Continuous Functions

Transcription

1 Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio Defiitio- Let P= ( a, f ( a )) be a poit o the curve f We say a fuctio f is cotiuous at a poit P if ad oly if the followig relatioship is satisfied f ( ) = f( a) a The it eists ad is equal to the value of the fuctio at a This really meas the followig relatioships are satisfied f ( ) = f( ) = f( a ) a a Cosider the followig curve ad aswer the questios below a li m ) a li m f ( f ( ) f ( ) + a a f ( a ) Cotiuous Summary - Yes No - + DNE Ud No - Yes - -8 DNE - No Ud No Ud No f is cotiuous at =- f is discotiuous at = -4 f is discotiuous at = - f is cotiuous at = - f is discotiuous at = f is discotiuous at = f is discotiuous at = Q What does it mea for a fuctio to be cotiuous everywhere? Aswer-A cotiuous everywhere fuctio (aka a cotiuous fuctio) is a fuctio that has NO poits of discotiuity The previous eample is ot a cotiuous everywhere fuctio as it has discotiuities at =-4, =-, =, =, ad = Defiitio- Let I be a ope iterval We say f is a cotiuous o I, if f is a cotiuous for every poit i I

2 Eample- Cosider the fuctio above ad the ope iterval (-,) The fuctio f has o discotiuities i the iterval (-,), thus we ca say f is cotiuous over the iterval (-,) But, it is ot a cotiuous everywhere fuctio as there eists poits of discotiuity Q What does a cotiuous everywhere fuctios look like? Aswer- Ay fuctio that has o poits of discotiuity, ad there are may Ituitively, cotiuous everywhere fuctios are fuctios that ca be draw without havig to lift up a pecil! You try the graphs listed below a f a ( ) f ( a ) Cotiuous Note- I order to draw this curve you must lift up your pecil This is a discotiuous fuctio 4 a - - f a f ( a ) Cotiuous Note- I order to draw this curve you do ot have to lift up your pecil This is a cotiuous everywhere fuctio

3 Properties of Cotiuous Fuctios Let f ad g be two cotiuous at =a, ad let c be a costat, The : cf is cotiuous at a (see proof i class) f + g is cotiuous at a (see proof i class) f g is cotiuous at a (see proof i class) f g is cotiuous at a (see proof i class) f is cotiuous at a, if ga 0( see proof i class) g Fact- The costat fuctio These properties are eeded to prove the followig facts f ( ) = c is cotiuous Proof- Let c be some costat such that f ( ) c The f ( ) = c= c by properties of its a a But f ( a) = c as f is the costat fuctio Thus f ( ) = f( a) ad f is cotiuous at a a Sice a is arbitrary, f is cotiuous everywhere = ad a a arbitrary value Eample- f ( ) = F act- The idetity fuctio f ( ) Proof: Let f ( ) = f ( ) = a a a = is cotiuous be the idetity fuctio ad a some arbitrary value = by the defiitio of a it But f ( a) = a Thus f ( ) f( a) a as f is the idetity fuctio, = ad f is cotiuous at a Sice a is arbitrary, f is cotiuous everywhere Graph Fact- The fuctio Proof: Let f ( ) = is cotiuous ad a a arbitrary value The f ( ) = = = = a a= fuctios f ( ) = a a a a a But f ( a) = a, hece f ( ) f( a) a = Sice a is arbitrary, f is cotiuous everywhere a by properties of cotiues

4 This ca be geeralized to show that f ( ) = for ay atural umber is cotiuous everywhere 4 7 Fact- Polyomial fuctios are cotiuous Proof: Let P be a polyomial fuctio such that P = a + a + + a+ a0 for some atural umber, ad a a arbitrary umber The p( ) = a + a + + a + a = a + a + + a + a ( 0) ( ) ( ) ( ) ( ( ) ( 0) a a a a a a = a + a + + a + a by properties of its a a a a aa a a = aa+ a 0 as is cotiuous for all But P a == a a + a a + + aa+ a 0, h ece P = Pa Sice a is arbitrary, P is cotiuous everywhere Fact- Ratioal fuctios are cotiuous Proof: Let R be a ratioal fuctio such that R( ) a arbitrary umber such that Q( a) 0 P P P a a The R = = = a a Q Q Q a a polyomial fuctios are cotiuous everywhere P a But R( a) =, hece R( ) = R( a) Q a a Sice a is arbitrary, R is cotiuous everywhere a P = where P ad Q are polyomials, a Q F act- If g is cotiuous ad f ( ) = g( ), the f is cotiuous for > 0 Proof: Let f be cotiuous ad a a arbitrary value The f ( ) g( ) g( ) = = = a a a the fact t hat g is a cotiuous fuctio Sice a is arbitrary, f is cotiuous everywhere by properties of its ad the fact that f g a = a by properties of its, ad Fact- If h = f g such that f is cotiuous at b ad b= g, the h is cotiuous at a That is, h h a ; f g = f g a ; f g f g a a a = a Proof: See proof i class F act- If h( ) f g( ) a = = such that g is cotiuous at a, ad f is cotiuous at g(a), the h is cotiuous at a That is, a cotiuous fuctio of a cotiuous fuctio is a cotiuous fuctio Proof: See proof i class 4 0 )

5 ample- Let f ad g be cotiuous fuctios with f()=-7 ad [ f g ] E =, what is g()? Use properties of cotiuity [ f( ) g( ) ] = f ( ) g( ) = f ( ) g( ) = by properties of cotiuity f g 7 g = g = = = by the defiitio of cotiuity Usig algebra, g()=-/ Eample- Use cotiuity to evaluate the followig i cos cos t ( ) π { cos cos( ) } cos cos( ) cos cos( ) π = π = π π π π π = cos 0 = cos = cos = 0 = 0 Eample- Eplai why the fuctio f( ) = + + is discotiuous at =- + Upo first glace, oe otices a ratioal fuctio, which may be require more care whe graphig However, whe you simplify this ratioal fuctio a liear fuctio is revealed ( + )( + ) + + f( ) = = = This is the reaso you see the graph of a liear fuctio However, closer look at the lie represetig this graph a HOLE at =- is revealed Eample- Determie whether the followig fuctio f is cotiuous + + for f( ) = + 4 for = Usig the defiitio of cotiuity, f ( ) = f( ) = f( a) a a Whe, we have a Ratioal fuctio We kow Ratioal fuctios are cotiuous everywhere they are defied Therefore, the top ratioal fuctio is cotiuous ecept at Whe =, we have a costat fuctio We kow costat fuctios are cotiuous everywhere B ut to use the defiitio, f ( ) f ( ) while We ca coclude that f is discotiuous at =- = = Eample- Determie whether the followig fuctio f is cotiuous > f ( ) = < Usig the defiitio of cotiuity, f ( ) = f( ) = f( a) a a f = 4 Whe > we have a polyomial fuctio which is cotiuous everywhere Therefore, f is cotiuous for > Whe we have a costat fuctio which is cotiuous everywhere Therefore, f is cotiuous for Whe < we have a polyomial fuctio which is cotiuous everywhere Therefore, f is cotiuous for <

6 T he oly cocers we should have are at the poits = ad = By simply usig the defiitios, we ca determie cotiuity while f = 0 so that f is discotiuous at =- f ( ) = + = f 0 while f = 0, so that f is cotiuous at = Coclusio- The fuctio f is cotiuous ecept at =- Eample- Let f ( ) The graph reveals various asymptotes = Determie the values + cos for which f is discotiuous We ca determie these asymptotes by determiig what values for the deomiator are 0 + cos = 0whe cos = This happes whe =±, ±, ±, π π π Therefore, = = = = π+ cos π+ cos π+ cos While, = = = = π cos π cos π cos That is f is discotiuous at =± π, ± π, ± π, { } E ample- Let f ( ) = cos si cos State the domai of f The domai of cos( ) mai of is all real umbers The do si is all real umbers The r si cos is all real umbers refo e, the domai of s, t e domai of cos } Thu h {si c os is all real umbers Itermediate Value Theorem Let f be a cotiuous fuctio over a closed iterval [a,b] The there is a c i (a,b) such that f(a)<f(c)<f(b) Eample- Show that a root eists for the equatio f = L et f ( 0) = which is a positive umber also () By the Itermediate Value Theorem, there eists a umber c i That is, th f c = ere is a c such that 0 = 0i the iterval [0, ] f = which is a egative umber 0, such that f ( c)

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