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

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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. ONE-TO-ONE AND MANY-TO-ONE FUNCTIONS A one-to-one funtion is n funtion where: ² for eh there is onl one vlue of nd ² for eh there is onl one vlue of. One-to-one 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 one-to-one, it will hve n inverse funtion whih we denote f (). Funtions tht re not one-to-one re lled mn-to-one. 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 mn-to-one, it does not hve n inverse funtion. However, for mn-to-one funtion we n often define new funtion using the sme formul ut with restrited domin to mke it one-to-one 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. SELF-INVERSE FUNCTIONS An funtion whih hs n inverse, nd whose grph is smmetril out the line =, i s self-inverse funtion. For emple: ² The funtion f() = is the identif funtion, nd is lso self-inverse funtion. ² The reiprol funtion f() =, 6= 0, is lso self-inverse 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 NON-CALCULATOR 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 NON-CALCULATOR 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 self-inverse 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 one-to-one 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() =

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