Here are the graphs of some power functions with negative index y (x) =ax n = a n is a positive integer, and a 6= 0acoe±cient.

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1 BEE4 { Bsic Mthemtics for Economists BEE5 { Introduction to Mthemticl Economics Week 9, Lecture, Notes: Rtionl Functions, 26//2 Hint: The WEB site for the tetbook is worth look. Dieter Blkenborg Deprtment of Economics University of Eeter Objective To get better feel for the techniques nd terminology used so fr we discuss the grphs of some rtionl functions. The relevnt reding for this hndout is Chpter 3, Sections {3 in Ho mnn nd Brdley (2). 2 Horizontl nd verticl symptotes Here re the grphs of some power functions with negtive inde y () = n = n is positive integer, nd 6= coe±cient. n where 2 2 Fig. : y () = Fig. 2: y () = Fig. 3: y () = 2 Fig. 4: y () = 4 6 The grphs pproches the horizontl is when gets lrge in mgnitude, regrdless of the sign of ; without ctully ever touching it. We sy \the horizontl is y =is horizontl symptote" ndwritebrie y! n =!+ n =: The symbol for in nity is not number. It is just used s shortcut to indicte the behviour of function s gets lrger nd lrger in mgnitude or to describe the wy in which the vlues of function \eplode" ner to point where the function is not de ned.

2 Similrly, the verticl is is verticl symptote in the sense tht the grph pproches but never reches it when gets very smll. The vlues y () \eplode", i.e., get rbitrrily lrge in mgnitude, the closer gets to the number where the function is not de ned. Whether y () is thereby positive or negtive depends on whether we pproch = from the left (we write! to indicte this) or from the right (we write! +). The relevnt sign is determined by the sign of the coe±cient nd depends on whether n is odd or even. One hs =sign() ;! + n nd for odd n! = n! + n while for even n 2. Verticl symptotes Consider the rtionl function! = + n! + n y () = 2 ( ) : It is not de ned for =nd =. The vlues of the function re \eploding" in mgnitude ner to these two numbers. We hve two verticl symptotes there. For very close to (sy = or = )theterm will be very close to ; so our function will look here pproimtely like the grph the function = which hs 2 ( ) 2 shpe similr Figure 4 bove. We obtin y () =!! 2 = y () =! +! + = 2 When isclosetothen 2 will lso be close to nd our function y () will look similr to the function. (The error we re mking here is negligible since the term is eploding.) The grph of the function looks ner to = like the grph of the function ner =: 2 2 Fig. 6: The function Fig. 7: The function 2

3 Therefore y () =! +! + =!! + =+ y () =!! =!! = Our ndings re con rmed by the grph of our function y (): Fig. 8: y () = 2 ( ) The emple illustrtes the following generl point: Theorem Suppose rtionl function y () cn be written s y () = u () ( ) k v () where k is positive integer nd where u () nd v () re polynomils with u () 6= nd v () 6=. Then y () is not de ned for = nd the verticl line through = is verticl symptote. Moreover, u () y () = sign! + v () u () y () = ( )k sign! v () 2.2 Horizontl symptotes The behvior for very lrge positive or negtive of rtionl function y () = n n + n n + :::+ + m + m + :::+ b + b with n 6=nd 6= with n; m positive integers, n 6=nd 6=, is determined by the quotient of the leding terms n n m = n n m 3

4 Fig. 9: y () = Fig. : y () = Fig. : y () = Fig. 2: y () = Cse n<m: The horizontl is is horizontl symptote s in Figure 9, where n =, m =3. Cse 2 n = m: The horizontl line y = n y () = y () =!!+ is horizontl symptote. y () = y () = n!!+ In Figure one gets n = m =3, the horizontl symptote is y =5. Cse 3 n = m +: The grph hs n symptotic line with slope n.hence y () = y () =sign!!+ µ n : Thus the slope of the symptotic line is esy to nd by dividing the leding coe±cients. To nd the intercept of the symptotic line we must perform polynomil division. In the emple of Figure one gets = : 2 The symptotic line is y =5 +5. Cse 4 n>m+: The function hs only verticl symptotic lines. We deduce the behviour for very lrge positive or negtive vlues from the quotient of the leding term 4

5 n n m in the sme wy s we did it with the leding term of polynomil in the hndout on sign digrms. Nmely by µ n y () = sign!+ µ y () = n! ( )n m sign Actully, the function hs n \symptotic polynomil curve": For instnce, if we perform polynomil division for the emple in Figure 2 we get y () = = The remining \proper frction" { nd this is lwys the cse { hs the horizontl 3 2 is s n symptotic line. This mens tht the originl function y () hs for lrge positive or negtive vlues of lmost the sme vlues s the polynomil Thisis indicted in the grph. 3 Emple 3.7 from the tetbook y () = ( +) 2. The numertor hs degree, the denomintor degree 2. Hence the horizontl is is n symptote nd y () = y () =:!!+ 2. The denomintor is lwys non-negtive. Hence y () is positive or negtive ccording to whether is positive or negtive. 3. Thedenomintor is zero for =. The corresponding verticl line is n symptote nd y ()! + =! + ( +) 2 = y ()! =! ( +) 2 = Summrizing the informtion so fr: < = << = < + y () j j + 5

6 4. The rst derivtive is (using the quotient rule nd the generl power rule) which gives the sign digrm y () = + ( +) 3 < = << = < ( +) y () + 5. The second derivtive is y () = ( ) ( +)3 ( +) 3( +) ( +) 6 = ( +) 4 = 2 4 ( +) 4 which gives the sign digrm < = <<2 =2 2 < ( +) y () + Insted of giving summry sign digrm, let me describe verblly the informtion obtined by the bove clcultions: Coming from on the left the grph of the function is close to, but below the horizontl is. As increses it remins below it nd is decresing nd downwrdbowed (concve) until it \eplodes" towrds to the left of 2. To the right of 2 the function shoots up from nd is incresing nd downwrd-bowed. It crosses the horizontl is t =; but remins incresing until it reches locl mimum t =. From there onwrd it is decresing but it remins positive. There is n in ection point t = 2 where the grph turns from being downwrd bowed to being upwrd bowed. From there onwrds the grph pproches more nd more the horizontl is. We cn deduce ll this before turning on the grphic clcultor. Of course it is useful to drw the grph in order to see tht we re correct: Fig. 2: y () = 6 (+) 2

7 Eercise Red crefully through emple 3.8 of the tetbook. Verify tht the rst nd second derivtive re correct. Unluckily the in ection point cn only be found numericlly in this emple. References Hoffmnn, L. D., nd G. L. Brdley (2): Clculus for Business, Economics nd the Socil Sciences. McGrw Hill, Boston, 7th, interntionl edn. 7