HANDOUT 14. A.) Introduction: Many actions in life are reversible. * Examples: Simple One: a closed door can be opened and an open door can be closed.


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1 Inverse Funcions Reference Angles Inverse Trig Problems Trig Indeniies HANDOUT 4 INVERSE FUNCTIONS KEY POINTS A.) Inroducion: Many acions in life are reversible. * Examples: Simple One: a closed door can be opened and an open door can be closed. Complicaed one: an elecronic communicaion sends coded messages o a cell phone, he cell phone hen decodes he elecronic communicaion. In his case, i is ESSENTIAL ha he inverse funcion decodes he signal correcly. Mahemaics and inverse funcions are exremely imporan in he developmen of communicaion sysems. B.) Finding Inverses: Inverses in Mahemaics are associaed wih he basic concep of reversing a calculaion and arriving a an original resul. x +0 For example: In he expression,, we ook an x added 0 and divided he resul by. The inverse of his would be o ake x, muliply i by and subrac 0, i.e. x 0. The inverse UNDOES he operaion on x. C.) A New Symbol f  (x) The exponen on f does NOT mean over f. I is simply a symbol indicaing he inverse of f. Examples: f(x) = x + hen he inverse funcion is defined as f  (x) = x f(4) = 9 and f  (9) = 4 D.) To find an inverse from a se of poins: Inerchange he coordinaes of each ordered pair. Examples: Le f = he following se of ordered pairs (,), (,4), (,7). If we inerchange he x and y s, we ge (,), (4,), (7,). Since each x has one y, his new inverse funcion is a funcion. We say ha he original funcion is INVERTIBLE. However, look a he ordered pairs: (4,6), (, 6), (7, 9). If we inerchange hese, we ge (6, 4), (6, ) and (9,7). In his case each x does no have a unique y since if x= 6 we have wo yvalues. We say his funcion is NOT inverible. I does no have an inverse. The easies way o find ou if a funcion is inverible is o check o see if i passes he horizonal line es. * Some ideas and problems are from College Algebra by Mark Dugopolski. Reading, MA. 998 and College Algebra by Rockswold, Hornsby, and Lial. Reading, MA. 999.
2 E.) Consider he x funcion. Is i inverible? No. I does no pass he horizonal line es. However, his funcion is so imporan in mahemaics ha we need some way o make i inverible. There are also oher funcions in mahemaics ha we would like o have an inverse. (e.g rig funcions.) Therefore, mahemaicians have decided o RESTRICT he domain of hese imporan funcions Thus he domain for he x funcion is resriced o all posiive x and now i passes he horizonal line es F) More KEY POINTS: Funcions ha pass he horizonal line es are said o be oneoone. One o one funcions are inverible. Le f be a oneoone funcion. Then g is he INVERSE funcion of f if he following composiions occur: f(g(x)) = x for every x in he domain of g and g(f(x)) = x for every x in he domain of f. Then we give g he symbol, f  If f is he funcion and g is is inverse hen, he o he domain of f = he range of g and o he range of f = he domain of g This means ha if (a, b) is on f, hen (b, a) is on f  for every x. G.) Graphically, a funcion and is inverse are reflecions of each oher over he line y = x. Demonsraed in class.
3 Algebraic Mehod for Finding he Inverse Finding he Inverse Funcion: As we saw, o find he inverse, we simply swiched he x and y values. If we have an algebraic equaion represening a funcion, we can proceed o find is inverse by swiching he x and y in he equaion and solving for y. (I does no maer if he x and y are swiched a he beginning or a he end. I swich hem a he beginning so I won forge!!) + Example : Find he inverse funcion for f ( x) = x x +. Replace f(x) wih y y = y +. Swich he x and y x =. Solve for y x = y + x = y x = y y = x 4. Replace y by f ( x ) f ( x ) = x. Check domain of inverse. I should be he range of funcion. In his case he domain and range of he funcion and he inverse is all reals. Example : Find he inverse funcion for g ( x) = e x+. Replace g(x) wih y y = x+ e. Swich he x and y x = y+ e. Solve for y ln x = y + ln x = y ln x = y or ln x y = 4. Replace y by g ( x) ( x ln x g ) = Check domain and range: Domain of inverse is all real x > 0 Range of original funcion is all y > 0 Range of Inverse is all reals Domain of original funcion is all reals
4 Example : Find he inverse funcion for s( x) = + log x. Replace s(x) wih y. Swich he x and y y = + log x x = + log y. Solve for y x( + log y) = x + x log y = x log y = x x log y = x y = 0 x x x 4. Replace y by s  x (x) s ( x) = 0 Check domain and range. The range of he inverse funcion is found by checking i from he domain of he funcion. I can be seen graphically. Unforunaely, i requires a change in he window parameers which someimes requires a bi of effor. There is a horizonal asympoe a y =.0 for he inverse and a verical asympoe a x=.0 for he original funcion. Domain of inverse is all real x 0 Range of original funcion is all y 0 Range of Inverse is all reals y > 0 y.0 Domain of original funcion is all reals x > 0, x.0 Trigonomeric Inverses More laer Trig Inverse Funcions have resriced domains Inverse Cosine Funcion =Arccos or cos ( y ) = x if y = cos x and 0 x π Inverse Sine Funcion = Arcsin or Inverse Tan Funcion = Arcan or sin ( y ) = x an ( y ) = x if y = sin x and π/ x π/ if y = an x and π/ x π/ Example 4: Find a soluion o he equaion: Cos x = x π This says, find all he angles beween zero and Pi whose cosine is 0.6. We know ha here are wo angles in a circle where he cosine is posiive. The inverse cosine funcion will give us he angle in he firs quadran. The oher angle is in he 4 h quadran and we use reference angles o find ha angle. (Nex secion, bu we can figure i ou.) So cos  (0.6) = x Using he calculaor, we find x = 0.97 in degrees, his is. The angle in he 4 h quadran is π 0.97 =.6 in degrees,
5 Reference Angles Find reference angle A r o a given angle A. Wha is he reference angle o an angle in sandard posiion? If A is an angle in sandard posiion, is reference angle A r is he acue angle formed by he x axis and he erminal side of angle A. See figure below. Two or more coerminal angles have he same reference angle. Assume angle A is posiive and less han 60 o (π), we have 4 possible cases:. If angle A is in quadran I hen he reference angle A r = A.. If angle A is in quadran II hen he reference angle A r = 80 o A if A is given degrees or A r = π A if A is given in radians.. If angle A is in quadran III hen he reference angle A r = A 80 o if A is given degrees or A r = A π if A is given in radians. 4. If angle A is in quadran IV hen he reference angle A r = 60 o A if A is given degrees or A r = π A if A is given in radians. Adaped from: hp://
6 Example : Find he reference angle o angle A = 0 o. Soluion: Angle A is in quadran II and he reference angle is given by A r = 80 o 0 o = 60 o in Radians his would be π π/ π Example : Find he reference o angle A =. 4 Soluion: Firs find he quadran where his angle lands. The given angle is no posiive. Also noice ha more han one revoluion in he clockwise direcion is required o ge o he erminal side of his angle. (Tha is, we need o find ou wha quadran his angle is in.) Figure ou how many revoluions of posiive π in fourhs (since he denominaor is in 4hs) we need o go around o find he coerminal angle. 8π/4 is no enough. 6π/4 would pu us righ a he 0 or π posiion. So: π/4 + 6π/4 = π/4. This pus us in he s Quadran. Therefore he reference angle is π/4. Example : (# Page 9) This one is no quie as nice. Find he reference o angle A = 46π. 7 Firs find he quadran for his angle, A = 46π/7. Similar o Example above, find ou how many posiive revoluions of π s in sevenhs we need o ge o land a 0 again. π in sevenhs = 4π/7 he closes muliple of 4π/7 o 46π/7 is revoluions around = 4π/7. This lands us a he π posiion, hen we can figure ou how many 7 hs we have. 46π 4π 4π + = If we go in he clockwise direcion from he π posiion we land in he rd quadran. The measure in he posiive direcion is π/7. You could also reason ha 4 revoluions from he 0 posiion is 4*4π/7 = 6π/7. Now we have 6π/7 46π/7 = 0π/7. This again lands us in he rd quadran (7π /7 + π/7 = 0π/7). Again our reference angle is π/7. (Using he rule above for hird quadran angles, A = 0π/7 so A r = A π = 0π/7 7π/7) In order o find he coordinaes of 46π/7 we use he sine and cosine of π/7 making sure he signs fi wih rd quadran angles. 6
7 Using Reference angles wih Inverse Trig Problems wih Soluions Example : an θ = Find all he exac soluions for 0 θ π. Answer: Since he angen of some angle is negaive, ha angle mus be in he nd or 4 h quadran. Firs: ake he inverse angen. θ = an  ( ) This says, Thea is he angle in he nd or 4 h quadrans whose angen is negaive roo. The calculaor gives: θ = an  ( ) = This angle is in he 4 h quadran. We recognize ha he angle is a π/ angle. If we canno remember his, conver o degrees 80 using he conversion formula by muliplying by. We ge 60 and we see ha π Tan 60 = sin 60 /cos 60 = = /. We now have a reference angle we can use o find he exac angles. π/ is in he 4 h quadran. We wish o express our answers as posiive angles since we wan 0 θ π. So he wo angles in he second and 4 h quadrans ha have heir angens = ( ) are: θ = π/ and π/. How did we find hese. Since he reference angle is π/, and we need ha angle in he nd quadran, simply subrac π/ from π o ge π/. Similarly, in he 4 h quadran, go back π/ unis from π. Thus π π/ = π/ Our answer needs o be in radians since he quesion was o find 0 θ π. However, in degrees, hese are 0 and 00. We can also graph he angen funcion in he calculaor and look o see where i mees he line y =. We ge he same answers only we have numbers like.094 and.6. Noice hese are approximaions o π/ and π/. 7
8 Example : sin θ = 4 Find all he exac soluions for 0 θ π. Hin: There are 4 soluions for θ. Answer: Take he square roo of boh sides. We ge sin θ = ± so we need o find sin (+ ) and sin  ( ) We should recognize ha his is a π/6 angle. If no use he calculaor. sin (+ ) =.988 (.988 looks like i is in he firs quadran.) The angles beween 0 and π ha have a sin = posiive. are in he firs and second quadrans. Noice he reference angle for an angle wih a sin =. is π/6. Thus he angles we are looking for from he s and second quadrans are. π/6 and π π/6= π/6 Similarly, we find he oher angles in he rd and 4 h quadran, i.e. sin  ( ) =.988 = π/6 (his angle is in he 4 h quadran) so angle in rd quadran = π + π/6 = 7π/6 and π π/6 = π/6 in he 4 h. Summarizing: we ge π/6, π/6, 7π/6, π/6 These are all he values beween 0 and π ha have a sine equal o ± ½. You can also skech a graph and find hese values. 8
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