find the equtions of the smptotes of the funtion. Find n turning points of the funtion. d Sketh the funtion for Consider f() =. Use our lultor to help determine whether the following re true or flse: i As!,! 0. ii As!,! 0. iii The interept is. iv > 0 for ll. On the sme set of es, grph = f() nd = jf()j. Write down the eqution of n smpotes of = jf()j. 5 The grph of the funtion f() =( +) +4 is trnslted units to the right nd 4 units up. Find the funtion g() orresponding to the trnslted grph. Stte the rnge of f(). Stte the rnge of g(). 6 For eh of the following funtions: i Find = f(), the result when the funtion is trnslted. ii Sketh the originl funtion nd its trnslted funtion on the sme set of es. Clerl stte n smptotes of eh funtion. iii Stte the domin nd rnge of eh funtion. = = 7 Sketh the grph of f() = +, nd on the sme set of es sketh the grphs of: f() f() f()+ 8 Suppose f() = +. The funtion F is otined strething the funtion f vertill with sle ftor, then strething it horizontll with sle ftor, then trnslting it horizontll nd vertill. Find the funtion F (). Wht n e sid out the point (, ) under this trnsformtion? Wht hppens to the points (0, ) nd (, ) under this trnsformtion? d Show tht the points in lso lie on the grph of = F ().
22 9 The grph of = f() is given. On the sme set of es grph eh pir of funtions: = f() nd = f( ) + = f() nd = f() = f() nd = jf()j 0 Consider the funtion f() = +5. Find the smptotes of = f(). Disuss the ehviour of the grph ner these smptotes. Find the es interepts of = f(). d Sketh the grph of = f(). e Desrie the trnsformtions whih trnsform = into = f(). f Desrie the trnsformtions whih trnsform = f() into =. (,) = f() =4 Sketh the grph of f() = +, lerl showing the es interepts. Find the invrint points for the grph of = f(). Stte the eqution of the vertil smptote of = f() nd find its interept. d Sketh the grph of = f() on the sme es s in prt, showing lerl the informtion ou hve found. e On new pir of es, sketh the grphs of = jf()j nd = f(jj) showing lerl ll importnt fetures. REVIEW SET 5C If f() = 4, find in simplest form: f( 4) f() f Consider the grph of = f() shown. Use the grph to determine: i the oordintes of the turning point ii the eqution of the vertil smptote iii the eqution of the horizontl smptote iv the interepts. Grph the funtion g : 7! + on the sme set of es. Hene estimte the oordintes of the points of intersetion of = f() nd = g(). ³ = f()  d 4f( +)
23 Sketh the grph of f() =, nd on the sme set of es sketh the grph of: = f( ) = f() = f() d = f( ) 4 The grph of ui funtion = f() is shown longside. =f ( ) Sketh the grph of g() = f( ). Stte the oordintes of the turning points of = g(). ( 4, ) ( 0, ) 5 The grph of f() = is trnsformed to the grph of g() refletion nd trnsltion s illustrted. Find the formul for g() in the form g() = + +. V( , ) =f ( ) =g ( ) 6 Given the grph of = f(), sketh grphs of: f( ) f( +) f() =f ( ) (' 5 Ow)  7 The grph of f() = + +4 is trnslted to its imge = g() the vetor. Write the eqution of g() in the form g() = d. 8 Find the eqution of the line tht results when the line f() = + is trnslted: i units to the left ii 6 units upwrds. Show tht when the liner funtion f() = +, >0 is trnslted k units to the left, the resulting line is the sme s when f() is trnslted k units upwrds. 9 The funtion f() results from trnsforming the funtion = refletion in the is, then vertil streth with sle ftor, then trnsltion of. Find n epression for f(). Sketh = f() nd stte its domin nd rnge. Does = f() hve n inverse funtion? Eplin our nswer. d Is the funtion f selfinverse funtion? Give grphil nd lgeri evidene to support our nswer.
24 0 Consider = log 4. Find the funtion whih results from trnsltion of. Sketh the originl funtion nd the trnslted funtion on the sme set of es. d Stte the smptotes of eh funtion. Stte the domin nd rnge of eh funtion. The funtion g() results when = is trnsformed vertil streth with sle ftor, followed refletion in the is, followed trnsltion of units to the right. d Write n epression for g() in the form g() = + + d. Find the smptotes of = g(). Stte the domin nd rnge of g(). Sketh = g().
25 REVIEW SET 5A d d +5 g() = d f()   ( Q d_q_, ) 4 = = f() = Qe (4, ) (4, Q w) = f() 5 + = f() = f(  ) = f() = f( ) = f() = f( + ) = f() + 9, = f()  = 0 jf()j = j j, f(jj) = jj = f() = f() =0 > =f( ) = f() = = f( ) = f( ) = f( ) + REVIEW SET 5B = f() = ( + ) = g() = f()  V (,4)  8 A 0 (,) = = A (, 0) = A 0 (, )  f() = f() 4 = f() (0, ), (, ), ll points on =, [, ]  = f() = = f( + ) = f( +) = f( +)  no horizontl smptote =0, vertil smptote =0 min. turning point (:44, :88) d (4, 4) (, ) = (4, 0.056) (, ) (.44,.88) 4
26 6 6 4 i true ii flse iii flse iv true = f() =  9 = = f() (,) = f(  ) + (4,) horizontl smptote =0 5 g() =( ) +8 f j > 4g f j > 8g 6 i = ii = =4 =6 = f() (,Q w) = Ew = = = = f() (,) =4 = f() For =, VA is =0, HA is =0 For =, VA is =, HA is = = f() (, ) (,) 7 iii For =, domin is f j 6= 0g, rnge is f j 6= 0g For =, i = ii domin is f j =g, rnge is f j = g For =, HA is =0, no VA For =, HA is =, no VA iii For =, domin is f j R g, rnge is f j >0g For =, domin is f j R g, rnge is f j > g 8 F () =4 It is invrint (0, )! (, ) nd (, )! (0, ) d F ( )= nd F (0) = =   = 4  = = f() = + = f() = f() = f()+ 0 vertil smptote = 5 horizontl smptote = s! 5,! s! 5 +,!!, s! +!, s! interept is 5, interept is d, d =Te e Vertil streth with sle ftor 9, refletion in the is, µ 9 5 then trnslte. f Trnslte µ 5 sle ftor = f() Et 9 9. =4, reflet in is, then vertil streth with Qw = (, ) =We (, ) = f()
27 (, ) nd (, ) =, VA is e REVIEW SET 5C =f( )  8 d 0 + i (, :8) ii =0 iii = iv interepts : nd 0:75 = f()  = f() = g() =+ = f() 8 i = +8 ii = +8 f( + k) =( + k)+ = + + k = f()+k 9 f() = + = 4 = f() = Domin of f() is f j 6= g Rnge of f() is f j 6= g Yes, sine it is onetoone funtion (psses oth the vertil nd horizontl line tests). d Yes, sine f () =f() = +. Also, the grph of f() is smmetril out the line =. 0 = log 4 ( ) = =log 4 = 4 ( :65, :65), ( 0:8, 0:), nd (:55, :55) = f() =  = f() = f() = f() = f() 4 (, 4) nd = f() (4, 0) (, 4) (, 0) (4, 0) For = log 4, VA is =0, no HA For = log 4 ( ), VA is =, no HA d For = log 4, Domin is f j >0g, Rnge is f R g For = log 4 ( ), g() = 6 Domin is f j >g, Rnge is f R g HA is =0, VA is = Domin = f j 6= g, Rnge = f j 6= 0g d = Q =log 4 () (, 4) 5 g() = = g() = 6  (5' O w ) = f() = f() = f(+) = f() 7 g() =
How to Graphically Interpret the Complex Roots of a Quadratic Equation
Universit of Nersk  Linoln DigitlCommons@Universit of Nersk  Linoln MAT Em Epositor Ppers Mth in the Middle Institute Prtnership 7007 How to Grphill Interpret the Comple Roots of Qudrti Eqution Crmen
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More information1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.
. Definition, Bsi onepts, Types. Addition nd Sutrtion of Mtries. Slr Multiplition. Assignment nd nswer key. Mtrix Multiplition. Assignment nd nswer key. Determinnt x x (digonl, minors, properties) summry
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More information1. Area under a curve region bounded by the given function, vertical lines and the x axis.
Ares y Integrtion. Are uner urve region oune y the given funtion, vertil lines n the is.. Are uner urve region oune y the given funtion, horizontl lines n the y is.. Are etween urves efine y two given
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationPROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * challenge questions
PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * hllenge questions e The ll will strike the ground 1.0 s fter it is struk. Then v x = 20 m s 1 nd v y = 0 + (9.8 m s 2 )(1.0 s) = 9.8 m s 1 The speed
More informationRatio and Proportion
Rtio nd Proportion Rtio: The onept of rtio ours frequently nd in wide vriety of wys For exmple: A newspper reports tht the rtio of Repulins to Demorts on ertin Congressionl ommittee is 3 to The student/fulty
More informationExponential and Logarithmic Functions
Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define
More informationState the size of angle x. Sometimes the fact that the angle sum of a triangle is 180 and other angle facts are needed. b y 127
ngles 2 CHTER 2.1 Tringles Drw tringle on pper nd lel its ngles, nd. Ter off its orners. Fit ngles, nd together. They mke stright line. This shows tht the ngles in this tringle dd up to 180 ut it is not
More informationWords Symbols Diagram. abcde. a + b + c + d + e
Logi Gtes nd Properties We will e using logil opertions to uild mhines tht n do rithmeti lultions. It s useful to think of these opertions s si omponents tht n e hooked together into omplex networks. To
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationVolumes by Cylindrical Shells: the Shell Method
olumes Clinril Shells: the Shell Metho Another metho of fin the volumes of solis of revolution is the shell metho. It n usull fin volumes tht re otherwise iffiult to evlute using the Dis / Wsher metho.
More informationSection 55 Solving Right Triangles*
55 Solving Right Tringles 379 79. Geometry. The re of retngulr nsided polygon irumsried out irle of rdius is given y A n tn 80 n (A) Find A for n 8, n 00, n,000, nd n 0,000. Compute eh to five deiml
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationMATH PLACEMENT REVIEW GUIDE
MATH PLACEMENT REVIEW GUIDE This guie is intene s fous for your review efore tking the plement test. The questions presente here my not e on the plement test. Although si skills lultor is provie for your
More informationDensity Curve. Continuous Distributions. Continuous Distribution. Density Curve. Meaning of Area Under Curve. Meaning of Area Under Curve
Continuous Distributions Rndom Vribles of the Continuous Tye Density Curve Perent Density funtion f () f() A smooth urve tht fit the distribution 6 7 9 Test sores Density Curve Perent Probbility Density
More informationRadius of the Earth  Radii Used in Geodesy James R. Clynch Naval Postgraduate School, 2002
dius of the Erth  dii Used in Geodesy Jmes. Clynh vl Postgrdute Shool, 00 I. Three dii of Erth nd Their Use There re three rdii tht ome into use in geodesy. These re funtion of ltitude in the ellipsoidl
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationLesson 2.1 Inductive Reasoning
Lesson.1 Inutive Resoning Nme Perio Dte For Eerises 1 7, use inutive resoning to fin the net two terms in eh sequene. 1. 4, 8, 1, 16,,. 400, 00, 100, 0,,,. 1 8, 7, 1, 4,, 4.,,, 1, 1, 0,,. 60, 180, 10,
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationChapter. Contents: A Constructing decimal numbers
Chpter 9 Deimls Contents: A Construting deiml numers B Representing deiml numers C Deiml urreny D Using numer line E Ordering deimls F Rounding deiml numers G Converting deimls to frtions H Converting
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationIf two triangles are perspective from a point, then they are also perspective from a line.
Mth 487 hter 4 Prtie Prolem Solutions 1. Give the definition of eh of the following terms: () omlete qudrngle omlete qudrngle is set of four oints, no three of whih re olliner, nd the six lines inident
More informationSine and Cosine Ratios. For each triangle, find (a) the length of the leg opposite lb and (b) the length of the leg adjacent to lb.
 Wht You ll ern o use sine nd osine to determine side lengths in tringles... nd Wh o use the sine rtio to estimte stronomil distnes indiretl, s in Emple Sine nd osine tios hek Skills You ll Need for Help
More informationSOLVING EQUATIONS BY FACTORING
316 (560) Chpter 5 Exponents nd Polynomils 5.9 SOLVING EQUATIONS BY FACTORING In this setion The Zero Ftor Property Applitions helpful hint Note tht the zero ftor property is our seond exmple of getting
More informationMath Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.
Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while
More informationModule 5. Threephase AC Circuits. Version 2 EE IIT, Kharagpur
Module 5 Threehse A iruits Version EE IIT, Khrgur esson 8 Threehse Blned Suly Version EE IIT, Khrgur In the module, ontining six lessons (7), the study of iruits, onsisting of the liner elements resistne,
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls : The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N):  counting numers. {,,,,, } Whole Numers (W):  counting numers with 0. {0,,,,,, } Integers (I): 
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationVectors Summary. Projection vector AC = ( Shortest distance from B to line A C D [OR = where m1. and m
. Slr prout (ot prout): = osθ Vetors Summry Lws of ot prout: (i) = (ii) ( ) = = (iii) = (ngle etween two ientil vetors is egrees) (iv) = n re perpeniulr Applitions: (i) Projetion vetor: B Length of projetion
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationMaximum area of polygon
Mimum re of polygon Suppose I give you n stiks. They might e of ifferent lengths, or the sme length, or some the sme s others, et. Now there re lots of polygons you n form with those stiks. Your jo is
More informationSeeking Equilibrium: Demand and Supply
SECTION 1 Seeking Equilirium: Demnd nd Supply OBJECTIVES KEY TERMS TAKING NOTES In Setion 1, you will explore mrket equilirium nd see how it is rehed explin how demnd nd supply intert to determine equilirium
More informationTallahassee Community College. Simplifying Radicals
Tllhssee Communit College Simplifing Rdils The squre root of n positive numer is the numer tht n e squred to get the numer whose squre root we re seeking. For emple, 1 euse if we squre we get 1, whih is
More informationPractice Test 2. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn
Prtie Test 2 1. A highwy urve hs rdius of 0.14 km nd is unnked. A r weighing 12 kn goes round the urve t speed of 24 m/s without slipping. Wht is the mgnitude of the horizontl fore of the rod on the r?
More informationRightangled triangles
13 13A Pythgors theorem 13B Clulting trigonometri rtios 13C Finding n unknown side 13D Finding ngles 13E Angles of elevtion nd depression Rightngled tringles Syllus referene Mesurement 4 Rightngled tringles
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationSECTION 72 Law of Cosines
516 7 Additionl Topis in Trigonometry h d sin s () tn h h d 50. Surveying. The lyout in the figure t right is used to determine n inessile height h when seline d in plne perpendiulr to h n e estlished
More informationHeron s Formula for Triangular Area
Heron s Formul for Tringulr Are y Christy Willims, Crystl Holom, nd Kyl Gifford Heron of Alexndri Physiist, mthemtiin, nd engineer Tught t the museum in Alexndri Interests were more prtil (mehnis, engineering,
More informationReal Analysis HW 10 Solutions
Rel Anlysis HW 10 Solutions Problem 47: Show tht funtion f is bsolutely ontinuous on [, b if nd only if for eh ɛ > 0, there is δ > 0 suh tht for every finite disjoint olletion {( k, b k )} n of open intervls
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Alger. Opertions nd Epressions. Common Mistkes. Division of Algeric Epressions. Eponentil Functions nd Logrithms. Opertions nd their Inverses. Mnipulting
More informationModule Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials
MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic
More informationWarmup for Differential Calculus
Summer Assignment Wrmup for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationc b 5.00 10 5 N/m 2 (0.120 m 3 0.200 m 3 ), = 4.00 10 4 J. W total = W a b + W b c 2.00
Chter 19, exmle rolems: (19.06) A gs undergoes two roesses. First: onstnt volume @ 0.200 m 3, isohori. Pressure inreses from 2.00 10 5 P to 5.00 10 5 P. Seond: Constnt ressure @ 5.00 10 5 P, isori. olume
More informationClause Trees: a Tool for Understanding and Implementing Resolution in Automated Reasoning
Cluse Trees: Tool for Understnding nd Implementing Resolution in Automted Resoning J. D. Horton nd Brue Spener University of New Brunswik, Frederiton, New Brunswik, Cnd E3B 5A3 emil : jdh@un. nd spener@un.
More informationTHE RATIONAL NUMBERS CHAPTER
CHAPTER THE RATIONAL NUMBERS When divided by b is not n integer, the quotient is frction.the Bbylonins, who used number system bsed on 60, epressed the quotients: 0 8 s 0 60 insted of 8 s 7 60,600 0 insted
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationChapter. Fractions. Contents: A Representing fractions
Chpter Frtions Contents: A Representing rtions B Frtions o regulr shpes C Equl rtions D Simpliying rtions E Frtions o quntities F Compring rtion sizes G Improper rtions nd mixed numers 08 FRACTIONS (Chpter
More informationOUTLINE SYSTEMONCHIP DESIGN. GETTING STARTED WITH VHDL August 31, 2015 GAJSKI S YCHART (1983) TOPDOWN DESIGN (1)
August 31, 2015 GETTING STARTED WITH VHDL 2 Topdown design VHDL history Min elements of VHDL Entities nd rhitetures Signls nd proesses Dt types Configurtions Simultor sis The testenh onept OUTLINE 3 GAJSKI
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationQuick Guide to Lisp Implementation
isp Implementtion Hndout Pge 1 o 10 Quik Guide to isp Implementtion Representtion o si dt strutures isp dt strutures re lled Sepressions. The representtion o n Sepression n e roken into two piees, the
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More informationpq Theory Power Components Calculations
ISIE 23  IEEE Interntionl Symposium on Industril Eletronis Rio de Jneiro, Brsil, 911 Junho de 23, ISBN: 78379128 pq Theory Power Components Clultions João L. Afonso, Memer, IEEE, M. J. Sepúlved Freits,
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationCHAPTER 31 CAPACITOR
. Given tht Numer of eletron HPTER PITOR Net hrge Q.6 9.6 7 The net potentil ifferene L..6 pitne v 7.6 8 F.. r 5 m. m 8.854 5.4 6.95 5 F... Let the rius of the is R re R D mm m 8.85 r r 8.85 4. 5 m.5 m
More informationGeometry 71 Geometric Mean and the Pythagorean Theorem
Geometry 71 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the
More informationCalculating Principal Strains using a Rectangular Strain Gage Rosette
Clulting Prinipl Strins using Retngulr Strin Gge Rosette Strin gge rosettes re used often in engineering prtie to determine strin sttes t speifi points on struture. Figure illustrtes three ommonly used
More informationSquare Roots Teacher Notes
Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationAngles 2.1. Exercise 2.1... Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example
2.1 Angles Reognise lternte n orresponing ngles Key wors prllel lternte orresponing vertilly opposite Rememer, prllel lines re stright lines whih never meet or ross. The rrows show tht the lines re prllel
More information2 DIODE CLIPPING and CLAMPING CIRCUITS
2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of
More informationMATHEMATICS I & II DIPLOMA COURSE IN ENGINEERING FIRST SEMESTER
MATHEMATICS I & II DIPLOMA COURSE IN ENGINEERING FIRST SEMESTER A Plition nder Government of Tmilnd Distrition of Free Tetook Progrmme ( NOT FOR SALE ) Untohilit is sin Untohilit is rime Untohilit is inhmn
More informationReview. Scan Conversion. Rasterizing Polygons. Rasterizing Polygons. Triangularization. Convex Shapes. Utah School of Computing Spring 2013
Uth Shool of Computing Spring 2013 Review Leture Set 4 Sn Conversion CS5600 Computer Grphis Spring 2013 Line rsteriztion Bsi Inrementl Algorithm Digitl Differentil Anlzer Rther thn solve line eqution t
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationGRADE 4. Fractions WORKSHEETS
GRADE Frtions WORKSHEETS Types of frtions equivlent frtions This frtion wll shows frtions tht re equivlent. Equivlent frtions re frtions tht re the sme mount. How mny equivlent frtions n you fin? Lel eh
More informationNew combinatorial features for knots and virtual knots. Arnaud MORTIER
New omintoril fetures for knots nd virtul knots Arnud MORTIER April, 203 2 Contents Introdution 5. Conventions.................................... 9 2 Virtul knot theories 2. The lssil se.................................
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions
MSSCHUSES INSIUE OF ECHNOLOGY Deprtment of hysics 8.0 W02D3_0 Group roblem: ulleys nd Ropes Constrint Conditions Consider the rrngement of pulleys nd blocks shown in the figure. he pulleys re ssumed mssless
More information15.6. The mean value and the rootmeansquare value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style
The men vlue nd the rootmensqure vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time
More informationThe Pythagorean Theorem
The Pythgoren Theorem Pythgors ws Greek mthemtiin nd philosopher, orn on the islnd of Smos (. 58 BC). He founded numer of shools, one in prtiulr in town in southern Itly lled Crotone, whose memers eventully
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationAP STATISTICS SUMMER MATH PACKET
AP STATISTICS SUMMER MATH PACKET This pcket is review of Algebr I, Algebr II, nd bsic probbility/counting. The problems re designed to help you review topics tht re importnt to your success in the clss.
More informationReleased Assessment Questions, 2015 QUESTIONS
Relesed Assessmet Questios, 15 QUESTIONS Grde 9 Assessmet of Mthemtis Ademi Red the istrutios elow. Alog with this ooklet, mke sure you hve the Aswer Booklet d the Formul Sheet. You my use y spe i this
More informationLecture 15  Curve Fitting Techniques
Lecture 15  Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting  motivtion For root finding, we used given function to identify where it crossed zero where does fx
More informationLesson 4.1 Triangle Sum Conjecture
Lesson 4.1 ringle um onjecture Nme eriod te n ercises 1 9, determine the ngle mesures. 1. p, q 2., y 3., b 31 82 p 98 q 28 53 y 17 79 23 50 b 4. r, s, 5., y 6. y t t s r 100 85 100 y 30 4 7 y 31 7. s 8.
More informationA Note on Complement of Trapezoidal Fuzzy Numbers Using the αcut Method
Interntionl Journl of Applictions of Fuzzy Sets nd Artificil Intelligence ISSN  Vol.  A Note on Complement of Trpezoidl Fuzzy Numers Using the αcut Method D. Stephen Dingr K. Jivgn PG nd Reserch Deprtment
More information1 Fractions from an advanced point of view
1 Frtions from n vne point of view We re going to stuy frtions from the viewpoint of moern lger, or strt lger. Our gol is to evelop eeper unerstning of wht n men. One onsequene of our eeper unerstning
More informationVersion 001 CIRCUITS holland (1290) 1
Version CRCUTS hollnd (9) This printout should hve questions Multiplechoice questions my continue on the next column or pge find ll choices efore nswering AP M 99 MC points The power dissipted in wire
More informationFormal Languages and Automata Exam
Forml Lnguges nd Automt Exm Fculty of Computers & Informtion Deprtment: Computer Science Grde: Third Course code: CSC 34 Totl Mrk: 8 Dte: 23//2 Time: 3 hours Answer the following questions: ) Consider
More informationOn Equivalence Between Network Topologies
On Equivlene Between Network Topologies Tre Ho Deprtment of Eletril Engineering Cliforni Institute of Tehnolog tho@lteh.eu; Mihelle Effros Deprtments of Eletril Engineering Cliforni Institute of Tehnolog
More informationDATABASDESIGN FÖR INGENJÖRER  1056F
DATABASDESIGN FÖR INGENJÖRER  06F Sommr 00 En introuktionskurs i tssystem http://user.it.uu.se/~ul/tsommr0/ lt. http://www.it.uu.se/eu/course/homepge/esign/st0/ Kjell Orsorn (Rusln Fomkin) Uppsl Dtse
More information4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS
4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion 10 1. Review Theorem
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationCalculus of variations with fractional derivatives and fractional integrals
Anis do CNMAC v.2 ISSN 1984820X Clculus of vritions with frctionl derivtives nd frctionl integrls Ricrdo Almeid, Delfim F. M. Torres Deprtment of Mthemtics, University of Aveiro 3810193 Aveiro, Portugl
More informationLECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.
LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 6483.
More informationEXAMPLE EXAMPLE. Quick Check EXAMPLE EXAMPLE. Quick Check. EXAMPLE RealWorld Connection EXAMPLE
 Wht You ll Lern To use the Pthgoren Theorem To use the onverse of the Pthgoren Theorem... nd Wh To find the distne etween two doks on lke, s in Emple The Pthgoren Theorem nd Its onverse hek Skills You
More informationSection 1: Crystal Structure
Phsics 927 Section 1: Crstl Structure A solid is sid to be crstl if toms re rrnged in such w tht their positions re ectl periodic. This concept is illustrted in Fig.1 using twodimensionl (2D) structure.
More information